Please help with a question that I am working on just now...:)
If $z=2e^{i\theta}$ where $0<\theta<\pi$, how can I find the real and imaginary parts of $w=(z-2)/(z+2)$? Hence, how can I calculate w and deduce that this complex number always lies parallel to the real axis in the Argand plane?
The following information can be used:
$\cos(2\theta)=2\cos^2(\theta)-1$ and
$\sin(2\theta)=2\cos(\theta)\sin(\theta)$
Thanks for all your help!!
you can argue this geometrically as follows: i will work with $0 \le \theta \le \pi$ for $\theta$ in the lower half plane replace $\theta$ by $-\theta$
$z = 2e^{i\theta}$ lies on the circle with diameter $-2$ and $2$. that is the center of the circle is $0$ and the radius is $2$ now observe that the angle subtended by the diameter on the circle is $\pi/2$ this translates as $z - 2$ and $z + 2$ are orthogonal. we can say more $$z - 2 = 2 \sin(\theta/2)e^{i(\pi+\theta)/2}, \qquad z + 2 = 2\cos (\theta/2)e^{i\theta/2},$$ hence $$ w = \dfrac{z-2}{z+2} = i\tan (\theta/2) $$ so $w$ is a pure imaginary number therefore is parallel to imaginary axis.