Mixing
Characterizing critical points of a complex valued function.
$begingroup$
Consider a function $f : Bbb R^2 longrightarrow Bbb R^2$. A point $(x_0,y_0) in Bbb R^2$ is said to be a regular point of $f$ if $Df ((x_0,y_0)) : Bbb R^2 longrightarrow Bbb R^2$ is an invertible linear operator. A point $(x_0,y_0) in Bbb R^2$ is said to be a critical point of $f$ if it is not a regular point of $f$. Since $Bbb C$ is homeomorphic to $Bbb R^2$ so I was very eager to know the concept of critical points in the complex plane and I found that the critical points of a complex valued function $f$ over the complex plane are precisely those points $z_0 = x_0 + iy_0 in Bbb C$ such that $f'(z_0) = 0$. How do I relate these two concepts? In fact how do I proof the theorem which states that
"Let $f : Bbb C longrightarrow Bbb C$ be a differentiable function. Then a point $z_0 in Bbb C$ is a critical point of $f$ (in the multivariable sense) if and only if $f'(z_0) = 0$."
Please help me in proving this theorem. Then it will be really very helpful for me.
Thank you very much.
complex-analysis multivariable-calculus
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
$endgroup$
add a comment |
$begingroup$
Consider a function $f : Bbb R^2 longrightarrow Bbb R^2$. A point $(x_0,y_0) in Bbb R^2$ is said to be a regular point of $f$ if $Df ((x_0,y_0)) : Bbb R^2 longrightarrow Bbb R^2$ is an invertible linear operator. A point $(x_0,y_0) in Bbb R^2$ is said to be a critical point of $f$ if it is not a regular point of $f$. Since $Bbb C$ is homeomorphic to $Bbb R^2$ so I was very eager to know the concept of critical points in the complex plane and I found that the critical points of a complex valued function $f$ over the complex plane are precisely those points $z_0 = x_0 + iy_0 in Bbb C$ such that $f'(z_0) = 0$. How do I relate these two concepts? In fact how do I proof the theorem which states that
"Let $f : Bbb C longrightarrow Bbb C$ be a differentiable function. Then a point $z_0 in Bbb C$ is a critical point of $f$ (in the multivariable sense) if and only if $f'(z_0) = 0$."
Please help me in proving this theorem. Then it will be really very helpful for me.
Thank you very much.
complex-analysis multivariable-calculus
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
$endgroup$
add a comment |
$begingroup$
Consider a function $f : Bbb R^2 longrightarrow Bbb R^2$. A point $(x_0,y_0) in Bbb R^2$ is said to be a regular point of $f$ if $Df ((x_0,y_0)) : Bbb R^2 longrightarrow Bbb R^2$ is an invertible linear operator. A point $(x_0,y_0) in Bbb R^2$ is said to be a critical point of $f$ if it is not a regular point of $f$. Since $Bbb C$ is homeomorphic to $Bbb R^2$ so I was very eager to know the concept of critical points in the complex plane and I found that the critical points of a complex valued function $f$ over the complex plane are precisely those points $z_0 = x_0 + iy_0 in Bbb C$ such that $f'(z_0) = 0$. How do I relate these two concepts? In fact how do I proof the theorem which states that
"Let $f : Bbb C longrightarrow Bbb C$ be a differentiable function. Then a point $z_0 in Bbb C$ is a critical point of $f$ (in the multivariable sense) if and only if $f'(z_0) = 0$."
Please help me in proving this theorem. Then it will be really very helpful for me.
Thank you very much.
complex-analysis multivariable-calculus
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
$endgroup$
Consider a function $f : Bbb R^2 longrightarrow Bbb R^2$. A point $(x_0,y_0) in Bbb R^2$ is said to be a regular point of $f$ if $Df ((x_0,y_0)) : Bbb R^2 longrightarrow Bbb R^2$ is an invertible linear operator. A point $(x_0,y_0) in Bbb R^2$ is said to be a critical point of $f$ if it is not a regular point of $f$. Since $Bbb C$ is homeomorphic to $Bbb R^2$ so I was very eager to know the concept of critical points in the complex plane and I found that the critical points of a complex valued function $f$ over the complex plane are precisely those points $z_0 = x_0 + iy_0 in Bbb C$ such that $f'(z_0) = 0$. How do I relate these two concepts? In fact how do I proof the theorem which states that
"Let $f : Bbb C longrightarrow Bbb C$ be a differentiable function. Then a point $z_0 in Bbb C$ is a critical point of $f$ (in the multivariable sense) if and only if $f'(z_0) = 0$."
Please help me in proving this theorem. Then it will be really very helpful for me.
