Is $ f(z) = (\bar {z})^2 - 5z $ differentiable?

Question

Since the complex conjugate is not differentiable, can I immediately conclude that $ f= (\bar{z})^2 - 5z$ is also not differentiable at any point?

Ignoring the $5z$ for a moment, the formal definition of the derivative yields the following for $f(z) = (\bar{z})^2 $:

$\lim_{h\to0} \frac{f(z+h) - f(z)}{h} = \lim_{h\to0} \frac{ 2\bar{z}\bar{h} + (\bar{h})^2}{h}$

And if I'm not mistaken, the limit does not exist.

Answer 1

The limit indeed doesn't exist, because it depends on the path taken. Take $h=\epsilon$ and you get $2\bar{z}$, while for $h=i\epsilon$ you get $-2\bar{z}$. Note that it means that your function is differentiable at $z=0$.

Another way to prove the limit doesn't exist is to use the Cauchy–Riemann equations. It is essentially the same as taking different paths and requiring the results to be the same, yet it is a straight-forward way to show stuff without actually calculating limits.

Answer 2

To check for complex differentiability - especially when $\bar z$ occurs - you may use the Cauchy-Riemann equations:

• Write $f(x+iy)=u(x,y)+iv(x,y)$ and check where $u_x = v_y, u_y = -v_x$ are satisfied simultaneously.

$$\bar z^2-5z = ({x-iy})^2 - 5(x+iy) = x^2-5x-y^2 + i(-2xy-5)$$ $$\Rightarrow u(x,y) = x^2-5x-y^2, v(x,y)= -2xy-5y$$ $$\Rightarrow u_x = 2x-5, u_y=-2y,v_x = -2y,v_y=-2x-5$$ Now you get $$u_x=v_y \Leftrightarrow x=0; u_y=-v_x \Leftrightarrow y=0$$

It follows $f(z)\bar z^2-5z$ is complex differentiable only at $\boxed{z=0}$. Anywhere else the Cauchy-Riemann equations are not satisfied simultaneously.