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Complex Analysis question

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Jan21-14, 05:08 PM
P: 4

I'm sorry if I'm not posting this to the correct place - this is my first post on

My question regards derivatives of analytic functions. Here it goes:

w(z) = u(x,y) +iv(x,y)
be an analytic function,
z = x + iy,
for some x,y that are real numbers.

In order to find the derivative of this function, since it is analytic it does not matter from which direction I take the limit in the limiting process so I can easily derive that
(w(z))' = [itex]\frac{∂u(x,y)}{∂x}[/itex] +i[itex]\frac{∂v(x,y)}{∂x}[/itex]

So here is where my problem begins. I was doing some problems and then one of them asked me to find [itex]\frac{∂w(z)}{∂z}[/itex], which I believe should be exactly the same thing as the derivative above, but I tried to apply chain rule to it and thus:

[itex]\frac{∂w(z)}{∂z}[/itex] = [itex]\frac{∂u(x,y)}{∂x}[/itex][itex]\frac{∂x}{∂z}[/itex] +[itex]\frac{∂u(x,y)}{∂y}[/itex][itex]\frac{∂y}{∂z}[/itex] + i([itex]\frac{∂v(x,y)}{∂x}[/itex][itex]\frac{∂x}{∂z}[/itex] + [itex]\frac{∂v(x,y)}{∂y}[/itex][itex]\frac{∂y}{∂z}[/itex])

I get this to equal twice the initially mentioned derivative for all the functions I tried it on.
It seems that differentiating only the real or only the imaginary component (the latter multiplied by i) gives the derivative. I can't explain this to myself. I would be happy if someone points out where my error is.

Thanks in advance (apologies for my poor Latex use)
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Jan22-14, 07:32 AM
P: 1,666
Quote Quote by Natura View Post
Thanks in advance (apologies for my poor Latex use)
What's poor about it? Well, that (w(z))' thing is a little unclear. Would have been more clear to say


Now, if you did the differentiating correctly, then you should get the same results. So if you don't, then you won't right?

What exactly are all those [itex]\frac{dx}{dz}[/itex] and [itex]\frac{dy}{dz}[/itex]?
Jan22-14, 09:00 AM
Mandelbroth's Avatar
P: 615
Is it asking for the Wirtinger derivative? If so, you're actually looking to compute $$\frac{\partial w}{\partial z}=\frac{1}{2}\left(\frac{\partial w}{\partial x}-i\frac{\partial w}{\partial y}\right).$$

Jan22-14, 02:56 PM
P: 4
Complex Analysis question

Firstly, thank you for the responses.

I agree I wasn't clear enough in my initial post. I'll try to correct that now.

z = x + iy

We can rearrange to get
x = z -iy

[itex]\frac{∂x}{∂z}[/itex] = [itex]\frac{∂z}{∂z}[/itex] = 1

Similarly for y we get
[itex]\frac{∂y}{∂z}[/itex] = -i

Then using the Cauchy-Riemann relations to eliminate all of the y derivatives and substituting the above results for [itex]\frac{∂x}{∂z}[/itex] and [itex]\frac{∂y}{∂z}[/itex] I get that
[itex]\frac{∂w}{∂z}[/itex] = 2*[itex]\frac{∂w}{∂x}[/itex]

As for the Wirtinger derivative, it makes sense the way it is defined but I would like to see how it is derived because I don't see where the factor of (1/2) comes from which is apparently what I am missing.

Thanks again.
Jan22-14, 03:26 PM
P: 4
Nevermind, I can see that my expressions for [itex]\frac{∂x}{∂z}[/itex] and [itex]\frac{∂y}{∂z}[/itex] are wrong and are off by a factor of (1/2) ... Thanks again.
Jan23-14, 11:48 AM
P: 1,666
Quote Quote by Natura View Post
Nevermind, I can see that my expressions for [itex]\frac{∂x}{∂z}[/itex] and [itex]\frac{∂y}{∂z}[/itex] are wrong and are off by a factor of (1/2) ... Thanks again.
Natura, let me make sure you understand this ok?

We have [itex]w=f(z)=u(x,y)+iv(x,y)[/itex]



so that:


You got that right?
Jan25-14, 06:24 PM
P: 4
Yeah, I figured it out last time, but thanks for asking. Appreciate it. :)

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