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[SOLVED] Differentiability of a Complex Function

shen07

Member
Aug 14, 2013
54
\(\displaystyle f:\mathbb{C}\rightarrow\mathbb{C}
\\
f(z)=\left\{\begin{array} \frac{(\bar{z})^2}/ {z} \quad z\neq0 \\
0 \quad z=0
\end{array}
\right.\)

Show that f is differentiable at z=0, but the Cauchy Riemann Equations hold at z=0.

Well i have tried to start the first part but i am stuck, could you please help me out.

WORKING:
f is diff at z=0

if \(\displaystyle \lim_{z \to 0} \frac{f(z)-f(0)}{z-0}\; exists\\
\lim_{z\to0}\frac{ \frac{(\bar{z})^2}{z}-0}{z-0}=
\lim_{z\to0}\frac{(\bar{z})^2}{z^2}

\)

Now we get indeterminate form in the limit but how can we differentiate \(\displaystyle \bar{z}\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
\(\displaystyle f:\mathbb{C}\rightarrow\mathbb{C}
\\
f(z)=\left\{\begin{array} \frac{(\bar{z})^2}/ {z} \quad z\neq0 \\
0 \quad z=0
\end{array}
\right.\)

Show that f is not differentiable at z=0, but the Cauchy Riemann Equations hold at z=0.

Well i have tried to start the first part but i am stuck, could you please help me out.

WORKING:
f is diff at z=0

if \(\displaystyle \lim_{z \to 0} \frac{f(z)-f(0)}{z-0}\; exists\\
\lim_{z\to0}\frac{ \frac{(\bar{z})^2}{z}-0}{z-0}=
\lim_{z\to0}\frac{(\bar{z})^2}{z^2}

\)

Now we get indeterminate form in the limit but how can we differentiate \(\displaystyle \bar{z}\)
There is an important word missing from the statement of the problem. The result should say that f is not differentiable at 0, despite satisfying the C–R equations at that point.

You have correctly shown that the condition for f to be differentiable at 0 is that \(\displaystyle \lim_{z \to 0} \frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{(\bar{z})^2}{z^2}\) should exist. To see that the limit does not in fact exist, use the polar form $z = re^{i\theta}$, so that $\bar{z} = re^{-i\theta}$. You should find that the limit as $r\to0$ varies for different values of $\theta$, and is thus not uniquely defined.

To see that the C–R equations hold at 0, write $z=x+iy$, $\bar{z} = x-iy$ and find the partial derivatives of $f(x,y)$ at the origin.
 

shen07

Member
Aug 14, 2013
54
There is an important word missing from the statement of the problem. The result should say that f is not differentiable at 0, despite satisfying the C–R equations at that point.

You have correctly shown that the condition for f to be differentiable at 0 is that \(\displaystyle \lim_{z \to 0} \frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{(\bar{z})^2}{z^2}\) should exist. To see that the limit does not in fact exist, use the polar form $z = re^{i\theta}$, so that $\bar{z} = re^{-i\theta}$. You should find that the limit as $r\to0$ varies for different values of $\theta$, and is thus not uniquely defined.

To see that the C–R equations hold at 0, write $z=x+iy$, $\bar{z} = x-iy$ and find the partial derivatives of $f(x,y)$ at the origin.


Well you are right i just miss the not,

\(\displaystyle \lim_{z \to 0} \frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{(\bar{z})^2}{z^2}\)
now using $z = re^{i\theta}$
\(\displaystyle \lim_{z\to0}\frac{(\bar{z})^2}{z^2}=\lim_{r\to0} \frac{(re^{-i\theta})^2}{(re^{i\theta})^2}

\)
now the r's cancel out, i didn't understand what you tried to tell me, Could you Clarify Please.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
If $\theta = 0$ then

$\displaystyle \frac{e^{-i2\theta}}{e^{i2\theta}} = 1$

If $\theta = \frac{\pi}{4}$ then

$\displaystyle \frac{e^{-i2\theta}}{e^{i2\theta}} = \frac{1}{e^{i\pi}} = -1$.

These values are independent of $r$ (as you noted, the $r$'s cancel).

What is your conclusion?
 

shen07

Member
Aug 14, 2013
54
If $\theta = 0$ then

$\displaystyle \frac{e^{-i2\theta}}{e^{i2\theta}} = 1$

If $\theta = \frac{\pi}{4}$ then

$\displaystyle \frac{e^{-i2\theta}}{e^{i2\theta}} = \frac{1}{e^{i\pi}} = -1$.

These values are independent of $r$ (as you noted, the $r$'s cancel).

What is your conclusion?
Since $\theta$ varies, the function is not continuous at z=0 hence Not Differentiable. Right?

MultiVariable Function? is it?
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
If the limit DID exist, we should be able to choose $r > 0$ such that on a disk centered at the origin with radius $r$, the quantity:

$\displaystyle \frac{(re^{-i\theta})^2}{(re^{i\theta})^2}$ is always within $\epsilon$ of the limit, for ANY $\epsilon > 0$.

What happens if we choose $0 < \epsilon < 1$? Will ANY $r > 0$ work?