# Help with Complex Analysis

1. Apr 23, 2004

### 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

2. Apr 23, 2004

### MiGUi

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

$$\frac{2}{(z-i)^{2}}i$$

Last edited: Apr 23, 2004
3. Apr 23, 2004

### 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.

4. Apr 24, 2004

### Theelectricchild

no one is stupid here.

5. Apr 24, 2004

### 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?

6. Apr 25, 2004

### 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.