Mixing
Derivatives by complex number and conjugate
$begingroup$
Suppose we write a complex number $z=x+iy$. Then we have $overline{z}=x-iy$, and therefore $x=dfrac12(z+overline{z})$, $y=-dfrac12i(z-overline{z})$.
If the rules of calculus were applicable, we would obtain $$frac{partial{f}}{partial{z}}=frac12left(frac{partial{f}}{partial{x}}-ifrac{partial{f}}{partial{y}}right), frac{partial{f}}{partial{overline{z}}}=frac12left(frac{partial{f}}{partial{x}}+ifrac{partial{f}}{partial{y}}right)$$
How does this follow from the rules of calculus?
calculus complex-analysis
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
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add a comment |
$begingroup$
Suppose we write a complex number $z=x+iy$. Then we have $overline{z}=x-iy$, and therefore $x=dfrac12(z+overline{z})$, $y=-dfrac12i(z-overline{z})$.
If the rules of calculus were applicable, we would obtain $$frac{partial{f}}{partial{z}}=frac12left(frac{partial{f}}{partial{x}}-ifrac{partial{f}}{partial{y}}right), frac{partial{f}}{partial{overline{z}}}=frac12left(frac{partial{f}}{partial{x}}+ifrac{partial{f}}{partial{y}}right)$$
How does this follow from the rules of calculus?
calculus complex-analysis
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
$endgroup$
13
$begingroup$
Chain rule: $frac{partial f}{partial z} = frac{partial f}{partial x}cdotfrac{partial x}{partial z} + frac{partial f}{partial y}cdotfrac{partial y}{partial z}$.
$endgroup$
– Daniel Fischer♦
Aug 23 '13 at 18:04
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives.
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
$endgroup$
– Dèö
Dec 29 '18 at 17:42
add a comment |
$begingroup$
Suppose we write a complex number $z=x+iy$. Then we have $overline{z}=x-iy$, and therefore $x=dfrac12(z+overline{z})$, $y=-dfrac12i(z-overline{z})$.
If the rules of calculus were applicable, we would obtain $$frac{partial{f}}{partial{z}}=frac12left(frac{partial{f}}{partial{x}}-ifrac{partial{f}}{partial{y}}right), frac{partial{f}}{partial{overline{z}}}=frac12left(frac{partial{f}}{partial{x}}+ifrac{partial{f}}{partial{y}}right)$$
How does this follow from the rules of calculus?
calculus complex-analysis
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
$endgroup$
Suppose we write a complex number $z=x+iy$. Then we have $overline{z}=x-iy$, and therefore $x=dfrac12(z+overline{z})$, $y=-dfrac12i(z-overline{z})$.
If the rules of calculus were applicable, we would obtain $$frac{partial{f}}{partial{z}}=frac12left(frac{partial{f}}{partial{x}}-ifrac{partial{f}}{partial{y}}right), frac{partial{f}}{partial{overline{z}}}=frac12left(frac{partial{f}}{partial{x}}+ifrac{partial{f}}{partial{y}}right)$$
How does this follow from the rules of calculus?
calculus complex-analysis
calculus complex-analysis
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
asked Aug 23 '13 at 18:01
PJ MillerPJ Miller
2,57822769
2,57822769
13
$begingroup$
Chain rule: $frac{partial f}{partial z} = frac{partial f}{partial x}cdotfrac{partial x}{partial z} + frac{partial f}{partial y}cdotfrac{partial y}{partial z}$.
$endgroup$
– Daniel Fischer♦
Aug 23 '13 at 18:04
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives.
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
$endgroup$
– Dèö
Dec 29 '18 at 17:42
add a comment |
13
$begingroup$
Chain rule: $frac{partial f}{partial z} = frac{partial f}{partial x}cdotfrac{partial x}{partial z} + frac{partial f}{partial y}cdotfrac{partial y}{partial z}$.
$endgroup$
– Daniel Fischer♦
Aug 23 '13 at 18:04
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives.
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
$endgroup$
– Dèö
Dec 29 '18 at 17:42
13
13
$begingroup$
Chain rule: $frac{partial f}{partial z} = frac{partial f}{partial x}cdotfrac{partial x}{partial z} + frac{partial f}{partial y}cdotfrac{partial y}{partial z}$.
$endgroup$
– Daniel Fischer♦
Aug 23 '13 at 18:04
$begingroup$
Chain rule: $frac{partial f}{partial z} = frac{partial f}{partial x}cdotfrac{partial x}{partial z} + frac{partial f}{partial y}cdotfrac{partial y}{partial z}$.
