Say we have some entire function $f:\mathbb{C} \rightarrow \mathbb{C}$. Does this guarantee that the function $Re(f)$ will also be entire?
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$\begingroup$ It will be entire iff $f$ is constant. $\endgroup$– Severin SchravenMar 10, 2019 at 22:07
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$\begingroup$ Ah. Does this come from Cauchy Riemann equations or something different? $\endgroup$– D.DogMar 10, 2019 at 22:08
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$\begingroup$ 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. $\endgroup$– MarkMar 10, 2019 at 22:17
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$\begingroup$ Could you explain why we require that v is constant for u to be holomorphic? $\endgroup$– D.DogMar 10, 2019 at 22:18
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$\begingroup$ The real part of $f$ has $0$ imaginary part. Apply C-R to $\Re(f)$ $\endgroup$– saulspatzMar 10, 2019 at 22:20
2 Answers
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).
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$\begingroup$ I follow the first line but I don't understand why Re(f) being holomorphic requires that Im(f) is constant, could you explain? $\endgroup$– D.DogMar 10, 2019 at 22:16
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$\begingroup$ $Im(f)$ 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. $\endgroup$– MarkMar 10, 2019 at 22:19
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$\begingroup$ 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. $\endgroup$– MarkMar 10, 2019 at 22:23