TITLE: Complex Conjugation question
QUESTION [1 upvotes]: I had a complex analysis exam yesterday, and one of the questions is bothering me.
Suppose $f(z)$ is an entire function. Show that $g(z) = (f(z^*))^*$ is also entire. Here $^*$ indicates complex conjugate.
The problem said to "think about Cauchy-Riemann"
The next part asked to find $g(z)$ for each $f(z)$:
$f(z) = c_0 + c_1 z + c_2 z^2 + ... + c_n z^n$
$f(z) = \sin(z)$
Is it not true that $g(z)$ is just $f(z)$ in both of these cases?
Thanks in advance for the guidance.

REPLY [1 votes]: There are lots of ways to prove this theorem, but here is what I prefer to write down:
we know $f$ is entire is really the same as $f$ is just the Taylor series, which has radius of convergence infinity.
Now let $f(z)=a_0+a_1z+a_2z^2+a_3z^3+.....$
Then $g(z)$ is simply the power series with all the coefficients replaced by their complex conjugate.
Clearly $g(z)$, as a power series, have the same radius of convergence as $f$, so is entire.
This also helps with you to compute your examples.
Actually I thought the hint is "misleading" as it doesn't help compute the examples. But anyway, the hint is correct, you can directly verify the real and imaginary part satisfy the Cauchy-Riemann equation, therefore.......
I apologize for the poor notations, I'm new here and haven't figured out how to do Latex.
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