$endgroup$
– Daniel Fischer♦
Aug 23 '13 at 18:04
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives.
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives.
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$
$endgroup$
– Dèö
Dec 29 '18 at 17:41
$begingroup$
It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
$endgroup$
– Dèö
Dec 29 '18 at 17:42
$begingroup$
It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
$endgroup$
– Dèö
Dec 29 '18 at 17:42
add a comment |
3 Answers
3
active
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votes
$begingroup$
Let me expand Daniel Fischer's hint.
$z = x + iy rightarrow bar{z} = x - iy$ which gives $x = frac{1}{2}(z + bar{z})$ and $y = frac{1}{2i}(z - bar{z})$. So we may consider the function $f(x,y)$ as a function of $z$ and $bar{z}$.
Differentiating the relations we shall get $frac{partial{x}}{partial{z}} = frac{1}{2}$, $frac{partial{y}}{partial{z}} = frac{1}{2i}$, $frac{partial{x}}{partial{bar{z}}} = frac{1}{2}$ and $frac{partial{y}}{partial{bar{z}}} =frac{-1}{2i}$
Now use chain rule
$$frac{partial{f}}{partial{z}} = frac{partial{f}}{partial{x}}frac{partial{x}}{partial{z}} + frac{partial{f}}{partial{y}}frac{partial{y}}{partial{z}} = frac{1}{2}left(frac{partial{f}}{partial{x}} - ifrac{partial{f}}{partial{y}}right)$$
Similarly applying the chain rule for $bar{z}$ you shall get another result.
Here we shall assume that all the rules of calculus applied here is applicable.
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
$endgroup$
2
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
add a comment |
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives. One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$ It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
answered Dec 29 '18 at 12:41
DèöDèö
16113
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For the usual complex derivative df(z)/dz to exist, the Cauchy-Riemann equations must be satisfied.
————————————————————
We can deduce the Cauchy-Riemann equations heuristically based on the chain rule. Assuming that f'(z) exists, we have:
f'(z) = f'(x)x'(z)+f'(y)y'(z).
Since x=(z+z*)/2 and y=-i(z-z*)/2), we can do the derivations x'(z) and y'(z):
f'(z) = 1/2[f'(x)(1+z*')-if'(y)(1-z*')] (1)
Regarding the term z*', from the definition of the complex derivative, we have:
z*' = limdz->0 [(z+dz)*-z*]/dz
= limdz->0 dz*/dz
= limdz->0 (dx-idy)/(dx+idy).
If dy=0, this is 1. If dx=0, it is -1. If dx=dy, it is -i. If dx=-dy, it is i. Hence, this is a non-existing limit value, since it depends on the path dz traverses in order to approach 0.
For the derivative of f(z) to exist, we must require z*' to vanish from the expression (1) above. The part that involves z*' is:
1/2[f'(x)+if'(y)]z*'
By requiring it to be zero, we obtain:
f'(x)=-if'(y).
This is the Cauchy-Riemann equations in complex form.
If we insert f(z)=u(x,y)+iv(x,y), where u and v are the real and imaginary parts of f(z), respectively, we get:
u'(x)+iv'(x)=v'(y)-iu'(y).
By setting the real and imaginary parts on either side equal to each other, we obtain the Cauchy-Rieman equations, which must hold for f(z) to be differentiable at z:
u'(x)=v'(y),
u'(y)=-v'(x).
If we assume that those are satisfied, (1) reduces to:
f'(z)=1/2[f'(x)-if'(y)], (2)
which is what they wrote.
————————————————————
A non-heuristic way to derive the Cauchy-Riemann equations, and also equation (2), is as follows:
Again we assume thayt f'(z) exists.
We start with the definition of the complex derivative:
f'(z) = lim dz->0 [f(z+dz)-f(z)]/dz,
where dz=dx+idy.
This limit exists only if it is independent of which way dz approaches zero.
First, let dx=0, and derive the derivative. Then let, instead, dy=0 and redo the calculation. Finally, insert f(z)=u(x,y)+iv(x,y). We then get:
f'(z) = u'(x)+iv'(x) (3)
and
f'(z) = -iu'(y)+v'(y). (4)
When we equate the real and imaginary parts, we have the Cauchy-Riemann equations. Our result is, that if f(z) is differentiable at z, then they must be satisfied. We have also shown, that if f'(z) exists, the four partial derivatives in (3) and (4) must exist.
Assuming that f'(z) exists, and by writing f'(z)=1/2[f'(z)+f'(z)], from (3) and (4) we have:
f'(z) = 1/2[u'(x)+iv'(x)-iu'(y)+v'(y)]
= 1/2[u'(x)+iv'(x)-i(u'(y)+iv'(y))]
= 1/2[f'(x)-if'(y)].
