Verifying Whether A Complex Function Is Differentiable

  • #1
Bashyboy
1,421
5
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?
 
Physics news on Phys.org
  • #2
Ah, I think I may have figured it out. Let me type up my solution, so that it might be verified...
 
  • #3
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##.
 
  • #4
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.
 

Similar threads

Back
Top