Thank you very much.
complex-analysis multivariable-calculus
complex-analysis multivariable-calculus
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
edited Jan 26 at 6:07
Dbchatto67
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
asked Jan 25 at 8:15
Dbchatto67Dbchatto67
1,859219
1,859219
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To relate the two concepts, it is helpful to write $f$ in terms of its real and imaginary parts:
$$ f(x + iy) := u(x, y) + i v(x,y).$$
We can now construct a function $g : mathbb R^2 to mathbb R^2$ as follows:
$$ g : begin{bmatrix} x \ yend{bmatrix} mapsto begin{bmatrix} u(x,y) \ v(x,y) end{bmatrix}.$$
The claim is that $f'(x_0 + i y_0) = 0$ if and only if $Dg(x_0, y_0) $ is not invertible. (Of course, $f'$ denotes the complex derivative of $f : mathbb C to mathbb C$, whereas $Dg$ denotes the real multivariate derivative of $g : mathbb R^2 to mathbb R^2$.)
Now the derivative of $f$ is given by
$$ f' = frac{partial u}{partial x} + ifrac{partial v}{partial x}.$$
Meanwhile, the derivative of $g$ is given by
$$ Dg = begin{bmatrix} frac{partial u}{partial x} & frac{partial u }{partial y} \ frac{partial v}{partial x} & frac{partial v }{partial y} end{bmatrix},$$
and its determinant is
$$ det Dg = frac{partial u}{partial x}frac{partial v}{partial y} - frac{partial v }{partial x}frac{partial u }{partial y} = left( frac{partial u }{partial x}right)^2 + left( frac{partial v}{partial x}right)^2$$
[To derive the second expression, I used the Cauchy-Riemann equations $frac{partial v}{partial y } = frac{partial u }{partial x}$ and $frac{partial u}{partial y } = - frac{partial v }{partial x}$]
Thus
$$ |f'|^2 = det Dg.$$
which means that $f'$ is zero if and only if $Dg$ is singular.
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
$endgroup$
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
add a comment |
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$begingroup$
To relate the two concepts, it is helpful to write $f$ in terms of its real and imaginary parts:
$$ f(x + iy) := u(x, y) + i v(x,y).$$
We can now construct a function $g : mathbb R^2 to mathbb R^2$ as follows:
$$ g : begin{bmatrix} x \ yend{bmatrix} mapsto begin{bmatrix} u(x,y) \ v(x,y) end{bmatrix}.$$
The claim is that $f'(x_0 + i y_0) = 0$ if and only if $Dg(x_0, y_0) $ is not invertible. (Of course, $f'$ denotes the complex derivative of $f : mathbb C to mathbb C$, whereas $Dg$ denotes the real multivariate derivative of $g : mathbb R^2 to mathbb R^2$.)
Now the derivative of $f$ is given by
$$ f' = frac{partial u}{partial x} + ifrac{partial v}{partial x}.$$
Meanwhile, the derivative of $g$ is given by
$$ Dg = begin{bmatrix} frac{partial u}{partial x} & frac{partial u }{partial y} \ frac{partial v}{partial x} & frac{partial v }{partial y} end{bmatrix},$$
and its determinant is
$$ det Dg = frac{partial u}{partial x}frac{partial v}{partial y} - frac{partial v }{partial x}frac{partial u }{partial y} = left( frac{partial u }{partial x}right)^2 + left( frac{partial v}{partial x}right)^2$$
[To derive the second expression, I used the Cauchy-Riemann equations $frac{partial v}{partial y } = frac{partial u }{partial x}$ and $frac{partial u}{partial y } = - frac{partial v }{partial x}$]
Thus
$$ |f'|^2 = det Dg.$$
which means that $f'$ is zero if and only if $Dg$ is singular.
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
$endgroup$
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
add a comment |
$begingroup$
To relate the two concepts, it is helpful to write $f$ in terms of its real and imaginary parts:
$$ f(x + iy) := u(x, y) + i v(x,y).$$
We can now construct a function $g : mathbb R^2 to mathbb R^2$ as follows:
$$ g : begin{bmatrix} x \ yend{bmatrix} mapsto begin{bmatrix} u(x,y) \ v(x,y) end{bmatrix}.$$
The claim is that $f'(x_0 + i y_0) = 0$ if and only if $Dg(x_0, y_0) $ is not invertible. (Of course, $f'$ denotes the complex derivative of $f : mathbb C to mathbb C$, whereas $Dg$ denotes the real multivariate derivative of $g : mathbb R^2 to mathbb R^2$.)
