Derivative of complex function using definition

**Glitch** · April 9th 2011, 04:40 AM

The question

Show that the definition of $\frac{d}{dz}(\frac{1}{1 - z}) = \frac{1}{(1 - z)^2}$

My attempt
I tried using:

$\lim_{\delta z \to 0} \frac{f(z + \delta z) - f(z)}{\delta z}$

But having fractions in the numerator in this case appears difficult to simplify. I tried substituting z = x + iy, but I ended up with a massive mess which still results in a division by 0. Is there a good method of solving this problem?

Thanks.

**Plato** · April 9th 2011, 04:48 AM

Originally Posted by Glitch

Show that the definition of $\frac{d}{dz}(\frac{1}{1 - z}) = \frac{1}{(1 - z)^2}$
$\lim_{\delta z \to 0} \frac{f(z + \delta z) - f(z)}{\delta z}$
I ended up with a massive mess

WHY?

$\dfrac{\frac{1}{1-z-dz}-\frac{1}{1-z}}{dz}=\dfrac{(1-z)-(1-z-dz)}{dz[(1-z)(1-z-dz)}$

**chisigma** · April 9th 2011, 04:49 AM

Originally Posted by Glitch

The question

Show that the definition of $\frac{d}{dz}(\frac{1}{1 - z}) = \frac{1}{(1 - z)^2}$

My attempt
I tried using:

$\lim_{\delta z \to 0} \frac{f(z + \delta z) - f(z)}{\delta z}$

But having fractions in the numerator in this case appears difficult to simplify. I tried substituting z = x + iy, but I ended up with a massive mess which still results in a division by 0. Is there a good method of solving this problem?

Thanks.

With simple algebraic steps You obtain...

$\displaystyle \frac{f(z + \delta)- f(z)}{\delta} = \frac{1}{\delta} \ \{\frac{1}{1-z-\delta} - \frac{1}{1-z}\} =$

$\displaystyle = \frac{1}{(1-z-\delta)\ (1-z)}$ (1)

Now what is the $\lim \delta \rightarrow 0$ of the expression (1)?...

Kind regards

$\chi$ $\sigma$

**Glitch** · April 9th 2011, 04:55 AM

Originally Posted by Plato

WHY?

$\dfrac{\frac{1}{1-z-dz}-\frac{1}{1-z}}{dz}=\dfrac{(1-z)-(1-z-dz)}{dz[(1-z)(1-z-dz)}$

I'm such an idiot sometimes. Thanks.

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