Differentiability of a Complex Function

In summary: 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 $\epsilon > 0$.
  • #1
shen07
54
0
\(\displaystyle f:\mathbb{C}\rightarrow\mathbb{C}
\\
f(z)=\left\{\begin{array} \frac{(\bar{z})^2}/ {z} \quad z\neq0 \\
0 \quad z=0
\end{array}
\right.\)

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 \(\displaystyle \bar{z}\)
 
Physics news on Phys.org
  • #2
shen07 said:
\(\displaystyle f:\mathbb{C}\rightarrow\mathbb{C}
\\
f(z)=\left\{\begin{array} \frac{(\bar{z})^2}/ {z} \quad z\neq0 \\
0 \quad z=0
\end{array}
\right.\)

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.
 
  • #3
Opalg said:
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.
 
  • #4
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?
 
  • #5
Deveno said:
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:
  • #6
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?
 

FAQ: Differentiability of a Complex Function

What is differentiability of a complex function?

Differentiability of a complex function refers to the ability of a function to have a derivative at a given point in the complex plane. In other words, it measures how smooth and continuous a function is at a specific point in the complex plane.

How is differentiability different in the complex plane compared to the real plane?

In the complex plane, differentiability is much more restrictive compared to the real plane. In the real plane, a function only needs to be continuous to be differentiable, but in the complex plane, a function must satisfy the Cauchy-Riemann equations to be differentiable.

What is the significance of differentiability of a complex function?

Differentiability of a complex function is important because it allows us to define the concept of a derivative in the complex plane. This allows us to study and analyze complex functions using techniques from calculus, such as finding critical points, local extrema, and inflection points.

How can I determine if a complex function is differentiable?

To determine if a complex function is differentiable, you must check if it satisfies the Cauchy-Riemann equations at a given point in the complex plane. These equations involve taking partial derivatives of the function with respect to the real and imaginary parts of the complex variable.

Can a function be differentiable at some points but not others in the complex plane?

Yes, a function can be differentiable at some points but not others in the complex plane. This is because the Cauchy-Riemann equations only need to be satisfied at a specific point for a function to be differentiable at that point. The function may not be differentiable at other points where the Cauchy-Riemann equations are not satisfied.

Back
Top