Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a function $f(\bar{z},z)$ mapping from $\mathbb{C}^n \times \mathbb{C}^n \rightarrow \mathbb{C}^n$, which I would like to find the roots of numerically.

Since it is nicely formulated in terms of $\bar{z}, z$, is it safe to use Wirtinger calculus $\partial_z, \partial_{\bar{z}}$, extract the partial derivatives of real and imaginary part after evaluation by $\partial_x=\partial_z+\partial_{\bar{z}}$ and $\partial_y=i(\partial_z-\partial_{\bar{z}})$ and run a real-valued Newton-like algorithm?

If I am overlooking something here, can somebody point me to a reference helping me out?

EDIT: I reworked this question with a more evolved example:

consider the function $f(\vec{z}) = z_1^2 + \bar{z}_2^3 + \bar{z}_1z_2z_3 + \bar{z}_3 + i$ and I am trying to find the zero of its first derivative.

Let's work out the first derivative $F$ in terms of the three component vector $\vec{z}$:

$F=\begin{pmatrix} \partial_{z_1} f \\ \partial_{z_2} f \\ \partial_{z_3} f \\ \partial_{\bar{z}_1} f \\ \partial_{\bar{z}_2} f \\ \partial_{\bar{z}_3} f \end{pmatrix} = \begin{pmatrix} 2z_1\\ \bar{z}_1z_3\\\bar{z}_1z_2 \\ z_2z_3 \\ 3\bar{z}_2^2 \\ 1\end{pmatrix} $

I evaluate this function with my starting point $(z_0, \bar{z}_0)$, corresponding to a starting point $(x_0, y_0)$, and map it to the real domain using the linear transformation stated above.

$F=\begin{pmatrix} \partial_{x_1} f \\ \partial_{y_1} f \\ \partial_{x_2} f \\ \partial_{y_z} f \\ \partial_{x_3} f \\ \partial_{y_3} f \end{pmatrix}_{x_0,y_0} = \vec{U_F} + i \vec{V_F} = \begin{pmatrix} U_F \\ V_F \end{pmatrix} $

Which is suddenly a 12dim vector, yielding two equations for each real variable? I am not sure where I should proceed from here...

share|improve this question

1 Answer 1

Yes, this is a reasonable approach. Essentially, you are using the Newton algorithm over real numbers, with complex variables appearing only as a formal device for calculating real derivatives.

For example, suppose you want to solve $$\begin{split} z_1^2+\bar z_2^3+\bar z_1 z_2+i&=0 \\ z_1-z_2^2 +\bar z_1\bar z_2-3 &= 0 \end{split}$$ (I'm using bars instead of stars for complex conjugation.)

The Wirtinger derivatives are $$ \frac{\partial f}{\partial z_1}=\begin{pmatrix} 2z_1 \\ 1 \end{pmatrix},\quad \frac{\partial f}{\partial \bar z_1}=\begin{pmatrix} z_2 \\ \bar z_2 \end{pmatrix},\quad \frac{\partial f}{\partial z_2}=\begin{pmatrix} \bar z_1 \\ 2z+2 \end{pmatrix},\quad \frac{\partial f}{\partial \bar z_1}=\begin{pmatrix} 3\bar z_2^2 \\ \bar z_1 \end{pmatrix} $$ These can be converted to real derivatives in $x_1,x_2,y_1,y_2$. Note that you will also be separating real and imaginary parts in the target space, e.g., $w_1=u_1+iv_1$, $w_2=u_2+iv_2$. So, for example, $$ \frac{\partial f}{\partial x_1}=\begin{pmatrix} 2z_1+z_2 \\ \bar z_2 + 1 \end{pmatrix} =\begin{pmatrix}2x_1+x_2 \\ 2y_1+y_2 \\ x_2+1 \\ -y_2 \end{pmatrix} $$ The real Jacobian matrix in this example has size $4\times 4$, and you would use it exactly as Newton method says.

share|improve this answer
    
Thank you for your answer! P.S.: I think you might rework in your answer the index of the last Wirtinger derivative and the second entry of the third. –  LostInComplexPlane Jun 16 at 6:27
    
One further question, by "separating real and imaginary parts in the target space" you mean $f_j = w_j = u_j + iv_j$ ? –  LostInComplexPlane Jun 16 at 6:48
    
And what happens if I had for example six variables? Wouldn't the Jacobian then be a $4\times 6$ matrix? What do I do then? –  LostInComplexPlane Jun 16 at 9:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.