How to solve ${\rm Ln}({\rm cosecant}(z)) = {\rm Ln}(2) + iz$?

Question

$\newcommand{\Ln}{\operatorname{Ln}}$ $$\Ln(\csc(z)) = \Ln(2) + iz$$ I need to know what complex number $z$ is

I haven't tried it but you can write the cosecant as exponentials — H. Gutsche, Sep 27 '19 at 15:12

score 1 · Accepted Answer · answered Sep 27 '19 at 15:16

1

Hint

$$\dfrac{\csc(z)}2=e^{iz}$$

If $e^{iz}=a,$

$2a=\dfrac{2i}{a-1/a}$ as $a\ne0$

$a^2-1=i$

$a^2=\sqrt2e^{i\pi/4}$

answered Sep 27 '19 at 15:16

lab bhattacharjee

You need to add the condition that the Ln on the left side and thus also the right side has imaginary part in $[-\pi,\pi]$, limiting the real part of $z$. – Lutz Lehmann Sep 27 '19 at 15:20

1 Answers1