Verifying Whether A Complex Function Is Differentiable

[Bashyboy]

The problem is to determine whether the function

$f(z) = \left\{\begin{array}{l} \frac{\overline{z}^2}{z}~~~if~~~z \ne 0 \\ 0 ~~~if~~~z=0 \end{array}\right.$

is differentiable at the point $z=0$. My two initial thoughts were to show that the function was not continuous at the point $z=0$, which is may very well be (I don't know as of yet), or to show that the limit

$\frac{df}{dz} (z_0) = \lim\limits_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0}$

exists or does not. However, I am not sure of how to deal with the fact that $f(z)$ is a multi-part function.

Any suggestions?

 

[Bashyboy]

Ah, I think I may have figured it out. Let me type up my solution, so that it might be verified...

 

[Bashyboy]

Using the definition of the derivative given in the 1st post,

$\lim\limits_{z \rightarrow 0 } \frac{f(z) - f(z_0)}{z-z_0}$

Because $z$ only closely approaches 0, but is never actually zero, we know that $f(z)$ is evaluated by the first part of the definition:

$\lim\limits_{z \rightarrow 0 } \frac{\frac{\overline{z}^2}{z} - 0}{z-0} \iff$

$\lim\limits_{z \rightarrow 0 } \frac{\overline{z}^2}{z^2} \iff$

$\lim\limits_{z \rightarrow 0 } \left(\frac{\overline{z}}{z}\right)^2$

If we let $z = r e^{i \theta}$, then

$\lim\limits_{z \rightarrow 0 } \left( \frac{re^{i \theta}}{r e^{- i \theta}} \right)^2 \iff$

$\lim\limits_{z \rightarrow 0 } e^{4 i \theta} \iff$

This limit does not exist, because the it has a different value depending upon which line of angle $\theta$ you travel towards $0$. Therefore, the derivative does not exist at $z=0$.

 

[Mark44]

Minor point: Don't use $\iff$ to connect expressions that have the same value. Use = for these situations. $\iff$ is used between statements (usually equation or inequalities) that are equivalent; i.e., that have the same truth values.