# For distinct non-zero complex $z_1$, $z_2$, $z_3$ satisfying $|z-1|=1$ and $z_2^2=z_1z_3$, which of the following are true?

$$z_1, z_2, z_3$$ are three non zero distinct points satisfying $$|z-1|=1 \space \& \space z_2^2=z_1 z_3$$ then

$$\qquad$$ (A) $$\displaystyle\frac{z_3-z_2}{z_2+z_3-2}$$ is purely imaginary

$$\qquad$$ (B) $$\displaystyle\operatorname{Arg}\left(\frac{z_2-1}{z_1-1}\right)=2\operatorname{Arg}\left(\frac{z_3}{z_2}\right)$$

$$\qquad$$ (C) $$\displaystyle\operatorname{Arg}\left(\frac{z_2-1}{z_1-1}\right)=2 \operatorname{Arg}\left(\frac{z_3}{z_1}\right)$$

$$\qquad$$ (D) $$\left|\frac{1}{z_2}-\frac{1}{z_3}\right|+\left|\frac{1}{z_1}-\frac{1}{z_2}\right|=\left|\frac{1}{z_1}-\frac{1}{z_3}\right|$$

My approach is as follows:

$$z$$ represent the points in the circle $$(x-1)^2+y^2=1$$

Hence the parametric points are $$x=1+\cos\theta \,$$ & $$\,y=\sin\theta$$

Not able to approach as it is become complex in nature and not able to solve.

Let $$z_1 = e^{i\theta_1}+1 = 2e^\frac{i\theta_1}{2}cos\frac{\theta_1}{2}$$

So $$arg(z_1) = \frac{\theta_1}{2}$$

$$z_2 = e^{i\theta_2}+1$$

$$z_3 = e^{i\theta_3}+1$$

So $$arg(z_i)= \frac{\theta_i}{2}, i =1,2,3$$

So $$\frac{z_3-z_2}{z_3+z_2-2}$$ becomes

$$= \frac{e^{i(\theta_3 -\theta_2 )}-1}{ e^{i(\theta_3 - \theta_2)}+1}$$ = $$itan(\frac{\theta_3 - \theta_2}{2})$$

Which is purely imaginary as $$\theta_3 \ne \theta_2$$

Now using the 2nd relation: $$2arg(z_2) = arg(z_1)+ arg(z_3)$$

So

$$\theta_2 = \frac{\theta_1 + \theta_3}{2}$$. This relation is satisfied by equation in option (B) but not (C).

Now applying triangle inequality on option (D).

We find

$$|\frac{1}{z_2} - \frac{1}{z_3}|+|\frac{1}{z_1} - \frac{1}{z_2}| \ge |\frac{1}{z_1} - \frac{1}{z_3}|$$ , equality is achieved only when

$$\frac{1}{z_2}-\frac{1}{z_3} = \frac{1}{z_1}- \frac{1}{z_2}$$

Which on simplifying gives $$z_1 , z_2 , z_3$$ to be in AP. But they are already in GP.This is possible only when they are equal which is restricted in the question.

So only (A) and (B) are correct.

It required a few key incites, and it was definitely fun, so let's see it:

First of all, let's observe that $$(1+i,2i,1-i)$$ is a solution and see that options A and C don't work here. Now, we have to choose between B and D. I chose D because it made more sense, and it worked out.

Now, let's set: $$a=\frac1{z_1}, b=\frac1{z_2}, c=\frac1{z_1}$$ $$A,B,C$$ are the points representing $$a,b,c$$

Now, the statement is that $$|a-b|+|b-c|=|a-c|$$, which says that $$|AB|+|BC|=|AC|$$, which means that $$A,B,C$$ are in a line. If you take the last example, we can see that the line should be $$x=\frac12$$

Statement 1:the function $$f(z)=\frac1z$$ turns the circle $$|z-1|=1$$ into the line $$\Re(z)=\frac12$$

Proof: Let's say $$|z-1|=1$$, therefore $$z=x+yi$$ and $$(x-1)^2+y^2=1$$ so $$x^2+y^2=2x$$. Now: $$\Re(\frac1z)=\Re(\frac1{x+yi})=\Re(\frac{x-yi}{x^2+y^2})=\Re(\frac{x-yi}{2x})=\frac x{2x} = \frac12$$

So now we need to prove that $$B$$ is between $$A$$ and $$C$$, or in other words $$\arg(b)\in [\arg(a),\arg(c)]$$*

*Assuming $$\arg(a)<\arg(c)$$

$$\arg(a)+\arg(c)=\arg(ac)=\arg(b^2)=2\arg(b)$$ $$\arg(b)=\frac{\arg(a)+\arg(c)}2$$ so $$\arg(b)$$ is between $$\arg(a)$$ and $$\arg(c)$$

Q.E.D