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Pyrokenesis
Apr23-04, 10:04 AM
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
Apr23-04, 10:21 AM
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

Pyrokenesis
Apr23-04, 09:58 PM
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
Apr24-04, 06:57 PM
no one is stupid here.

Ebolamonk3y
Apr24-04, 10:07 PM
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
Apr25-04, 06:09 PM
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.