Proving imaginary number lies parallel to the real axis.
Define $$z \equiv 2e^{i\theta}$$We are to obtain the imaginary and real parts of $$w=\frac{z-2}{z+2}$$ I ended up getting $w=\frac{i\sin\theta}{1+\cos\theta}$ I got this by multiplying by $\frac{e^{-i\theta}+1}{e^{-i\theta}+1}$ I suspect this is wrong as it failed to provide a valid solution for the next part which says to prove $\ln{w}$ will always lie on a line parallel to the real axis. I got: $$\ln{w} =\ln{\frac{\sin\theta}{1+\cos\theta}} + i\left(\frac{\pi}{2}+2\pi n\right)$$ For $n\in\mathbb{Z}$ Which implies $\ln{w}$ lies perpendicular to the real and parallel to the imaginary. Many thanks in advance.
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$\begingroup$You should double check your definition of $w$ after plugging in the definition of $z$; I found:
$w = i\tan\frac{\theta}{2}$.
In any case, this isn't the big issue! You simply misinterpreted your result of $\ln w$. If you think of the real and imaginary parts of $\ln w$ as an $(x,y)$ coordinate, you'll find:
$\left(\ln\left|\tan\frac{\theta}{2}\right|,\frac{\pi}{2}+2\pi n\right)$
This has a constant $y$ coordinate and a varying $x$ coordinate---this is by definition a line parallel to the $x$, or equivalently the real axis!
So, ultimately, you had the right mathematical answer all along, just the wrong interpretation.
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