If $\frac{z-\alpha}{z+\alpha},(\alpha \in R)$ is a purely imaginary number and $|z|=2$, can we find value of $\alpha$ geometrically? If $\dfrac{z-\alpha}{z+\alpha},(\alpha \in R)$ is a purely imaginary number and $|z|=2$, then find value of $\alpha$.
Now I took $\dfrac{z-\alpha}{z+\alpha}=t$ and as t is purely imaginary, and use the fact that $t+ \bar{t}=0$ and obtained the answer $\alpha = \pm2$.
But I was wondering that if there is any way to think about the answer more directly using geometry of complex numbers given that $z$ lies on a circle centered at origin having radius $2$.
 A: If $\displaystyle{z-\alpha\over z+\alpha} = i\lambda\  $ for some $\lambda\in\mathbb R$ then $z-\alpha$ and $z+\alpha$ are perpendicular. 
But those two complex numbers can be seen as the diagonals of the parallelogram with vertices at $0, z, \alpha$ and $z+\alpha$, and if the diagonals of a parallelogram are perpendicular, all sides are equal, which is possible in this case only when $\alpha=\pm 2$
A: For now, let's say $\alpha$ may be any complex no. $z_1$ and $(-\alpha)$ be another complex no. $z_2$

Here consider such a circular arc passing through $z_1$, $z_2$ and another complex no. $z_o$
From the property of circles, angle (a) between $(z_1-z_o)$ and $(z_2-z_o)$ will remain constant wherever $z_o$ moves on the arc.
We can write this as (using rotation theorem) : 
$$\frac{z_1-z_o}{|z_1-z_o|} =\frac{z_2-z_o}{|z_2-z_o|} e^{ia}$$
$$\to \frac{z_1-z_o}{z_2-z_o} =\frac{|z_1-z_o|}{|z_2-z_o|} e^{ia}$$
Taking argument of both the sides:
$$\arg \left( \frac{z_1-z_o}{z_2-z_o} \right) = a$$
So we can draw an analogy here that:

For any two fixed $z_1$ and $z_2$, if $$\arg \left( \frac{z_1-z_o}{z_2-z_o} \right) = a \ (constant)$$ then the locus of $z_o$ will be an. arc passing through $z_1$, $z_2$ and $z_o$


Consider this:
$$\frac{z-\alpha}{z+\alpha} = bi$$
$b \in \mathbb{R}$ and $\alpha \in \mathbb{R}$
Since these two complex numbers are equal, their principal argument must also be equal.
$$\arg \left( \frac{z-\alpha}{z+\alpha} \right) = \arg (bi)$$
$$\arg \left( \frac{z-\alpha}{z+\alpha} \right) = \frac{\pi}{2}$$
Since $\alpha$ and $(-\alpha)$ are fixed complex no. (on the Real axis since both are real no.), the locus of $z$ will be an arc as mentioned above.
Moreover, the arc will be a semicircle as the angle is $\frac{\pi}{2}$. (Another property of circles)
SO,
 
$z$ will have to lie on such a $\color{red}{semicircle}$.
As we increase $\alpha$, the radius of this $\color{red}{semicircle}$ will increase. (See Here for visualisation (vary the slider))
But we know that $z$ has two lie on the circle centred at $0$ and of radius $2$. 
So the points where this circle meets the Real Axis have to be $(\pm2,0)$ and these endpoints of the semicircle are nothing but $(\pm \alpha,0)$
So,
$$\color{green}{\alpha = \pm 2}$$
NOTE:
The block-quoted matter which defines the locus of the above conditions can be used in similar questions, just take in mind:

The locus of $z_o$ will be :
$\bullet$ An arc if $a \in (0,\pi)$
$\bullet$ A line segment if $a = \pi$
$\bullet$ A pair of rays if $a=0$

