Determine where $f'(z)$ exists and find its value at those points. 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.  
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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) $f(z) = x^3 + i(1-y)^3$
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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.
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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!!
 A: (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 $f^\prime(i)=0$.
A: 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".)
