Differentiability of a Complex Function

[shen07]

$$ f:\mathbb{C}\rightarrow\mathbb{C}
\\
f(z)=\left\{\begin{array} \frac{(\bar{z})^2}/ {z} \quad z\neq0 \\
0 \quad z=0
\end{array}
\right.$$

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 $$ \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 $$\bar{z}$$

 

[Opalg]

$$ f:\mathbb{C}\rightarrow\mathbb{C}
\\
f(z)=\left\{\begin{array} \frac{(\bar{z})^2}/ {z} \quad z\neq0 \\
0 \quad z=0
\end{array}
\right.$$

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 $$ \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 $$\bar{z}$$

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 $$ \lim_{z \to 0} \frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{(\bar{z})^2}{z^2}$$ 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]

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 $$ \lim_{z \to 0} \frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{(\bar{z})^2}{z^2}$$ 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,

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

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]

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?

 

[Deveno]

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?