Complex cosine equation (complex analysis)

[malawi_glenn]

Homework Statement

Solve $$cosz = 2i , z\in \mathbb{C}$$

The Attempt at a Solution

$$e^{iz}+e^{-iz} = 4i$$

$$t=e^{-z}$$

$$t+t^{-1}=4i \Rightarrow t^{2}-4it+1=0$$

$$t = (2 \pm \sqrt{5})i$$

$$log(e^{-z}) = logt$$

$$z = x + yi;x,y \in \mathbb{R}$$

$$log(e^{-z}) = log(e^{-y+ix}) = -y +xi + 2\pi ni; n\in\mathbb{Z}$$

$$logt = log((2 \pm \sqrt{5})i) = ln|(2 \pm \sqrt{5})| \pm \frac{\pi}{2} + 2\pi hi, h \in \mathbb{Z}$$

$$\Rightarrow y= -ln|(2 \pm \sqrt{5})|$$

$$\Rightarrow x= \pm \frac{\pi}{2} + 2\pi p, p \in \mathbb{Z}$$

$$z= \pm \frac{\pi}{2} + 2\pi p -iln|(2 \pm \sqrt{5})|$$

answer in book:
$$z = \pm\lbrace \frac{\pi}{2} -iln(2 + \sqrt{5})\rbrace+2\pi n, n\in \mathbb{Z}$$

Were did I do wrong? :S

 

[Dick]

You didn't go wrong. The book just found a sneakier way to write the answer. Since (sqrt(5)+2)*(sqrt(5)-2)=1, they are actually reciprocals. So their logs are negatives of each other. Thus you can sneak the log under the same +/- as the pi/2.

 

[malawi_glenn]

aa, it is almost always the most simplest of solutions :) I shall check it out tomorrow. It is very late now in Sweden hehe. This is a summer course I am taking now, Complex analysis with applications.

thanx!