# 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 $$2$$.

• Do you know the rotation theorem? Jun 1 '20 at 19:01
• @SaketGurjar Yes. Jun 1 '20 at 19:02
• Ok good....because my answer makes use of it.. Jun 1 '20 at 19:32

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 $$a=0$$

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$$