Mixing
Slight generalization of differentiation so that Re{z} is differentiable in the complex plane?
If I make a small change $delta z$ to a complex number $z$, then $mathbb{R}mathrm{e}{z}rightarrow mathbb{R}mathrm{e} {z+delta z} = mathbb{R}mathrm{e} {z}+mathbb{R}mathrm{e} {delta z}$. Differentiation is closely related to finding linear approximations to functions about points. There is no constant $alpha$ so that $alpha delta z=mathbb{R}mathrm{e}{delta z}$, so conventionally we would say that no derivative could exist.
However, if I map $z$ to the real-valued vector $hat z=[mathrm{Re}{z},mathrm{Im}{z}]^T$, then the real-part function becomes the projecting matrix $R=bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. Clearly, the derivative with respect to $hat z$ of $Rhat z$ is $bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. This implies that there is something nearby to complex numbers where the derivative of $mathrm{Re}{ z}$ is defined. What am I dealing with here?
linear-algebra complex-analysis
asked Nov 20 '18 at 16:50
Display Name
1988
add a comment |
If I make a small change $delta z$ to a complex number $z$, then $mathbb{R}mathrm{e}{z}rightarrow mathbb{R}mathrm{e} {z+delta z} = mathbb{R}mathrm{e} {z}+mathbb{R}mathrm{e} {delta z}$. Differentiation is closely related to finding linear approximations to functions about points. There is no constant $alpha$ so that $alpha delta z=mathbb{R}mathrm{e}{delta z}$, so conventionally we would say that no derivative could exist.
However, if I map $z$ to the real-valued vector $hat z=[mathrm{Re}{z},mathrm{Im}{z}]^T$, then the real-part function becomes the projecting matrix $R=bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. Clearly, the derivative with respect to $hat z$ of $Rhat z$ is $bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. This implies that there is something nearby to complex numbers where the derivative of $mathrm{Re}{ z}$ is defined. What am I dealing with here?
linear-algebra complex-analysis
asked Nov 20 '18 at 16:50
Display Name
1988
The function $(x,y) mapsto x$ is differentiable, the function $z mapsto operatorname{re} z$ is not. Differentiability with respect to a complex scalar imposes more constraints that differentiability with respect to two real variables. Analytic functions are open maps.
– copper.hat
Nov 20 '18 at 16:59
add a comment |
If I make a small change $delta z$ to a complex number $z$, then $mathbb{R}mathrm{e}{z}rightarrow mathbb{R}mathrm{e} {z+delta z} = mathbb{R}mathrm{e} {z}+mathbb{R}mathrm{e} {delta z}$. Differentiation is closely related to finding linear approximations to functions about points. There is no constant $alpha$ so that $alpha delta z=mathbb{R}mathrm{e}{delta z}$, so conventionally we would say that no derivative could exist.
However, if I map $z$ to the real-valued vector $hat z=[mathrm{Re}{z},mathrm{Im}{z}]^T$, then the real-part function becomes the projecting matrix $R=bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. Clearly, the derivative with respect to $hat z$ of $Rhat z$ is $bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. This implies that there is something nearby to complex numbers where the derivative of $mathrm{Re}{ z}$ is defined. What am I dealing with here?
linear-algebra complex-analysis
asked Nov 20 '18 at 16:50
Display Name
1988
If I make a small change $delta z$ to a complex number $z$, then $mathbb{R}mathrm{e}{z}rightarrow mathbb{R}mathrm{e} {z+delta z} = mathbb{R}mathrm{e} {z}+mathbb{R}mathrm{e} {delta z}$. Differentiation is closely related to finding linear approximations to functions about points. There is no constant $alpha$ so that $alpha delta z=mathbb{R}mathrm{e}{delta z}$, so conventionally we would say that no derivative could exist.
However, if I map $z$ to the real-valued vector $hat z=[mathrm{Re}{z},mathrm{Im}{z}]^T$, then the real-part function becomes the projecting matrix $R=bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. Clearly, the derivative with respect to $hat z$ of $Rhat z$ is $bigl( begin{smallmatrix}1 & 0\ 0 & 0end{smallmatrix}bigr)$. This implies that there is something nearby to complex numbers where the derivative of $mathrm{Re}{ z}$ is defined. What am I dealing with here?
linear-algebra complex-analysis
linear-algebra complex-analysis
asked Nov 20 '18 at 16:50
Display Name
1988
asked Nov 20 '18 at 16:50
Display Name
1988
asked Nov 20 '18 at 16:50
Display Name
1988
asked Nov 20 '18 at 16:50
Display Name
1988
asked Nov 20 '18 at 16:50
Display Name
1988
1988
The function $(x,y) mapsto x$ is differentiable, the function $z mapsto operatorname{re} z$ is not. Differentiability with respect to a complex scalar imposes more constraints that differentiability with respect to two real variables. Analytic functions are open maps.