We have shown that if we assume that f'(z) exists, then the Cauchy-Riemann equations will be satisfied, and (2) will hold.
————————————————————
It is possible to show that if two real-valued continuous functions u(x,y) and v(x,y) have continuous first-order partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the complex function f(z)=u+iv is analytic (defined and differentiable) in D.
————————————————————
An example where (2) does not hold, is when f(z)=|z|^2. We then have:
f(z) = |z|^2
= x^2+y^2
= zz*.
By assuming that f'(z) exists and using the product rule, we have:
f'(z) = 1z*+zz*'
= z*+zz*'.
We see that unless z=0, this involves z*', that does not exist. We can therefore conclude that the function |z|^2 is a function that is not differentiable at any point, except at z=0. That is because its Cauchy-Riemann equations are not satisfied, except at that point. Equation (2), which would give f'(z) equal to z*, is therefore not valid.
Although it's derivative exists in z=0, it is not analytic there, because it is not analytic in a neighborhood of that point. The function |z|^2 is therefore not analytic.
————————————————————
Finally, it can be mentioned that although the ordinary complex derivative requires the Cauchy-Riemann equations to hold, the Wirtinger derivatives do not. These derivatives are what Daniel Fisher defined above. The Wirtinger derivative of |z|^2 is z*, and the Wirtinger derivative of z* is zero.
Wirtinger calculus is outlined here:
https://onlinelibrary.wiley.com/doi/pdf/10.1002/0471439002.app1
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
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3 Answers
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3 Answers
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$begingroup$
Let me expand Daniel Fischer's hint.
$z = x + iy rightarrow bar{z} = x - iy$ which gives $x = frac{1}{2}(z + bar{z})$ and $y = frac{1}{2i}(z - bar{z})$. So we may consider the function $f(x,y)$ as a function of $z$ and $bar{z}$.
Differentiating the relations we shall get $frac{partial{x}}{partial{z}} = frac{1}{2}$, $frac{partial{y}}{partial{z}} = frac{1}{2i}$, $frac{partial{x}}{partial{bar{z}}} = frac{1}{2}$ and $frac{partial{y}}{partial{bar{z}}} =frac{-1}{2i}$
Now use chain rule
$$frac{partial{f}}{partial{z}} = frac{partial{f}}{partial{x}}frac{partial{x}}{partial{z}} + frac{partial{f}}{partial{y}}frac{partial{y}}{partial{z}} = frac{1}{2}left(frac{partial{f}}{partial{x}} - ifrac{partial{f}}{partial{y}}right)$$
Similarly applying the chain rule for $bar{z}$ you shall get another result.
Here we shall assume that all the rules of calculus applied here is applicable.
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
$endgroup$
2
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
add a comment |
$begingroup$
Let me expand Daniel Fischer's hint.
$z = x + iy rightarrow bar{z} = x - iy$ which gives $x = frac{1}{2}(z + bar{z})$ and $y = frac{1}{2i}(z - bar{z})$. So we may consider the function $f(x,y)$ as a function of $z$ and $bar{z}$.
Differentiating the relations we shall get $frac{partial{x}}{partial{z}} = frac{1}{2}$, $frac{partial{y}}{partial{z}} = frac{1}{2i}$, $frac{partial{x}}{partial{bar{z}}} = frac{1}{2}$ and $frac{partial{y}}{partial{bar{z}}} =frac{-1}{2i}$
Now use chain rule
$$frac{partial{f}}{partial{z}} = frac{partial{f}}{partial{x}}frac{partial{x}}{partial{z}} + frac{partial{f}}{partial{y}}frac{partial{y}}{partial{z}} = frac{1}{2}left(frac{partial{f}}{partial{x}} - ifrac{partial{f}}{partial{y}}right)$$
Similarly applying the chain rule for $bar{z}$ you shall get another result.
Here we shall assume that all the rules of calculus applied here is applicable.
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
$endgroup$
2
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
add a comment |
$begingroup$
Let me expand Daniel Fischer's hint.
$z = x + iy rightarrow bar{z} = x - iy$ which gives $x = frac{1}{2}(z + bar{z})$ and $y = frac{1}{2i}(z - bar{z})$. So we may consider the function $f(x,y)$ as a function of $z$ and $bar{z}$.
