all complex solutions of $z\sin(z)=1$?

Question

A possibly easy question,

Can we find all complex solutions of $z\sin(z)=1$ ?

My try:

Let $$\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$ so we have $$ z\frac{e^{iz} - e^{-iz}}{2i}=1 $$

Not sure how can I continue ?

Thanks

EDIT : I have posted a related question here.

Answer 1

As shown elsewhere, $z\sin z=1$, with $z\in$ C, has only real solutions. So we are ultimately left with solving $x\sin x=1$, with $x\in$ R, which, unlike the solution to your other question, does not possess a closed form, not even in terms of the special Lambert W function. However, what we can say for certain, is that it has exactly two solutions in each interval of the form $\Big[2n\pi,(2n+1)\pi\Big]$ for $n\ge0$ and $\Big[(2k-1)\pi,2k\pi\Big]$ for $k\le0$, whose values get closer to the two extremities of the interval as x increases in absolute value. The only way to compute them is by using numerical methods, such as Newton's, for instance. Here are the first few, with a precision of $45$ decimals:

$$\pm1.11415714087193008730052517816920390395410138$$ $$\pm2.77260470826599123395356972149927927932229123$$

$$\pm6.43911723841724646172451484031087947865696505$$ $$\pm9.31724294141480961860128851356951156244980218$$

Answer 2

See the following paper:

Charles Edward Siewert and Ernest Edmund Burniston, Exact analytical solutions of $ze^{z} = a,$ Journal of Mathematical Analysis and Applications 43 #3 (September 1973), 626-632.

Author's Abstract: By means of the theory of complex variables, the solutions of $ze^{z} = a,$ where $a$ is in general complex, are established analytically, and thereby reduced to elementary quadratures.

Answer 3

With Lagrange reversion:

$$x\sin(x)=1\iff x=\{2\pi k+\csc^{-1}(x),(2k+1)\pi-\csc^{-1}(x)\}$$

Therefore:

$$x_{2k}=2k\pi+\sum_{n=1}^\infty\frac1{n!} \left.\frac{d^{n-1}}{dw^{n-1}}\csc^{-1}(w)^n\right|_{2k\pi} \\x_{2k+1}=(2k+1)\pi+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}\csc^{-1}(w)^n\right|_{(2k+1)\pi}$$

with examples shown here. The next step could be expanding $\ln^n(x)$

Answer 4

If we look at $\frac{\sin x}{x}=0$, we can factor \begin{equation} \frac{\sin x}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\ldots =\prod_{k=1}^{\infty}\left(1-\frac{x}{k\pi}\right)\left(1+\frac{x}{k\pi}\right)\end{equation} What about $$x\sin x=x^2\prod_{k=1}^{\infty}\left(1-\frac{x^2}{(k\pi)^2}\right)=1$$ $$2\ln x + \sum_{n=1}^{\infty}\ln \left(1-\frac{x^2}{(n\pi)^2}\right)=0$$ Maybe it helps somehow.