Determine where $f'(z)$ exists and find its value at those points.

Question

I am revising complex analysis for an upcoming test and I am finding it hard to finish off certain questions. I feel I start well but cannot remember how to wrap up the answers in proper form.

$--------------------------------------$ Letting $z = x + iy$, for each of the following functions determine where $f'(z)$ exists and find its value at those points.

(a) $f(z) = z$ Im$(z)$

(b) $f(z) = x^3 + i(1-y)^3$

$--------------------------------------$

For part (a) I am pretty certain $f'(z)$ does not exist:

Let $f(z) = $Im$(z)$, then for $z, h \in \Bbb C,$ with $h \ne 0,$ we have

$\frac{Im(z+h) - Im(z)}{h}$ = $\frac{Im(z)+Im(h)-Im(z)}{h}$ = $\frac{Im(h)}{h}$.

Now, if $h \rightarrow 0 $ through real values, Im$(h)=0$, and

$\lim \limits_{h \to 0} {\frac{Im(z+h)-Im(z)}{h}}$ = $\lim \limits_{h \to 0}{\frac{Im(h)}{h}}$.

At this point I think I need to mention imaginary values and conclude that $f'(z)$ doesn't exist by using the real and imaginary values, but my notes are giving me limited assistance in how to do this correctly.

$--------------------------------------$

For part (b) I had done a very similar question before, but again, I struggle to conclude my answers in a mathematical way.

If $f(z) = x^3 + i(1-y)^3$ , then $u(x,y)=x^3$ and $v(x,y)=(1-y)^3$ , so that

$\frac{\partial u}{\partial x}$=$3x^2$, $\frac{\partial v}{\partial y} =-3(1-y)^2$

$\frac{\partial u}{\partial y}=0$, $\frac{\partial v}{\partial x}=0$.

The Cauchy-Riemann equations become $3x^2 +3(1-y)^2=0$

How do I then find the points at which $f'(z)$ exists?

Any help is much appreciated!!

Answer 1

(a) $f(z)= z \text{ Im } z=(x+iy)y=xy + iy^2$

Hence $u(x,y)=xy$ and $v(x,y)=y^2$.

$\frac{\partial u}{\partial x} = y, \ \frac{\partial u}{\partial y} =x$ and $\frac{\partial v}{\partial x} =0, \ \frac{\partial v}{\partial y} =2y$.

Cauchy-Riemann equations are valid if and only if

$$y=2y \text{ and } x=0$$ The only solution is $z=0$. Also $$\left\vert \frac{f(z)-f(0)}{z-0} \right\vert \le \vert z \vert$$ as $\vert \text{Im }z\vert \le \vert z \vert$. Hence $\lim\limits_{z \to 0} \frac{f(z)-f(0)}{z-0}=0$ and $f^\prime(0)=0$.

(b) you found the correct equation $$3x^2 +3(1-y)^2=0$$

whose only solution is $x=0$ and $y=1$ as a sum of positive real numbers only vanishes if the real numbers are equal to zero. In that case, $f$ is holomorph only at $z=i$.

To find $f^\prime(i)$, you have to notice that $f(i)=0$ and $$\left\vert \frac{f(z)-f(i)}{z-i} \right\vert = \left\vert \frac{x^3+i(1-y)^3}{x+i(y-1)}\right\vert \le \vert x \vert^2 +\vert y-1\vert^2$$

Proving that $f^\prime(i)=0$.

Answer 2

You can use Cauchy-Riemann in part (a) as well. In part (a), you can write $f(x + iy) = (x + iy)y$.

For part (b):

If the Cauchy-Riemann equations do not hold at a certain $(x,y)$, you are guaranteed that $f$ is not differentiable at $x + iy$.

The converse is more complicated. One theorem states that if the partial derivatives $\partial u / \partial x$, $\partial u / \partial y$, $\partial v / \partial x$ and $\partial v / \partial y$ all exist in an open neighbourhood of $(x,y)$, and are continuous at $(x,y)$, and satisfy the Cauchy-Riemann equations at $(x,y)$, then $f$ is differentiable at $x + iy$.

If these criteria hold, then the derivative at $x + iy$ is $$ f'(x + iy) = \frac{\partial u}{\partial x} (x,y) + i \frac{\partial v}{\partial x} (x,y) = -i\frac{\partial u}{\partial y} (x,y) + \frac{\partial v}{\partial y} (x,y).$$

In your example, the point $(x,y) = (0,1)$ is the only point where the Cauchy-Riemann equations are satisfied. This is the point $z = i$ on the complex plane. So $f$ can only possibly be differentiable at $i$; it cannot be differentiable anywhere else.

In fact, $f$ is differentiable at $i$. The partial derivatives are certainly defined in an open neighbourhood of $i$. Furtherfore, the partial derivatives are continuous at $i$ and obey the Cauchy-Riemann equations at $i$, so the conditions of the theorem are met. Finally, the partial derivatives at $z = i$ are all zero, so $f'(i) = 0$.

If you prefer instead to prove $f'(i) = 0$ from first principles, it may help to write $$ h = k + il,$$ where $k, l \in \mathbb R$. Then $$ \frac{f(i + h) - f(i)}{h} - 0 = \frac{k^3 - il^3}{k + il} = k^2 - i kl - l^2.$$ You should then ask yourself: given an $\epsilon > 0$, does there exist a $\delta > 0$ such that $$ \sqrt{ k^2 + l^2 } < \delta \implies |k^2 - ikl - l^2 | < \epsilon ?$$ (The answer, of course, is "yes".)