Non-Differentiability of a complex function
Sophia Terry 
$$f(z)=\left\{ \begin{array}{ll} \frac{z^5}{\vert z^4\vert} &, z\neq 0 \\ 0 &, z=0\end{array}\right.$$ Proof that this function is not differentiable in $z_0=0$, but partial differentiable. Also proof that the Cauchy-Riemann differential equations are valid. Why this isn't a contradiction?
So we defined the differentiability with this:
Let $G\subset \mathbb{C}$ a domain and $f:G\to\mathbb{C}$ a function with $z_0\in G$. f is called differentiable if $\exists \Delta:G\to\mathbb{C}$ so that:
- $f(z)=f(z_0)+(z-z_0)\Delta(z)\,\forall z\in G$
- $\Delta$ is continuous in $z_0$.
So I have to proof the differentiability for $z_0=0$. There I get for the first condition: $$f(z)=0+z\cdot \Delta(z)$$
Therefore $\Delta(z)$ has to be 0. But now $\Delta$ is continuous and that can't be right.
The CR-DE are valid, because $f_x(0)=0=f_y(0)$. The differentiability in $\mathbb{R}$ I can proof with the calculation of an existing limit $$\lim\limits_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0}$$.
Where are my mistakes? Or do you have any hints for the rest of my attempts?
$\endgroup$ 111 Answer
$\begingroup$Assume that $f$ is differentiable in $0$. There exists then a continuous $\Delta$ defined around $0$ such that $\dfrac {z^5} {|z|^4} = z \Delta (z)$, which for $z \ne 0$ is equivalent to $\dfrac {z^4} {|z|^4} = \Delta (z)$. Let us show that this $\Delta$ is not continuous in $0$, as it should.
If $\Delta$ were continuous in $0$, then we would be able to speak about $\lim \limits _{z \to 0} \Delta (z)$. But, on the one hand,
$$\lim _{z \to 0} \Delta (z) = \lim _{x \to 0, x > 0} \frac {x^4} {|x|^4} = \lim _{x \to 0, x > 0} \frac {x^4} {x^4} = 1$$
and, on the other hand, if $r$ is a primitive root of $x^8 + 1 = 0$,
$$\lim _{z \to 0} \Delta (z) = \lim _{rx \to 0, x > 0} \frac {(r x)^4} {|r x|^4} = \lim _{x \to 0, x > 0} \frac {r^4 x^4} {x^4} = r^4 = -1$$
which shows that $\Delta$ does not have a limit in $0$, therefore cannot be continuous there.
To show that $f$ admits partial derivatives in $0$ we just use the definition.
$$\frac {\partial f} {\partial x} (0,0) = \lim _{x \to 0} \frac {f(x,0) - f(0,0)} {x-0} = \lim _{x \to 0} \frac {\frac {x^5} {|x|^4} - 0} x = \lim _{x \to 0} \frac {x^4} {|x|^4} = \lim _{x \to 0} \frac {x^4} {x^4} = 1 \\ \frac {\partial f} {\partial y} (0,0) = \lim _{y \to 0} \frac {f(0,y) - f(0,0)} {y-0} = \lim _{y \to 0} \frac {\frac {(\Bbb i y)^5} {|\Bbb i y|^4} - 0} y = \lim _{y \to 0} \frac {y^4} {|y|^4} \Bbb i ^5 = \lim _{y \to 0} \frac {y^4} {y^4} \Bbb i = \Bbb i.$$
Checking the Cauchy-Riemann equations in Cartesian coordinates is very annoying in this case (try it). It is much more convenient to use the following trick. If $\Re$ and $\Im$ denote the real and the imaginary part, and $f = u + \Bbb i v$, it is easy to show that
$$\frac {\partial u} {\partial x} (0,0) = \frac {\partial (\Re f)} {\partial x} (0,0) = \Re \left( \frac {\partial f} {\partial x} (0,0) \right) = \Re (1) = 1 \\ \frac {\partial u} {\partial y} (0,0) = \frac {\partial (\Re f)} {\partial y} (0,0) = \Re \left( \frac {\partial f} {\partial y} (0,0) \right) = \Re (\Bbb i) = 0 \\ \frac {\partial v} {\partial x} (0,0) = \frac {\partial (\Im f)} {\partial x} (0,0) = \Im \left( \frac {\partial f} {\partial x} (0,0) \right) = \Im (1) = 0 \\ \frac {\partial v} {\partial y} (0,0) = \frac {\partial (\Im f)} {\partial y} (0,0) = \Im \left( \frac {\partial f} {\partial y} (0,0) \right) = \Im (\Bbb i) = 1$$
from which it is easy to show that the Cauchy-Riemann equations hold at $0$.
But if the Cauchy-Riemann equations hold at a point, doesn't it mean that $f$ is $\Bbb C$-differentiable at that point? No, it doesn't, because $f$ is not $\Bbb R$-differentiable at $0$ - if it were, then the following limit would be $0$:
$$\lim _{(x,y) \to (0,0)} \frac {f(x,y) - \frac {\partial f} {\partial x} (0,0) (x-0) - \frac {\partial f} {\partial y} (0,0) (y-0)} {\| (x,y) \|} = \lim _{z \to 0} \frac {\frac {z^5} {|z|^4} - x - \Bbb i y} {|z|} = \lim _{z \to 0} \frac z {|z|} \left( \frac {z^4} {|z|^4} - 1 \right) .$$
Assuming that this limit exists, let us compute it along the trajectories $t \mapsto t$ and $t \mapsto rt$ with $r$ defined several paragraphs above and $t>0$:
$$\lim _{z \to 0} \frac z {|z|} \left( \frac {z^4} {|z|^4} - 1 \right) = \lim _{t \to 0, t>0} \frac t {|t|} \left( \frac {t^4} {|t|^4} - 1 \right) = 1 \cdot 0 = 0 \\ \lim _{z \to 0} \frac z {|z|} \left( \frac {z^4} {|z|^4} - 1 \right) = \lim _{rt \to 0, t>0} \frac {rt} {|rt|} \left( \frac {(rt)^4} {|rt|^4} - 1 \right) = r \cdot (r^4 - 1) = -2r$$
which shows that the limit does not exist, therefore $f$ is not differentiable at $0$, so one hypothesis needed for the equivalence between holomorphy and the Cauchy-Riemann equations is not satisfied, therefore there is no contradiction between the fact that $f$ is not $\Bbb C$-differentiable at $0$ and the fact that the Cauchy-Riemann equations are satisfied at $0$.
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