# Solving sin z = -z in reals

How do we prove that z=0 is the only real solution? I tried examining cases and quarters, but not sure how it is rigorously proven.

For $|z|>1$, you cannot have an equality since $|\sin z|\leq 1$. Inside the circle of radius 1, you notice that when $-z<0$, $\sin z$ is positive, and viceversa when $-z$ is positive $\sin z<0$. Then there is no chance then that the two values coincide.
Since $—1 \le \sin(z) \le 1$, we only need to look at $z \in [-1, 1]$.
If $z \ne 0$, the sign of $\sin(z)$ is the same as the sign of $z$.
Hint: suppose a second solution $z_1$. Apply Rolle's theorem (where?).
Notice $\sin z+z$ has derivative $1+\cos z$ which is nonnegative for all $z\in\Bbb R$ and which has isolated zeroes. So our function is increasing, and as $1+\cos 0>0$ the derivative is positive in a neighborhood of $0$ and so the function has a unique zero.
• Did you mean by "our function" $z\mapsto-\sin z-z$, or did you mean to say "weakly increasing" instead of "weakly decreasing"? And in any case your argument shows strict rather than weak monotonicity. Dec 2 '14 at 13:05