Derivative of Complex Variables

EngWiPy
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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^*?

Thanks in advance

PS.: All variables and constants are complex.
 
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Is this a homework question?
 
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
 
Mark44 said:
Is this a homework question?

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.
 
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
 
lurflurf said:
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

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
 
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
 
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