# Derivative of Complex Variables

1. Oct 19, 2011

### S_David

Hi,

What is the following derivative:

$$\frac{\partial}{\partial x}|b-ax|^2$$?

Now I know that $$|b-ax|^2=(b-ax)(b^*-a^*x^*)$$, so how to do the differentiation with respect to $$x^*$$?

PS.: All variables and constants are complex.

2. Oct 20, 2011

### Staff: Mentor

Is this a homework question?

3. Oct 20, 2011

### lurflurf

It is not clear what that derivative means.
A guess would be that it is a Wirtinger derivative in which case we have
$$\frac{\partial x^*}{\partial x}=0$$

4. Oct 20, 2011

### S_David

No it is not.

So lurflurf, are you saying that the derivative will be:

$$-a(b-ax)^*$$

I thought it will be like:

$$-(b-ax)a^*$$

but I couldn't prove it.

5. Oct 20, 2011

### lurflurf

Again it is not clear what that derivative means, but a good guess would be -a(b-ax)*.
Then we have
d|b-a x|^2=-a(b-a x)* dx+-a*(b-a x) dx*

notice that |b-a x|^2 is definitely not complex differentiable as it depends upon x and x* rather than upon x alone

6. Oct 20, 2011

### S_David

No, I just need to the derivative $$\frac{\partial}{\partial x}$$, where the derivative is partial with respect to x. I see some books writing x as a+jb, and then compute the derivatives with respect to a and b. I was just wondering if there is another way to do this.

Thanks

7. Oct 20, 2011

### lurflurf

There are different ways because a complex variable can change in more ways than a real variable. We can work with different coordinates or none. So a function of a complex variable can be described in diferent ways with two variables
|x| and arg(x)
Re(x) and Im(x)
x and x*
and so on

since you are interested in the x partial x and x* are natural, but you could use any set and the chain rule to find the x partial.
$$\frac{\partial}{\partial x}=\frac{\partial u}{\partial x}\frac{\partial}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial}{\partial v}$$
However you want to choose u and v