If $f(z) = u(z)+iv(z)$ is analytic, are both of $u(z)$ and $v(z)$ analytic?

Question

I think the title says it all, if $f(z) = u(z)+iv(z)$ is analytic, are both of $u(z)$ and $v(z)$ analytic?

I betting that this is not the case and that there are some obvious counterexamples, but I couldn't think of any. Thanks in advance!

Answer 1

Try $$u(z) = 1+e^{-\frac1{z^2}} \\ v(z) = i e^{-\frac1{z^2}} \\ f(z) = u(z) + iv(z) = 1 $$ $u(z)$ and $v(z)$ each have an essential singularity at $z=0$ wile $f(z)$ is analytic (and finite!) everywhere.

Answer 2

Here is an answer for the special case of an entire function $f$, that is analytic on whole of $\mathbb{C}$:

Let $u$ and $v$ be the real and imaginary part of $f$. Then $u$ and $v$ are both real.

By Picard's little theorem an entire non-constant function takes all values of $\mathbb{C}$ except possibly one. Since That is clearly not the case for real-valued functions, $u$ and $v$ cannot be entire.

Answer 3

The function $f(x+iy) = x+iy$ is analytic everywhere, but $u(x+iy) = x$ is nowhere complex-differentiable. Proof: Fix $z_0=x_0+iy_0.$ Then check that

$$\lim_{h\to 0} \frac{u(z_0+h) - u(z_0)}{h}$$

is $1$ as $h\to 0$ through real values, while the limit is $0$ if $h\to 0$ through imaginary values. Therefore the complex derivative $u'(z_0)$ fails to exist.