Zeros of a function

Question

Show that all zeros of $$f(z)=\sin z +z\cos z$$ are real.

I tried to use zeros of $\sin z$ and $\cos z$ are real even though I couldn't get any ideal.

Answer 1

The equation $f(z) = 0$ is equivalent to $z = -\tan z$. Writing this out in real and imaginary parts, where $z = x+iy$, and $-\tan z = u + iv$ you get $$ u = -\frac{\sin x \cos x}{\cos^2 x + \sinh^2 y} \quad \text{ and } \quad v = -\frac{\sinh y \cosh y}{\cos^2 x + \sinh^2 y} $$ and you are looking for the solutions of $u=x$, $v=y$. Now if $y \ne 0$, then $$ \left| \frac{yu}{xv} \right| = \left| \frac{y \sin x \cos x} { x \sinh y \cosh y} \right| < \left| \frac{\cos x}{\cosh y } \right| < 1 $$ since $\left|\frac{y}{\sinh y}\right| < 1$, $\left|\frac{\sin x}{x}\right| \le 1$, and $|\cos x| \le 1 < |\cosh y|$. This shows that $x \ne u$ or $y \ne v$.

The only case for which this argument does not work is $x=0$, since then we would be dividing by zero. However, in that case the equation $v = y$ becomes $y = - \tanh y$ which only has the trivial solution $y=0$.