Is $f$ is constant ? Yes/No

Question

Is the following statement is true/false

let $f$ be entire function and if $\;\operatorname{Re}f(z) = <\operatorname{Im}f(z)$ then $f$ is constant .

My attempts : No . I take $f(z) = z$ put $z = x +iy$ again take $x=y$

now we have $f(z) = y +iy$ but $f(z)$ is not constant

Is it true?

Answer 1

Yes, it is true. Let $g(z)=(1+i)f(z)$. Then, if $f(x+yi)=u(x,y)-v(x,y)i$, you have$$g(x+yi)=u(x,y)+v(x,y).$$Therefore, $\operatorname{Im}g(z)=0$ It is easy now to deduce from the Cauchy-Riemann equations that $g$ is constant. Since, $f=\frac g{1+i}$, $f$ is constant too.

Answer 2

It is apparent that, in general, $\Re z\ne\Im z$.

The set $\{z\in\Bbb C\,:\, \Re z=\Im z\}$ has empty interior in $\Bbb C$, whereas holomorphic non-constant functions are open. So $f$ must be constant.

Answer 3

Cauchy - Riemann equation immediately tell you that all first partial derivatives of the real and imaginary parts are $0$, so $f$ is a constant.