Differentiating the relations we shall get $frac{partial{x}}{partial{z}} = frac{1}{2}$, $frac{partial{y}}{partial{z}} = frac{1}{2i}$, $frac{partial{x}}{partial{bar{z}}} = frac{1}{2}$ and $frac{partial{y}}{partial{bar{z}}} =frac{-1}{2i}$
Now use chain rule
$$frac{partial{f}}{partial{z}} = frac{partial{f}}{partial{x}}frac{partial{x}}{partial{z}} + frac{partial{f}}{partial{y}}frac{partial{y}}{partial{z}} = frac{1}{2}left(frac{partial{f}}{partial{x}} - ifrac{partial{f}}{partial{y}}right)$$
Similarly applying the chain rule for $bar{z}$ you shall get another result.
Here we shall assume that all the rules of calculus applied here is applicable.
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
$endgroup$
Let me expand Daniel Fischer's hint.
$z = x + iy rightarrow bar{z} = x - iy$ which gives $x = frac{1}{2}(z + bar{z})$ and $y = frac{1}{2i}(z - bar{z})$. So we may consider the function $f(x,y)$ as a function of $z$ and $bar{z}$.
Differentiating the relations we shall get $frac{partial{x}}{partial{z}} = frac{1}{2}$, $frac{partial{y}}{partial{z}} = frac{1}{2i}$, $frac{partial{x}}{partial{bar{z}}} = frac{1}{2}$ and $frac{partial{y}}{partial{bar{z}}} =frac{-1}{2i}$
Now use chain rule
$$frac{partial{f}}{partial{z}} = frac{partial{f}}{partial{x}}frac{partial{x}}{partial{z}} + frac{partial{f}}{partial{y}}frac{partial{y}}{partial{z}} = frac{1}{2}left(frac{partial{f}}{partial{x}} - ifrac{partial{f}}{partial{y}}right)$$
Similarly applying the chain rule for $bar{z}$ you shall get another result.
Here we shall assume that all the rules of calculus applied here is applicable.
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
answered Aug 24 '13 at 1:36
DuttaDutta
3,78852243
3,78852243
2
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
add a comment |
2
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
2
2
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
$begingroup$
Any rules that you learned in calculus about derivatives of functions of a single variable, or derivatives of functions of two variables, apply to analytic functions in the complex plane. You can apply the rules to f(z) where z is a complex number, or to f(z) = u(z) + iv(z), or to f(x + iy). Things are simpler in the complex plane however because if f'(a) exists, f is analytic in some disk around a.
$endgroup$
– Betty Mock
Aug 24 '13 at 2:01
add a comment |
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives. One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$ It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
answered Dec 29 '18 at 12:41
DèöDèö
16113
$endgroup$
add a comment |
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives. One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$ It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
answered Dec 29 '18 at 12:41
DèöDèö
16113
$endgroup$
add a comment |
$begingroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives. One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$ It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
answered Dec 29 '18 at 12:41
DèöDèö
16113
$endgroup$
This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives. One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$ It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
answered Dec 29 '18 at 12:41
DèöDèö
16113
answered Dec 29 '18 at 12:41
DèöDèö
16113
answered Dec 29 '18 at 12:41
DèöDèö
16113
answered Dec 29 '18 at 12:41
DèöDèö
16113
16113
add a comment |
add a comment |
$begingroup$
For the usual complex derivative df(z)/dz to exist, the Cauchy-Riemann equations must be satisfied.
————————————————————
We can deduce the Cauchy-Riemann equations heuristically based on the chain rule. Assuming that f'(z) exists, we have:
f'(z) = f'(x)x'(z)+f'(y)y'(z).
Since x=(z+z*)/2 and y=-i(z-z*)/2), we can do the derivations x'(z) and y'(z):
f'(z) = 1/2[f'(x)(1+z*')-if'(y)(1-z*')] (1)
Regarding the term z*', from the definition of the complex derivative, we have:
z*' = limdz->0 [(z+dz)*-z*]/dz
= limdz->0 dz*/dz
= limdz->0 (dx-idy)/(dx+idy).
If dy=0, this is 1. If dx=0, it is -1. If dx=dy, it is -i. If dx=-dy, it is i. Hence, this is a non-existing limit value, since it depends on the path dz traverses in order to approach 0.
For the derivative of f(z) to exist, we must require z*' to vanish from the expression (1) above. The part that involves z*' is:
1/2[f'(x)+if'(y)]z*'
By requiring it to be zero, we obtain:
f'(x)=-if'(y).
This is the Cauchy-Riemann equations in complex form.
If we insert f(z)=u(x,y)+iv(x,y), where u and v are the real and imaginary parts of f(z), respectively, we get:
u'(x)+iv'(x)=v'(y)-iu'(y).
By setting the real and imaginary parts on either side equal to each other, we obtain the Cauchy-Rieman equations, which must hold for f(z) to be differentiable at z:
u'(x)=v'(y),
u'(y)=-v'(x).
