Compute the Jacobian matrix of $f(z,w)=(ze^{i\alpha},\ w)$

Question

I have a question:

Let $f$ be a holomorphic and $f(z,w)=(ze^{i \alpha},\ w)$ with $z,\ w \in \Bbb C$. Compute its Jacobian matrix.

I remember that:

For a holomorphic function $f$, its Jacobian matrix is $$\left(\begin{array}{cc}\mathrm{Re}~f' & -\mathrm{Im}~f'\\\mathrm{Im}~f' &\mathrm{Re}~f' \end{array}\right)$$

Is is correct? And How do we calculate the Jacobian matrix of $f$?

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Any help will be appreciated! Thanks!

Answer 1

Sort of shooting in the dark but I think I understand the main point

The Jacobian for a multivariable function f: from $R^n \rightarrow R^m$ with arguments $u_1 ... u_n$ is defined as

$$\begin{bmatrix} \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_1}{\partial u_1} ... \frac{\partial F_1}{\partial u_n} \\ \frac{\partial F_2}{\partial u_2 } \ \frac{\partial F_2}{\partial u_1} ... \frac{\partial F_2}{\partial u_n} \\ \vdots \ \ \ \ \ \ \ \vdots \ \ \ \ \ddots \ \ \ \ \vdots \\ \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_m}{\partial u_m} ... \frac{\partial F_m}{\partial u_n} \end{bmatrix} $$

Now you have an issue here in that there are two definitions of the Jacobian we are dealing with.

An Alternative Interpretation is to treat the entire Jacobian as the derivative of the function

and so the answer should take the form

$$\begin{bmatrix} Re\begin{pmatrix} \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_1}{\partial u_1} ... \frac{\partial F_1}{\partial u_n} \\ \frac{\partial F_2}{\partial u_2 } \ \frac{\partial F_2}{\partial u_1} ... \frac{\partial F_2}{\partial u_n} \\ \vdots \ \ \ \ \ \ \ \vdots \ \ \ \ \ddots \ \ \ \ \vdots \\ \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_m}{\partial u_m} ... \frac{\partial F_m}{\partial u_n} \end{pmatrix} && -Im\begin{pmatrix} \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_1}{\partial u_1} ... \frac{\partial F_1}{\partial u_n} \\ \frac{\partial F_2}{\partial u_2 } \ \frac{\partial F_2}{\partial u_1} ... \frac{\partial F_2}{\partial u_n} \\ \vdots \ \ \ \ \ \ \ \vdots \ \ \ \ \ddots \ \ \ \ \vdots \\ \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_m}{\partial u_m} ... \frac{\partial F_m}{\partial u_n} \end{pmatrix} \\ Im\begin{pmatrix} \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_1}{\partial u_1} ... \frac{\partial F_1}{\partial u_n} \\ \frac{\partial F_2}{\partial u_2 } \ \frac{\partial F_2}{\partial u_1} ... \frac{\partial F_2}{\partial u_n} \\ \vdots \ \ \ \ \ \ \ \vdots \ \ \ \ \ddots \ \ \ \ \vdots \\ \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_m}{\partial u_m} ... \frac{\partial F_m}{\partial u_n} \end{pmatrix} && Re\begin{pmatrix} \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_1}{\partial u_1} ... \frac{\partial F_1}{\partial u_n} \\ \frac{\partial F_2}{\partial u_2 } \ \frac{\partial F_2}{\partial u_1} ... \frac{\partial F_2}{\partial u_n} \\ \vdots \ \ \ \ \ \ \ \vdots \ \ \ \ \ddots \ \ \ \ \vdots \\ \frac{\partial F_1}{\partial u_2 } \ \frac{\partial F_m}{\partial u_m} ... \frac{\partial F_m}{\partial u_n} \end{pmatrix}\end{bmatrix} $$

Both answers carry the same data, I'm not exactly sure which is correct... Talk to your prof about this. My gut feeling now is that the latter convention seems more natural

Using the latter convention \begin{bmatrix} \begin{matrix} Re\left(e^{i \alpha} \right) & 0 \\ 0 & 1 \end{matrix} & \begin{matrix} -Im\left((e^{i \alpha}) \right) & 0 \\ 0 & 0 \end{matrix} \\ \begin{matrix} Im\left((e^{i \alpha}) \right) & 0 \\ 0 & 0 \end{matrix} & \begin{matrix} Re\left(e^{i \alpha} \right) & 0 \\ 0 & 1 \end{matrix} \end{bmatrix}

Answer 2

Just compute it directly:

$D f((z,w)) = \begin{bmatrix} { \partial f_1(z,w) \over \partial z} & { \partial f_1(z,w) \over \partial w} \\ { \partial f_2(z,w) \over \partial z} & { \partial f_2(z,w) \over \partial w} \end{bmatrix} = \begin{bmatrix} e^{i \alpha} & 0 \\ 0 & 1 \end{bmatrix}$