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Using the definition: $f'(z) = \lim_\limits{h \to 0} \frac {f(z+h)-f(z)}{h}$,

show $f(z) = \bar z$ is not differentiable if $f: \Bbb C \to \Bbb C$.

I know $\bar z = x-iy$ and i tried plugging it into the definition but I think I am doing something wrong. Thank You.

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The limit becomes $\frac{\overline{h}}{h}$ which if we let $h=x+yi$ is equal to $$\frac{x^2-y^2-2xyi}{x^2+y^2}$$ and its well known from real analysis that the limits of the real and imaginary parts dont exist.

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  • $\begingroup$ i got the limit as $\bar h / h$ but why would we let $h=x+iy$ again? $\endgroup$ – ISuckAtMathPleaseHELPME Feb 2 '17 at 22:26
  • $\begingroup$ To find the real and imaginary parts. $\endgroup$ – Rene Schipperus Feb 2 '17 at 22:27
  • $\begingroup$ after we conver h into x+iy form does (x,y) go to (0,0)? $\endgroup$ – ISuckAtMathPleaseHELPME Feb 3 '17 at 4:26
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We have $$ \frac{f(z+h)-f(z)}{h}=\frac{\bar{h}}{h} $$ If $h$ goes to zero along the real axis, the limit is $1$; if $h$ goes to zero along the imaginary axis, the limit is $-1$. So the limit doesn't exist.

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Here is another approach if you're interested in this one as well:

Let $z_0\in\mathbb C$ be arbitrary with $z_n=z_0+1/n$ as well as $y_n=z_0+\mathrm i/n$. Obviously $z_n\longrightarrow z_0~(n\to\infty)$ and $y_n\longrightarrow z_0~(n\to\infty)$. However

$$ \lim_{n\to\infty}\frac{f(z_n)-f(z_0)}{z_n-z_0} = 1\neq -1 = \lim_{n\to\infty}\frac{\overline{z_0+\mathrm i/n} - \overline{z_0}}{z_0+\mathrm i/n-z_0} = \lim_{n\to\infty}\frac{f(y_n)-f(z_0)}{y_n-z_0}. $$

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