$z\sin z=a$ has only real solutions

Consider \(f(z) \da z\sin(z) - a\).

Big: \(M(z) \da z\sin(z)\). Small: \(m(z) \da -a\).

Use the following estimate:

\[ \abs{z\sin(z)}^2 &= \abs{z\over 2}^2 \abs{e^{iz} - e^{-iz}}^2 \\ &\geq \abs{z\over 2}^2 \abs{ \abs e^{iz} } - \abs{ e^{-iz} } ^2 \\ &= \abs{z\over 2}^2 \abs{e^{-\Im(z)} - e^{\Im(z)} } \\ &\convergesto{\Im(z)\to\infty} \infty ,\] and so in particular a radius \(R\) can be chosen large enough so that \(\abs{z\sin(z)} > a\) for any \(a\). Thus for \(\abs{z} = R\), \[ \abs{m(z)} = \abs{a} \leq \abs{z\sin(z)} < \abs{M(z)} \implies \size Z_{M} = \size Z_{M+m} = \size Z_f .\] To count the number of zeros of \(z\sin(z)\), note that this equals zero at \(z=0\) with multiplicity two and \(z= k\pi\) for \(k\in \ZZ\). Choosing \(R = {\pi \over 2} + n\pi\) for \(n\) large enough, there are exactly \(2n+2\) such zeros (with multiplicity) to \(z\sin(z)\), and thus \(2n+2\) zeros to \(z\sin(z) - a\). Now using that \(z\sin(z) - a\) has exactly \(2n+2\) real roots (??), this must be all of them.

The count of real roots remains unproved in this argument.