I am going to solve the equation: $$\sin(z)\cdot \cos(z) = i$$ in my mathematics class. I can get pretty far by using eulers formula to manipulate the expression, and end up where i have $$e^{2iz} = -2 \pm \sqrt5$$ but from here i dont get it anymore.
The solution I have been provided does somehow make the root/r equal 3 - and end up with an expression like this for when you plus the root of five part:
$$3 \cdot e^{\arctan \dfrac{\sqrt5}{-2} + 2 \pi n} $$ But both root of five and $2$ is a part of the real part (not the imaginary) so i dont understand how one can use it like this both to get the 3 (r) in front and also in the arctg part to get the angle.
Thanks for help!
The equation being $$\sin(2z)=2i$$ let $z=a+I b$ and expand to have two equations corresponding to the real and imaginary parts. $$\sin (2 a) \cosh (2 b)=0\tag 1$$ $$\cos (2 a) \sinh (2 b)=2\tag 2$$
Since $\cosh(2b)$ is always larger than $1$, the solution for $a$ correspond to $$\sin(2a)=0 \qquad \implies a= k \frac \pi 2$$
Knowing $a$, the second equation write $$\sinh(2b)=\pm 2$$ Using the expoential representation $$\frac{e^{2b}-e^{-2b}}2=\pm 2 $$ which are quadratic polynomials in $e^{2b}$. So the roots $$b_{\pm}=\pm\frac{1}{2} \log \left(2+\sqrt{5}\right)$$