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Show that the function $f(z) = \overline{z}$ is Not differentiable anywhere in the $z$-plane.

I am thinking about Cauchy-Riemann theorem $ f(z) = x-iy $

Here $$ \begin{cases} u(x,y) = x \\ v(x,y) = -y \end{cases} $$ and both of them are differentiable and

\begin{align} \frac{\partial u}{\partial x} = 1 \ne \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} = 0 = \frac{\partial v}{\partial x} \end{align}

I really can't give an example to prove that the function isn't differentiable anywhere.

4 Answers 4

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In your example, $v(x,y) = -y$, not $y$.

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    Ohh thanks a lot ..and so soory to this question – 2017-01-25
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The fact that the Cauchy Riemann equation isn't verified is quite enough.

For a direct proof,

$$f'(z):=\lim_{h\to0}\frac{\overline{z+h}-\overline z}h={}\lim_{\epsilon\to0}\frac{\overline{z+\epsilon e^{i\theta}}-\overline z}{\epsilon e^{i\theta}}=\frac{e^{-i\theta}}{e^{i\theta}}=e^{-i2\theta},$$ which depends on $\theta$.

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    What should be the limit to be f(z) iis defrentiable – 2017-01-25
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    @user401187: independent of $\theta$. – 2017-01-25
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You could also use the definition of a complex derivative and let $h$ go to $0$ on the real axis, and then on the imaginary axis:

  • For $h \space \epsilon \space \mathbb{R}$ is $\frac{f(z + h) - f(z)}{h}$ equal to $\frac{2h-h}{h} = 1$

  • For $h \space \epsilon \space \mathbb{C}$ is $\frac{f(z + h) - f(z)}{h}$ equal to $\frac{- 2ih +ih}{ih} = -1$

Contradiction.

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    What should be the answer for f(z) to be differentiable.? – 2017-01-25
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    They should both be equal. – 2017-01-25
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    The terminology "complex axis" is misleading, I think you mean "imaginary axis". Also, $\mathbb{C}$ is the notation for the whole complex plane, not for the imaginary axis. – 2017-01-25
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    Yes, sorry. I'm still a math rookie so I make a lot of mistakes on notations. – 2017-01-25
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We have $f(z)=\bar z$

\begin{align} f'(z)&=\lim_{{\Delta z}\to0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}\\ &=\lim_{{\Delta z}\to0}\frac{\overline {z_0+\Delta z}-\overline z_0}{\Delta z}\\ &=\lim_{{\Delta z}\to0}\frac{\overline z_0+\overline\Delta z-\overline z_0}{\Delta z}\\ &=\lim_{{\Delta z}\to0}\frac{\overline\Delta z}{\Delta z}\\ &\\ &\text{If $\Delta z\to0$ through $(\Delta x,0)$,then}\\ &\\ f'(z)&=\lim_{(\Delta x,0)\to(0,0)}\frac{\Delta x}{\Delta x}=1\\ &\\ &\text{If $\Delta z\to0$ through $(0,\Delta y)$,then}\\ &\\ f'(z)&=\lim_{(0,\Delta y)\to(0,0)}\frac{-i\Delta y}{i\Delta y}=-1\\ \end{align}

i.e $f(z)=\bar z$ is nowhere differentiable in complex plane.