How to prove that $\sin(x+iy) = x+iy$ has infinitely many solutions? I know how to prove that $\sin(x) = x$ has only one solution, but I do not know how to extend this to complex analysis.
Complex Analysis – Does sin(x+iy) = x+iy Have Infinitely Many Solutions?
complex numberscomplex-analysistrigonometry
Best Answer
The great Picard theorem says that $f(z) = \sin(z)-z = c \in \Bbb C$ infinitely often for all but possibly one value of $c$. Note that $f(z+2\pi) = f(z) - 2\pi$. Suppose some $c_0$ is not hit infinitely many times; then $c_0+2\pi$ is, say $f(z_k) = c_0+2\pi$ for some infinite sequence $(z_k)$. By the above functional equation, then, $f(z_k+2\pi) = c_0$, providing an infinite sequence that maps to $c_0$. So $f^{-1}(c)$ is infinite for every $c \in \Bbb C$; in particular, this is the case for $c=0$ as desired.
Note that the same argument shows that for any holomorphic function $g$ with a sort of periodicity (i.e., there's some $c, d \in \Bbb C$ such that for all $z \in \Bbb C$, $g(z+c) = g(z)+d$), $g^{-1}(z)$ is infinite for any $z \in \Bbb C$. (The great Picard theorem demands that $g$ not be polynomial; but the above periodicity phenomenon precludes this from being possible, lest it have infinitely many roots.)