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

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?

• Let $f(x)=x+\sin \pi x$ then $f'(x) = 1+ \pi \cos \pi x$ and so... – complexmanifold Jan 30 '18 at 11:16
• I think you should have a plus sign between the exponential functions in the latter equation. After all $$\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y.$$ – Jyrki Lahtonen Jan 30 '18 at 12:42
• I am fairly sure that there is no "explicit" formula for the solutions, but that there are infinitely many. It might be possible to give an asymptotic expression. – Martin R Feb 1 '18 at 19:25

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.

• Plot is not a proof! You need inequalities or calculus. – gimusi Jan 30 '18 at 11:34
• Plotting graphs would only give hints but not any rigorous results. – Szeto Jan 30 '18 at 11:38
• If $a$ is just below $2n-1/2$, then $b$ will be roughly $\frac1\pi \ln(4n-1)$ – Michael Jan 30 '18 at 12:49
• Note, @gimusi The asker never asked for a rigorous proof; besides, this gives an intuitive approach, that should inspire the asker to think for themself and move forward. This site is about helping users to solve their own work, as much as possible, and so I think this is far more valuable than some supposed "rigorous proof" that, as Szeto clearly noted, fails to consider complex solutions. – Namaste Feb 1 '18 at 19:31
• @Szeto, hints, e.g., via an informative graph, does give good hints, and good hints often make the best answers. And, in any case, you never asked for rigorous results. – Namaste Feb 1 '18 at 19:32

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.

• However, I am looking for complex $x$. Your first line already implies that you are assuming $x$ to be real. – Szeto Jan 30 '18 at 11:34