Complex sine

Thank you to Barinder Singh Banwait, Langley Grammar School; Angus Balkham from Bexhill College and Derek Wan for your excellent solutions.

As $$\sin z=\frac{1}{2i}(e^{iz}-e^{-iz})=2$$ then, substituting $w=e^{iz}$, we have $$w-\frac{1}{w}=4i.$$ So $w^2 - 4iw -1 = 0$ and the solutions of this quadratic equation are: $$w=e^{iz}=\frac{4i\pm \sqrt{-12}}{2}=i(2\pm \sqrt{3}).$$ Taking logarithms gives $$iz={\log }_{e}i(2\pm \sqrt{3})={\log }_{e}i+{\log }_e(2\pm \sqrt{3})=i\frac{\pi }{2}+{\log }_e(2\pm \sqrt{3}).$$ Dividing by $i$ gives the solutions $z = {\pi \over 2} -i \log_e \left(2\pm \sqrt 3\right)$ but since $\sin z$ is periodic with period $2\pi$ the set of all solutions is given by $$z=\frac{\pi }{2}-i\log (2\pm \sqrt{3})+2n\pi .$$

The same method works to give solutions for $\sin z = a$ where $a$ is any complex number.

The step in the solution above $\log_e i = {\pi \over 2}$ follows from the definition of the complex logarithm function using $|i|=1$ and $\arg i = {\pi \over 2}$. The logarithms of $z$ are the numbers $\lambda$ such that $e^{\lambda} = z$, that is: $$\lambda =\log |z|+i(\arg z+2\pi n)$$ for integer $n$ . It is easy to check: $$e^{\log |z|+i(\arg z+2\pi n)}=e^{\log |z|}\times e^{i\arg z}\times e^{2\pi ni}=|z|e^{i\arg z}=z$$ giving the complex number $z$ in modulus-argument form. Every complex number has infinitely many complex logarithms each differing by an integer multiple of $2\pi i$ because $e^{2\pi ni}=1$.