If we assume that those are satisfied, (1) reduces to:
f'(z)=1/2[f'(x)-if'(y)], (2)
which is what they wrote.
————————————————————
A non-heuristic way to derive the Cauchy-Riemann equations, and also equation (2), is as follows:
Again we assume thayt f'(z) exists.
We start with the definition of the complex derivative:
f'(z) = lim dz->0 [f(z+dz)-f(z)]/dz,
where dz=dx+idy.
This limit exists only if it is independent of which way dz approaches zero.
First, let dx=0, and derive the derivative. Then let, instead, dy=0 and redo the calculation. Finally, insert f(z)=u(x,y)+iv(x,y). We then get:
f'(z) = u'(x)+iv'(x) (3)
and
f'(z) = -iu'(y)+v'(y). (4)
When we equate the real and imaginary parts, we have the Cauchy-Riemann equations. Our result is, that if f(z) is differentiable at z, then they must be satisfied. We have also shown, that if f'(z) exists, the four partial derivatives in (3) and (4) must exist.
Assuming that f'(z) exists, and by writing f'(z)=1/2[f'(z)+f'(z)], from (3) and (4) we have:
f'(z) = 1/2[u'(x)+iv'(x)-iu'(y)+v'(y)]
= 1/2[u'(x)+iv'(x)-i(u'(y)+iv'(y))]
= 1/2[f'(x)-if'(y)].
We have shown that if we assume that f'(z) exists, then the Cauchy-Riemann equations will be satisfied, and (2) will hold.
————————————————————
It is possible to show that if two real-valued continuous functions u(x,y) and v(x,y) have continuous first-order partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the complex function f(z)=u+iv is analytic (defined and differentiable) in D.
————————————————————
An example where (2) does not hold, is when f(z)=|z|^2. We then have:
f(z) = |z|^2
= x^2+y^2
= zz*.
By assuming that f'(z) exists and using the product rule, we have:
f'(z) = 1z*+zz*'
= z*+zz*'.
We see that unless z=0, this involves z*', that does not exist. We can therefore conclude that the function |z|^2 is a function that is not differentiable at any point, except at z=0. That is because its Cauchy-Riemann equations are not satisfied, except at that point. Equation (2), which would give f'(z) equal to z*, is therefore not valid.
Although it's derivative exists in z=0, it is not analytic there, because it is not analytic in a neighborhood of that point. The function |z|^2 is therefore not analytic.
————————————————————
Finally, it can be mentioned that although the ordinary complex derivative requires the Cauchy-Riemann equations to hold, the Wirtinger derivatives do not. These derivatives are what Daniel Fisher defined above. The Wirtinger derivative of |z|^2 is z*, and the Wirtinger derivative of z* is zero.
Wirtinger calculus is outlined here:
https://onlinelibrary.wiley.com/doi/pdf/10.1002/0471439002.app1
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
$endgroup$
add a comment |
$begingroup$
For the usual complex derivative df(z)/dz to exist, the Cauchy-Riemann equations must be satisfied.
————————————————————
We can deduce the Cauchy-Riemann equations heuristically based on the chain rule. Assuming that f'(z) exists, we have:
f'(z) = f'(x)x'(z)+f'(y)y'(z).
Since x=(z+z*)/2 and y=-i(z-z*)/2), we can do the derivations x'(z) and y'(z):
f'(z) = 1/2[f'(x)(1+z*')-if'(y)(1-z*')] (1)
Regarding the term z*', from the definition of the complex derivative, we have:
z*' = limdz->0 [(z+dz)*-z*]/dz
= limdz->0 dz*/dz
= limdz->0 (dx-idy)/(dx+idy).
If dy=0, this is 1. If dx=0, it is -1. If dx=dy, it is -i. If dx=-dy, it is i. Hence, this is a non-existing limit value, since it depends on the path dz traverses in order to approach 0.
For the derivative of f(z) to exist, we must require z*' to vanish from the expression (1) above. The part that involves z*' is:
1/2[f'(x)+if'(y)]z*'
By requiring it to be zero, we obtain:
f'(x)=-if'(y).
This is the Cauchy-Riemann equations in complex form.
If we insert f(z)=u(x,y)+iv(x,y), where u and v are the real and imaginary parts of f(z), respectively, we get:
u'(x)+iv'(x)=v'(y)-iu'(y).
By setting the real and imaginary parts on either side equal to each other, we obtain the Cauchy-Rieman equations, which must hold for f(z) to be differentiable at z:
u'(x)=v'(y),
u'(y)=-v'(x).
