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- Thread starter suvadip
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- Jan 17, 2013

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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 .

- Jan 26, 2012

- 236

The mistake is that you are confusingI 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?

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,

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.