$f$ is uniformly continuous $\Leftrightarrow$ $\mathrm{Re}f$ and $\mathrm{Im}f$ are uniformly continuous.

Question

Let $f$ be a complex-valued function.
In many complex analysis textbook, I have just found that $f$ is continuous $\Leftrightarrow$ $\mathrm{Re}f$ and $\mathrm{Im}f$ are continuous.

I wonder that whether $f$ is uniformly continuous $\Leftrightarrow$ $\mathrm{Re}f$ and $\mathrm{Im}f$ are uniformly continuous.

Do you know any book contain this statement? I would very much appreciate any help or guidance you are able to give me.

Answer 1

Hint
On $\mathbb C$ $|\Re z|+|\Im z|$ is an equivalent norm to the usual $|z|$. Use this to show that your conjecture is true.

Answer 2

Write f = g + ih, where g = Re(f) and h = Im(f). Now, both |g(x) - g(y)| and |h(x) - h(y)| are <= ||f(x) - f(y)||, showing the "only if" direction. For the "if" direction, notice that ||f(x) - f(y)|| <= |g(x) - g(y)| + |h(x) - h(y)|. This not only shows your result, but also the fact that f is Lipschitz if and only if both Re(f) and Im(f) are Lipschitz.