Finding all solutions of $x$ in $\sin(\pi x)+x=0$.

Question

Recently I am stuck on finding all solutions of the equation, $$\sin(\pi x)+x=0.$$

Nota Bene: I am searching for not only real solutions but also the complex ones.

Obviously $x=0+0i$ is a solution. But I am not sure whether there are any other solutions.

I have tried rewriting the equation by substituting $x=a+bi$ and by equating the real and imaginary parts respectively, obtained the simultaneous real equations: $$\begin{align} (e^{-b\pi}-e^{b\pi})\cos(\pi a)&=2b; \text{ and}\\ (e^{-b\pi}-e^{b\pi})\sin(\pi a)&=-2a.\end{align}$$

These equations are hard to solve. So, is there any method that I can find the solutions other than $x=0$, or instead proving $x=0$ is the only solution?

Answer 1

For $0\leq x\leq1$ we see that $\sin\pi x\geq0\geq-x$, where the equality occurs for $x=0$ only.

For $x>1$ we obtain $$x>1\geq-\sin\pi x$$ and since $f(x)=\sin\pi x+x$ is an odd function, we see that $0$ is an unique root.

If you mean to solve it in $\mathbb C$ then we have $$\tan{\pi a}=-\frac{a}{b},$$ which gives $$e^{\frac{\pi a}{tan\pi a}}-e^{-\frac{\pi a}{tan\pi a}}=-\frac{2a}{\sin\pi a}$$ and we got infinitely many solutions.

Answer 2

I think you have lost a sign in your real/imag parts. I get a sum of exponentials to go with $\sin\pi a$

One hint that my graph shows is a possible solution for $a$ just below $2n-\frac12$, and $b$ somewhere around 1.