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

**liquidFuzz** · September 21st 2010, 04:04 AM

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!

**Defunkt** · September 21st 2010, 12:54 PM

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'.

**liquidFuzz** · September 21st 2010, 01:49 PM

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

**HallsofIvy** · September 22nd 2010, 03:08 AM

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 $\lim_{z\to a} f(z)= f(a)$ .

**liquidFuzz** · September 22nd 2010, 03:32 AM

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

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