Now the derivative of $f$ is given by
$$ f' = frac{partial u}{partial x} + ifrac{partial v}{partial x}.$$
Meanwhile, the derivative of $g$ is given by
$$ Dg = begin{bmatrix} frac{partial u}{partial x} & frac{partial u }{partial y} \ frac{partial v}{partial x} & frac{partial v }{partial y} end{bmatrix},$$
and its determinant is
$$ det Dg = frac{partial u}{partial x}frac{partial v}{partial y} - frac{partial v }{partial x}frac{partial u }{partial y} = left( frac{partial u }{partial x}right)^2 + left( frac{partial v}{partial x}right)^2$$
[To derive the second expression, I used the Cauchy-Riemann equations $frac{partial v}{partial y } = frac{partial u }{partial x}$ and $frac{partial u}{partial y } = - frac{partial v }{partial x}$]
Thus
$$ |f'|^2 = det Dg.$$
which means that $f'$ is zero if and only if $Dg$ is singular.
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
$endgroup$
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
add a comment |
$begingroup$
To relate the two concepts, it is helpful to write $f$ in terms of its real and imaginary parts:
$$ f(x + iy) := u(x, y) + i v(x,y).$$
We can now construct a function $g : mathbb R^2 to mathbb R^2$ as follows:
$$ g : begin{bmatrix} x \ yend{bmatrix} mapsto begin{bmatrix} u(x,y) \ v(x,y) end{bmatrix}.$$
The claim is that $f'(x_0 + i y_0) = 0$ if and only if $Dg(x_0, y_0) $ is not invertible. (Of course, $f'$ denotes the complex derivative of $f : mathbb C to mathbb C$, whereas $Dg$ denotes the real multivariate derivative of $g : mathbb R^2 to mathbb R^2$.)
Now the derivative of $f$ is given by
$$ f' = frac{partial u}{partial x} + ifrac{partial v}{partial x}.$$
Meanwhile, the derivative of $g$ is given by
$$ Dg = begin{bmatrix} frac{partial u}{partial x} & frac{partial u }{partial y} \ frac{partial v}{partial x} & frac{partial v }{partial y} end{bmatrix},$$
and its determinant is
$$ det Dg = frac{partial u}{partial x}frac{partial v}{partial y} - frac{partial v }{partial x}frac{partial u }{partial y} = left( frac{partial u }{partial x}right)^2 + left( frac{partial v}{partial x}right)^2$$
[To derive the second expression, I used the Cauchy-Riemann equations $frac{partial v}{partial y } = frac{partial u }{partial x}$ and $frac{partial u}{partial y } = - frac{partial v }{partial x}$]
Thus
$$ |f'|^2 = det Dg.$$
which means that $f'$ is zero if and only if $Dg$ is singular.
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
$endgroup$
To relate the two concepts, it is helpful to write $f$ in terms of its real and imaginary parts:
$$ f(x + iy) := u(x, y) + i v(x,y).$$
We can now construct a function $g : mathbb R^2 to mathbb R^2$ as follows:
$$ g : begin{bmatrix} x \ yend{bmatrix} mapsto begin{bmatrix} u(x,y) \ v(x,y) end{bmatrix}.$$
The claim is that $f'(x_0 + i y_0) = 0$ if and only if $Dg(x_0, y_0) $ is not invertible. (Of course, $f'$ denotes the complex derivative of $f : mathbb C to mathbb C$, whereas $Dg$ denotes the real multivariate derivative of $g : mathbb R^2 to mathbb R^2$.)
Now the derivative of $f$ is given by
$$ f' = frac{partial u}{partial x} + ifrac{partial v}{partial x}.$$
Meanwhile, the derivative of $g$ is given by
$$ Dg = begin{bmatrix} frac{partial u}{partial x} & frac{partial u }{partial y} \ frac{partial v}{partial x} & frac{partial v }{partial y} end{bmatrix},$$
and its determinant is
$$ det Dg = frac{partial u}{partial x}frac{partial v}{partial y} - frac{partial v }{partial x}frac{partial u }{partial y} = left( frac{partial u }{partial x}right)^2 + left( frac{partial v}{partial x}right)^2$$
[To derive the second expression, I used the Cauchy-Riemann equations $frac{partial v}{partial y } = frac{partial u }{partial x}$ and $frac{partial u}{partial y } = - frac{partial v }{partial x}$]
Thus
$$ |f'|^2 = det Dg.$$
which means that $f'$ is zero if and only if $Dg$ is singular.
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
answered Jan 25 at 8:37
Kenny WongKenny Wong
19.1k21441
19.1k21441
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
add a comment |
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
$begingroup$
Very fantastic proof. +1 for it.
$endgroup$
– Dbchatto67
Jan 25 at 9:46
add a comment |
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