If we assume that those are satisfied, (1) reduces to:
f'(z)=1/2[f'(x)-if'(y)], (2)
which is what they wrote.
————————————————————
A non-heuristic way to derive the Cauchy-Riemann equations, and also equation (2), is as follows:
Again we assume thayt f'(z) exists.
We start with the definition of the complex derivative:
f'(z) = lim dz->0 [f(z+dz)-f(z)]/dz,
where dz=dx+idy.
This limit exists only if it is independent of which way dz approaches zero.
First, let dx=0, and derive the derivative. Then let, instead, dy=0 and redo the calculation. Finally, insert f(z)=u(x,y)+iv(x,y). We then get:
f'(z) = u'(x)+iv'(x) (3)
and
f'(z) = -iu'(y)+v'(y). (4)
When we equate the real and imaginary parts, we have the Cauchy-Riemann equations. Our result is, that if f(z) is differentiable at z, then they must be satisfied. We have also shown, that if f'(z) exists, the four partial derivatives in (3) and (4) must exist.
Assuming that f'(z) exists, and by writing f'(z)=1/2[f'(z)+f'(z)], from (3) and (4) we have:
f'(z) = 1/2[u'(x)+iv'(x)-iu'(y)+v'(y)]
= 1/2[u'(x)+iv'(x)-i(u'(y)+iv'(y))]
= 1/2[f'(x)-if'(y)].
We have shown that if we assume that f'(z) exists, then the Cauchy-Riemann equations will be satisfied, and (2) will hold.
————————————————————
It is possible to show that if two real-valued continuous functions u(x,y) and v(x,y) have continuous first-order partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the complex function f(z)=u+iv is analytic (defined and differentiable) in D.
————————————————————
An example where (2) does not hold, is when f(z)=|z|^2. We then have:
f(z) = |z|^2
= x^2+y^2
= zz*.
By assuming that f'(z) exists and using the product rule, we have:
f'(z) = 1z*+zz*'
= z*+zz*'.
We see that unless z=0, this involves z*', that does not exist. We can therefore conclude that the function |z|^2 is a function that is not differentiable at any point, except at z=0. That is because its Cauchy-Riemann equations are not satisfied, except at that point. Equation (2), which would give f'(z) equal to z*, is therefore not valid.
Although it's derivative exists in z=0, it is not analytic there, because it is not analytic in a neighborhood of that point. The function |z|^2 is therefore not analytic.
————————————————————
Finally, it can be mentioned that although the ordinary complex derivative requires the Cauchy-Riemann equations to hold, the Wirtinger derivatives do not. These derivatives are what Daniel Fisher defined above. The Wirtinger derivative of |z|^2 is z*, and the Wirtinger derivative of z* is zero.
Wirtinger calculus is outlined here:
https://onlinelibrary.wiley.com/doi/pdf/10.1002/0471439002.app1
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
$endgroup$
add a comment |
$begingroup$
For the usual complex derivative df(z)/dz to exist, the Cauchy-Riemann equations must be satisfied.
————————————————————
We can deduce the Cauchy-Riemann equations heuristically based on the chain rule. Assuming that f'(z) exists, we have:
f'(z) = f'(x)x'(z)+f'(y)y'(z).
Since x=(z+z*)/2 and y=-i(z-z*)/2), we can do the derivations x'(z) and y'(z):
f'(z) = 1/2[f'(x)(1+z*')-if'(y)(1-z*')] (1)
Regarding the term z*', from the definition of the complex derivative, we have:
z*' = limdz->0 [(z+dz)*-z*]/dz
= limdz->0 dz*/dz
= limdz->0 (dx-idy)/(dx+idy).
If dy=0, this is 1. If dx=0, it is -1. If dx=dy, it is -i. If dx=-dy, it is i. Hence, this is a non-existing limit value, since it depends on the path dz traverses in order to approach 0.
For the derivative of f(z) to exist, we must require z*' to vanish from the expression (1) above. The part that involves z*' is:
1/2[f'(x)+if'(y)]z*'
By requiring it to be zero, we obtain:
f'(x)=-if'(y).
This is the Cauchy-Riemann equations in complex form.
If we insert f(z)=u(x,y)+iv(x,y), where u and v are the real and imaginary parts of f(z), respectively, we get:
u'(x)+iv'(x)=v'(y)-iu'(y).
By setting the real and imaginary parts on either side equal to each other, we obtain the Cauchy-Rieman equations, which must hold for f(z) to be differentiable at z:
u'(x)=v'(y),
u'(y)=-v'(x).