– copper.hat
Nov 20 '18 at 16:59
add a comment |
The function $(x,y) mapsto x$ is differentiable, the function $z mapsto operatorname{re} z$ is not. Differentiability with respect to a complex scalar imposes more constraints that differentiability with respect to two real variables. Analytic functions are open maps.
– copper.hat
Nov 20 '18 at 16:59
The function $(x,y) mapsto x$ is differentiable, the function $z mapsto operatorname{re} z$ is not. Differentiability with respect to a complex scalar imposes more constraints that differentiability with respect to two real variables. Analytic functions are open maps.
– copper.hat
Nov 20 '18 at 16:59
The function $(x,y) mapsto x$ is differentiable, the function $z mapsto operatorname{re} z$ is not. Differentiability with respect to a complex scalar imposes more constraints that differentiability with respect to two real variables. Analytic functions are open maps.
– copper.hat
Nov 20 '18 at 16:59
add a comment |
1 Answer
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Consider $phi:mathbb R^2tomathbb C:(x,y)mapsto x+iy$ and its inverse $phi^{-1}:mathbb Ctomathbb R^2:zmapsto(mathop{mathrm{Re}}z,mathop{mathrm{Im}}z)$.
If $f:mathbb Ctomathbb C$ is holomorphic, then $phi^{-1}circ fcircphi:mathbb R^2tomathbb R^2$ is differentiable and its differential at any point is of the form $begin{pmatrix}a&-b\b&aend{pmatrix}$, but the converse is not true. The function $f=mathop{mathrm{Re}}$ provides an example.
answered Nov 20 '18 at 17:10
Federico
4,689514
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
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votes
Consider $phi:mathbb R^2tomathbb C:(x,y)mapsto x+iy$ and its inverse $phi^{-1}:mathbb Ctomathbb R^2:zmapsto(mathop{mathrm{Re}}z,mathop{mathrm{Im}}z)$.
If $f:mathbb Ctomathbb C$ is holomorphic, then $phi^{-1}circ fcircphi:mathbb R^2tomathbb R^2$ is differentiable and its differential at any point is of the form $begin{pmatrix}a&-b\b&aend{pmatrix}$, but the converse is not true. The function $f=mathop{mathrm{Re}}$ provides an example.
answered Nov 20 '18 at 17:10
Federico
4,689514
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
add a comment |
Consider $phi:mathbb R^2tomathbb C:(x,y)mapsto x+iy$ and its inverse $phi^{-1}:mathbb Ctomathbb R^2:zmapsto(mathop{mathrm{Re}}z,mathop{mathrm{Im}}z)$.
If $f:mathbb Ctomathbb C$ is holomorphic, then $phi^{-1}circ fcircphi:mathbb R^2tomathbb R^2$ is differentiable and its differential at any point is of the form $begin{pmatrix}a&-b\b&aend{pmatrix}$, but the converse is not true. The function $f=mathop{mathrm{Re}}$ provides an example.
answered Nov 20 '18 at 17:10
Federico
4,689514
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
add a comment |
Consider $phi:mathbb R^2tomathbb C:(x,y)mapsto x+iy$ and its inverse $phi^{-1}:mathbb Ctomathbb R^2:zmapsto(mathop{mathrm{Re}}z,mathop{mathrm{Im}}z)$.
If $f:mathbb Ctomathbb C$ is holomorphic, then $phi^{-1}circ fcircphi:mathbb R^2tomathbb R^2$ is differentiable and its differential at any point is of the form $begin{pmatrix}a&-b\b&aend{pmatrix}$, but the converse is not true. The function $f=mathop{mathrm{Re}}$ provides an example.
answered Nov 20 '18 at 17:10
Federico
4,689514
Consider $phi:mathbb R^2tomathbb C:(x,y)mapsto x+iy$ and its inverse $phi^{-1}:mathbb Ctomathbb R^2:zmapsto(mathop{mathrm{Re}}z,mathop{mathrm{Im}}z)$.
If $f:mathbb Ctomathbb C$ is holomorphic, then $phi^{-1}circ fcircphi:mathbb R^2tomathbb R^2$ is differentiable and its differential at any point is of the form $begin{pmatrix}a&-b\b&aend{pmatrix}$, but the converse is not true. The function $f=mathop{mathrm{Re}}$ provides an example.
answered Nov 20 '18 at 17:10
Federico
4,689514
answered Nov 20 '18 at 17:10
Federico
4,689514
answered Nov 20 '18 at 17:10
Federico
4,689514
answered Nov 20 '18 at 17:10
Federico
4,689514
4,689514
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
add a comment |
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
Just a note, the form of the differential makes it look like it belongs in $mathbb{R}^{2times 2}$.
– Display Name
Nov 20 '18 at 17:20
add a comment |
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The function $(x,y) mapsto x$ is differentiable, the function $z mapsto operatorname{re} z$ is not. Differentiability with respect to a complex scalar imposes more constraints that differentiability with respect to two real variables. Analytic functions are open maps.
– copper.hat
Nov 20 '18 at 16:59