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.
Solution 1:
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.
Solution 2:
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 $z$.