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differentiability of complex function

suvadip

Member
Feb 21, 2013
69
I have found a question
Prove that f(z)=Re(z) is not differentiable at any point.

According to me f(z)=Re(z)=Re(x+iy)=x which is differentiable everywhere. Then where is the mistake?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You have to differentiate between complex differentiable and the usual case of differetiability in real analysis.
If complex valued function is complex differentiable then it must satisfy the CR equations.

Or using the definition

\(\displaystyle \lim_{\Delta z \to 0} \frac{f(z+\Delta z )-f(z)}{\Delta z }= \lim _{ \Delta z \to 0}\frac{x + \Delta x - x }{\Delta z }=\lim _{ (\Delta x ,\Delta y ) \to 0}\frac{\Delta x }{\Delta x +i \Delta y }\)

Where the last limit doesn't exist .
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
I have found a question
Prove that f(z)=Re(z) is not differentiable at any point.

According to me f(z)=Re(z)=Re(x+iy)=x which is differentiable everywhere. Then where is the mistake?
The mistake is that you are confusing differenciable as used in real analysis and differenciable as used in complex analysis. It is the same word but it means different things!

Definition: A function $f:\mathbb{R}^n \to \mathbb{R}^m$ is called differenciable iff for any point $p\in \mathbb{R}^n$ there exists a linear map $L:\mathbb{R}^n \to \mathbb{R}^m$ such that $f(p+x) = f(p) + pL(x) + \varepsilon(x)$ where $\varepsilon(x)/|x| \to 0$ as $x\to 0$. This linear map $L$ is what we call the derivative of $f$ at $p$ and denote it by $Df(p)$.

In the special case when $n=m=1$, the definition of differenciable as a limit quotient coincides with this more general definition. So we rather adopt this new definition.

In complex analysis we define,

Definition: A function $f:\mathbb{C}\to \mathbb{C}$ is called differenciable iff for any point $p$ we have that the function $(f(z) - f(p))/(z-p)$ has a limit as $p\to z$. In this case we denote the derivative by $f'(p)$.

Any time we have a function $f:\mathbb{C}\to \mathbb{C}$ it induces a map $f_*:\mathbb{R}^2\to \mathbb{R}^2$. It is an exercise to show that if $f$ is differenciable in the complex sense then $f_*$ is differenciable in the real sense. But the converse is not true and your question is an example of how the converse may break down.