If we assume that those are satisfied, (1) reduces to:
f'(z)=1/2[f'(x)-if'(y)], (2)
which is what they wrote.
————————————————————
A non-heuristic way to derive the Cauchy-Riemann equations, and also equation (2), is as follows:
Again we assume thayt f'(z) exists.
We start with the definition of the complex derivative:
f'(z) = lim dz->0 [f(z+dz)-f(z)]/dz,
where dz=dx+idy.
This limit exists only if it is independent of which way dz approaches zero.
First, let dx=0, and derive the derivative. Then let, instead, dy=0 and redo the calculation. Finally, insert f(z)=u(x,y)+iv(x,y). We then get:
f'(z) = u'(x)+iv'(x) (3)
and
f'(z) = -iu'(y)+v'(y). (4)
When we equate the real and imaginary parts, we have the Cauchy-Riemann equations. Our result is, that if f(z) is differentiable at z, then they must be satisfied. We have also shown, that if f'(z) exists, the four partial derivatives in (3) and (4) must exist.
Assuming that f'(z) exists, and by writing f'(z)=1/2[f'(z)+f'(z)], from (3) and (4) we have:
f'(z) = 1/2[u'(x)+iv'(x)-iu'(y)+v'(y)]
= 1/2[u'(x)+iv'(x)-i(u'(y)+iv'(y))]
= 1/2[f'(x)-if'(y)].
We have shown that if we assume that f'(z) exists, then the Cauchy-Riemann equations will be satisfied, and (2) will hold.
————————————————————
It is possible to show that if two real-valued continuous functions u(x,y) and v(x,y) have continuous first-order partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the complex function f(z)=u+iv is analytic (defined and differentiable) in D.
————————————————————
An example where (2) does not hold, is when f(z)=|z|^2. We then have:
f(z) = |z|^2
= x^2+y^2
= zz*.
By assuming that f'(z) exists and using the product rule, we have:
f'(z) = 1z*+zz*'
= z*+zz*'.
We see that unless z=0, this involves z*', that does not exist. We can therefore conclude that the function |z|^2 is a function that is not differentiable at any point, except at z=0. That is because its Cauchy-Riemann equations are not satisfied, except at that point. Equation (2), which would give f'(z) equal to z*, is therefore not valid.
Although it's derivative exists in z=0, it is not analytic there, because it is not analytic in a neighborhood of that point. The function |z|^2 is therefore not analytic.
————————————————————
Finally, it can be mentioned that although the ordinary complex derivative requires the Cauchy-Riemann equations to hold, the Wirtinger derivatives do not. These derivatives are what Daniel Fisher defined above. The Wirtinger derivative of |z|^2 is z*, and the Wirtinger derivative of z* is zero.
Wirtinger calculus is outlined here:
https://onlinelibrary.wiley.com/doi/pdf/10.1002/0471439002.app1
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
$endgroup$
For the usual complex derivative df(z)/dz to exist, the Cauchy-Riemann equations must be satisfied.
————————————————————
We can deduce the Cauchy-Riemann equations heuristically based on the chain rule. Assuming that f'(z) exists, we have:
f'(z) = f'(x)x'(z)+f'(y)y'(z).
Since x=(z+z*)/2 and y=-i(z-z*)/2), we can do the derivations x'(z) and y'(z):
f'(z) = 1/2[f'(x)(1+z*')-if'(y)(1-z*')] (1)
Regarding the term z*', from the definition of the complex derivative, we have:
z*' = limdz->0 [(z+dz)*-z*]/dz
= limdz->0 dz*/dz
= limdz->0 (dx-idy)/(dx+idy).
If dy=0, this is 1. If dx=0, it is -1. If dx=dy, it is -i. If dx=-dy, it is i. Hence, this is a non-existing limit value, since it depends on the path dz traverses in order to approach 0.
For the derivative of f(z) to exist, we must require z*' to vanish from the expression (1) above. The part that involves z*' is:
1/2[f'(x)+if'(y)]z*'
By requiring it to be zero, we obtain:
f'(x)=-if'(y).
This is the Cauchy-Riemann equations in complex form.
If we insert f(z)=u(x,y)+iv(x,y), where u and v are the real and imaginary parts of f(z), respectively, we get:
u'(x)+iv'(x)=v'(y)-iu'(y).
By setting the real and imaginary parts on either side equal to each other, we obtain the Cauchy-Rieman equations, which must hold for f(z) to be differentiable at z:
u'(x)=v'(y),
u'(y)=-v'(x).
If we assume that those are satisfied, (1) reduces to:
f'(z)=1/2[f'(x)-if'(y)], (2)
which is what they wrote.
