$z_1, z_2, z_3$ 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.


2 Answers 2


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: $2arg(z_2) = arg(z_1)+ arg(z_3)$


$\theta_2 = \frac{\theta_1 + \theta_3}{2}$. 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 $\arg(c)$



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