# Thread:

1. ## f(z) is derivable - then it's continuous - Prove it!

Prove that if $f$ is derivable in $z_0$ then $f$ is continuous in $z_0$. $\Omega \in \mathbb{C}, ~f:\Omega \rightarrow \mathbb{C}, ~ z_0 \in \Omega$

I started with the definition.
$\lim_{z \to 0} \frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}$

Using Cauchys integral formula.

$\frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}$ =
$\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-(z_0+\Delta z)} \, dz -\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-z_0} \, dz$ =
$\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)} \, dz$ =
$\frac{1}{2\pi i } \oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \, dz$

Now comparing the result with Cauchy's Integral formula, first order.
$\oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \, dz - \oint_C \frac{f(z)}{(z-z_0)^2} \, dz$ = $\oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)^2} \, dz$

What happens with the right side if $\Delta z \rightarrow 0$?

I'm sort of in the dark here... I thought maybe I proved something by showing how Cauchy's and the definition of derivation dance so nicely together.

Help or any pointers would be appreciated!

2. You're overcomplicating it! Note that $lim_{z \to z_0} f(z)-f(z_0) = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}\cdot (z-z_0)$

Also, I think that the proper term is 'differentiable', not 'derivable'.

3. Well as far as languages, I like Math and my native tongue. Bare with me... Tomorrow I'll dig into the stuff you suggested.

4. Another person who wants us to "bare" with him! This website is getting too raw for me!

liquidFuzz, in order that a function be differentiable at z= a, you must have $\lim_{\Delta z\to 0}\frac{f(a+ \Delta z)- f(a)}{\Delta z}$ exist. Since that denominator obviously goes to 0, in order that the limit exist, the numerator must also exist (a necessary though not sufficient condition). That is, we must have $\lim_{\Delta z\to 0} f(a+ \Delta z}- f(a)= 0$, whence $\lim_{\Delta z\to 0} f(a+ \Delta z)= f(a)$. Letting $z= a+ \Delta z$ that is the same as $\lim_{z\to a} f(z)= f(a)$.

5. Mhmm, but by showing that the limit exists have I proved that a differentiable function is continuous?
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