Question: given $f(z)= e^z-2i$ then determine the number of zeros of $f$.
My attempt:
$e^z -2i=0$ ⇔$e^z=2i$
Now by taking natural log on both side we can see, $f$ has countably infinite number of zeros.
But, can't we see this using Laurent series?
$f(z)= e^z- 2i = (1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...)-2i$
$= 1-2i + z+\frac{z^2}{2!}+\frac{z^3}{3!}+...$
But here we can't determine $f$ has any zero? :-( or may be i am not able to see them.
I can remove case of uncountable number of zeros because. Let if $S =$ set of zeros of $f(z)$ is uncountable then it has limit point in $\mathbb{C}$. So by identity theorem $f(z)≡0$ for all $z$ in $\mathbb{C}$. So that $e^z=2i$ for all $z$ in $\mathbb{C}$. But it is contradiction, so our assumption that, it has uncountable number of zeros is wrong.
Can some corollary of great picard theorem, will help us to see $f(z)$ has infinite number of zeros?
Please Help me..
Yes, you can used Picard's great theorem to achieve this. Applying it to the exponential function, it tells us that, with, at most, one exception, for each complex number $w$, the equation $e^z=w$ has infinitely many solutions. But you already know an exception: $w=0$. Therefore, the equation $e^z=2i$ has intinitely many solutions.