What is the derivative of a complex function?

[Pyrokenesis]

I am having trouble with the following question, any help would be blinding.

Find the value of ther derivative of:

(z - i)/(z + i) at i.

I tried to use the fact that f'(z0) = lim z->z0 [f(z) - f(z0)]/z - z0. I also tried using the fact that z = x + iy and rationalising the denominator, but had no joy either way.

Probably just being stupid!

Dexter

 

[MiGUi]

Well:
[tex]
F'[\frac{f(x)}{g(x)}] = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^{2}}
[/tex]
so then...
[tex]
\frac{\partial}{\partial z} \left(\frac{(z - i)}{(z + i)} \right) = \frac{(z + i) - (z - i)}{(z-i)^{2}}
[/tex]
and if you order it...

[tex]
\frac{2}{(z-i)^{2}}i[/tex]

 

[Pyrokenesis]

Cheers

Thanks.

I was being stupid, that formula and fact that differentiation rules for real calculus and complex calculus is the same, was on the previous page to that question.

 

[Theelectricchild]

no one is stupid here.

 

[Ebolamonk3y]

MiGui... I am confused as to why it is (z-i)^2 and not (z+2)^2... because you set your g(x)=z+i... g(x)^2=(z+i)^2... why the negative?

 

[Pyrokenesis]

Thanks TheElectricChild.

Ebolamonk3y, I think MiGUi, just got the functions mixed up, an easy mistake to make. You are right, g(x)=z+i... g(x)^2=(z+i)^2, therefore, the answer is:

2i/(z + i)^2, which after substituting i for z, yields:

-i/2.