————————————————————
A non-heuristic way to derive the Cauchy-Riemann equations, and also equation (2), is as follows:
Again we assume thayt f'(z) exists.
We start with the definition of the complex derivative:
f'(z) = lim dz->0 [f(z+dz)-f(z)]/dz,
where dz=dx+idy.
This limit exists only if it is independent of which way dz approaches zero.
First, let dx=0, and derive the derivative. Then let, instead, dy=0 and redo the calculation. Finally, insert f(z)=u(x,y)+iv(x,y). We then get:
f'(z) = u'(x)+iv'(x) (3)
and
f'(z) = -iu'(y)+v'(y). (4)
When we equate the real and imaginary parts, we have the Cauchy-Riemann equations. Our result is, that if f(z) is differentiable at z, then they must be satisfied. We have also shown, that if f'(z) exists, the four partial derivatives in (3) and (4) must exist.
Assuming that f'(z) exists, and by writing f'(z)=1/2[f'(z)+f'(z)], from (3) and (4) we have:
f'(z) = 1/2[u'(x)+iv'(x)-iu'(y)+v'(y)]
= 1/2[u'(x)+iv'(x)-i(u'(y)+iv'(y))]
= 1/2[f'(x)-if'(y)].
We have shown that if we assume that f'(z) exists, then the Cauchy-Riemann equations will be satisfied, and (2) will hold.
————————————————————
It is possible to show that if two real-valued continuous functions u(x,y) and v(x,y) have continuous first-order partial derivatives that satisfy the Cauchy-Riemann equations in some domain D, then the complex function f(z)=u+iv is analytic (defined and differentiable) in D.
————————————————————
An example where (2) does not hold, is when f(z)=|z|^2. We then have:
f(z) = |z|^2
= x^2+y^2
= zz*.
By assuming that f'(z) exists and using the product rule, we have:
f'(z) = 1z*+zz*'
= z*+zz*'.
We see that unless z=0, this involves z*', that does not exist. We can therefore conclude that the function |z|^2 is a function that is not differentiable at any point, except at z=0. That is because its Cauchy-Riemann equations are not satisfied, except at that point. Equation (2), which would give f'(z) equal to z*, is therefore not valid.
Although it's derivative exists in z=0, it is not analytic there, because it is not analytic in a neighborhood of that point. The function |z|^2 is therefore not analytic.
————————————————————
Finally, it can be mentioned that although the ordinary complex derivative requires the Cauchy-Riemann equations to hold, the Wirtinger derivatives do not. These derivatives are what Daniel Fisher defined above. The Wirtinger derivative of |z|^2 is z*, and the Wirtinger derivative of z* is zero.
Wirtinger calculus is outlined here:
https://onlinelibrary.wiley.com/doi/pdf/10.1002/0471439002.app1
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
edited Jan 3 at 0:10
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
answered Dec 28 '18 at 22:19
Espen StensrudEspen Stensrud
12
12
add a comment |
add a comment |
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13
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Chain rule: $frac{partial f}{partial z} = frac{partial f}{partial x}cdotfrac{partial x}{partial z} + frac{partial f}{partial y}cdotfrac{partial y}{partial z}$.
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– Daniel Fischer♦
Aug 23 '13 at 18:04
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This idea is only $textit{heuristic and intuitive}$, mathematically it is incorrect. The moment you will change $z$, it's conjugate $bar{z}$ by the definition would automatically change, since $bar{z}=frac{|z|^2}{z}$. The variables $z$ and $bar{z}$ are not independent, and we can not perform the $textit{partial derivations}$, by keeping one as constant. So technically $frac{partial x}{partial z} neq frac{1}{2}$ and so as the other such partial derivatives.
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– Dèö
Dec 29 '18 at 17:41
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One has to think $z$ as a unit in the complex plane, where $z=x+iy$ is just a representation. Instead of justifying the above derivatives mathematically, one can simply define $$frac{partial f}{partial z}:= frac{1}{2}Bigg(frac{partial f}{partial x}-ifrac{partial f}{partial y}Bigg)text{ and } frac{partial f}{partial bar{z}}:= frac{1}{2}Bigg(frac{partial f}{partial x}+ifrac{partial f}{partial y}Bigg).$$
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– Dèö
Dec 29 '18 at 17:41
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It is just that the notation becomes handy and one can directily write the Cauchy-Riemann equations as $frac{partial f}{partial bar{z}} = 0$. One more place wehere one can find this notation to be handy is in the text $textit{Complex Analysis}$ by $textit{Lars V. Ahlfors}$, section $2.3. textit{Conformal Mapping.}$
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– Dèö
Dec 29 '18 at 17:42