# How to show that $f(z)=\frac{z^{5}}{|z|^4}$ satisfies the Cauchy Riemann equations at $z=0$ but not differentable at $z=0$?

$$f(z) = \left\{ \begin{array}{ll} \frac{z^{5}}{|z|^4} & \quad z \neq 0 \\ 0 & \quad z= 0 \end{array} \right.$$ I am trying to solve this past exam question. Similar question was asked in Show that $f(z)=\frac{z^5}{|z|^4}$ but has not been answered.

My attempt:

Let $z=x+iy$ then we can separate the real and imaginary parts of $f(z)$ as

$$f(z)=u+iv=\left(\frac{x^5-10 x^3 y^2+5 x y^4}{\left(x^2+y^2\right)^2}\right)+i\left(\frac{5 x^4 y-10 x^2 y^3+y^5}{\left(x^2+y^2\right)^2}\right)$$

$$\frac{\partial u}{\partial x}=\frac{5 x^4-30 x^2 y^2+5 y^4}{\left(x^2+y^2\right)^2}-\frac{4 x \left(x^5-10 x^3 y^2+5 x y^4\right)}{\left(x^2+y^2\right)^3}$$ $$\frac{\partial v}{\partial y}=\frac{5 x^4-30 x^2 y^2+5 y^4}{\left(x^2+y^2\right)^2}-\frac{4 y \left(5 x^4 y-10 x^2 y^3+y^5\right)}{\left(x^2+y^2\right)^3}$$ $$\frac{\partial u}{\partial y}=\frac{20 x y^3-20 x^3 y}{\left(x^2+y^2\right)^2}-\frac{4 y \left(x^5-10 x^3 y^2+5 x y^4\right)}{\left(x^2+y^2\right)^3}$$ $$\frac{\partial v}{\partial x}=\frac{20 x^3 y-20 x y^3}{\left(x^2+y^2\right)^2}-\frac{4 x \left(5 x^4 y-10 x^2 y^3+y^5\right)}{\left(x^2+y^2\right)^3}$$

I am not sure what to do next. If I substitute $z=0\rightarrow x=0,y=0$ in the equations above then they become infinite because the bottom term $(x^{2}+y^{2})$ becomes zero. So how can I show that $f(z)$ satisfies the Cauchy Riemann equations at $z=0$. Also, how can I show that $f(z)$ is not differentiable at $z=0$?

Thanks.

It must satisfy the CR condition in $0$, because they only involve partial derivatives in the $x-$ and $y-$ directions, and the function coincides with $f(z)=z$ both on $x$ and $y$-axis. (Note that $|ix|^4=x^4=(ix)^4$ for real $x$.)

• I understand that but In the exam I have to prove mathematically.
– gbd
Commented Jan 5, 2017 at 10:30
• This is a mathematical proof. A limit $\lim_{x\to 0} g(x)$ only depends on the values of $g(x)$ (and not on the values of a possible extension of $g$ to a higher-dimensional space). Commented Jan 5, 2017 at 10:32

The formulas you obtained are only valid for $(x,y)\neq (0,0)$. We have $u(0,0) = v(0,0) = 0$. Now:

$$\frac{\partial u}{\partial x}(0,0) = \lim_{h\to 0}\frac{u(h,0)-u(0,0)}{h} = \lim_{h\to 0} \frac{h}{h} =1$$

Etc.

We have:

$$g(z):=\frac{f(z)-f(0)}{z-0} = \frac{z^4}{|z|^4} = e^{4i\arg(z)}$$

If $\lim_{z\to 0} g(z)$ exists, say equals $c$, then $\lim_{z\to 0} g(h(z)) = c$ for any $h$ with $\lim_{z\to 0}h(z) = 0$. Now consider:

• $h_1(z) = (1+i)|z|$

• $h_2(z) = i|z|$

• How to show that $f(z)$ is not differentiable at $z=0$?
– gbd
Commented Jan 5, 2017 at 10:24
– user384138
Commented Jan 5, 2017 at 10:26

Simple You have written $$f(z)$$ as $$u(z)+iv(z)$$ now simply follow the definition of differentiation of $$u_x$$,$$v_x$$,$$u_y$$,$$v_y$$

for $$u_x$$ $$u_x(0+0i)=lim_{x\rightarrow 0}\frac {u(x,0)-u(0,0)}{x}$$ $$=lim_{x\rightarrow 0}\frac{\frac{x^5}{x^4}}{x}$$ $$=lim_{x\rightarrow 0}1$$ $$u_x(0)=1$$

Similarly calculate for $$v_x,u_y,v_y$$. We will get $$v_x(0)=0, u_y(0)=0,v_y(0)=1$$ and hence it satisfy the CR equations at $$z=0$$.

But do find $$u$$ and $$v$$ looks quite tedious process. Instead of this you could have found $$f_x(0)$$ and $$f_y(0)$$ where $$f_x=u_x+iv_x$$ and $$f_y=u_y+iv_y$$

i.e $$f_x(0)=\lim_{x\rightarrow 0}\frac{f(x+i0)-f(0+i0)}{x}$$ $$f_x(0)=\lim_{x\rightarrow 0}\frac{x^5}{x^5}=1$$ therefore $$u_x=1$$ and $$v_x=0$$

$$f_y(0)=\lim_{y\rightarrow 0}\frac{f(0+iy)-f(0+i0)}{y}$$ $$f_y(0)=\lim_{y\rightarrow 0}\frac{(iy)^5}{y^5}=i^5=i$$ therefore $$u_y=0$$ and $$v_y=1$$ and hence the CR equations.