# is the real part of a holomorphic function holomorphic?

Say we have some entire function $$f:\mathbb{C} \rightarrow \mathbb{C}$$. Does this guarantee that the function $$Re(f)$$ will also be entire?

• It will be entire iff $f$ is constant. Mar 10, 2019 at 22:07
• Ah. Does this come from Cauchy Riemann equations or something different? Mar 10, 2019 at 22:08
• Yes, it exactly follows from Cauchy Riemann equations. If $f=u+iv$ is defined in an open connected domain and $v$ is constant then the partial derivatives of $v$ are zero everywhere. By Cauchy Riemann we conclude that the partial derivatives of $u$ are zero as well. So $f$ is constant.
– Mark
Mar 10, 2019 at 22:17
• Could you explain why we require that v is constant for u to be holomorphic? Mar 10, 2019 at 22:18
• The real part of $f$ has $0$ imaginary part. Apply C-R to $\Re(f)$ Mar 10, 2019 at 22:20

If $$\mathfrak{R}(f)$$ is entire, then $$\exp(i\mathfrak{R}(f))$$ is also entire. But $$|\exp(i \mathfrak{R}(f))| \leq 1$$, so $$\exp(i\mathfrak{R}(f))$$ is constant (Liouville's theorem) and thus so is $$\mathfrak{R}(f)$$. Hence, $$f$$ has constant real part, and the C-R equations get you that $$f$$ has constant imaginary part as well.

The only entire holomorphic functions whose imaginary parts are constant are the constant functions. So $$\operatorname{Re}(f)$$ is holomorphic if and only if it is constant (which implies that $$f$$ itself is constant).

• I follow the first line but I don't understand why Re(f) being holomorphic requires that Im(f) is constant, could you explain? Mar 10, 2019 at 22:16
• $Im(f)$ is not necessary constant. But $Im(Re(f))$ is constant because $Re(f)$ is always a real number. This is why $Re(f)$ being holomorphic implies that it is constant.
– Mark
Mar 10, 2019 at 22:19
• ahhh of course. thanks for the explanation. Mar 10, 2019 at 22:20
• But yes, after we found out that $Re(f)$ is constant then since $f$ is holomorphic with constant real part we conclude that $f$ is constant. So in the end $Im(f)$ is constant as well. But we conclude that only after we proved $Re(f)$ is constant.
– Mark
Mar 10, 2019 at 22:23