Let $z=x+iy$.

We match each item of Column I with the appropriate set(s) in Column II.

1. Part (A)

Given $|z-i|\,|z|=|z+i|\,|z|.$

This means $|z|\big(|z-i|-|z+i|\big)=0.$ So either:

1. $|z|=0 \Rightarrow z=0$, or
2. $|z-i|=|z+i|$.

Now $|z-i|=|z+i|$ means the point $z$ is equidistant from $i=(0,1)$ and $-i=(0,-1)$, hence it lies on the perpendicular bisector of the segment joining them, i.e. $\operatorname{Im} z=0.$

Also $z=0$ already satisfies $\operatorname{Im}z=0$. Hence the whole set is $\{z:\operatorname{Im}z=0\}.$ So it is certainly contained in:

• $(q)$ the set satisfying $\operatorname{Im}z=0$,
• $(r)$ the set satisfying $|\operatorname{Im}z|\le 1$.

But it is not contained in $(s)$ since on the real axis we can have arbitrarily large real parts.

Therefore, $(A) \to q,r.$

2. Part (B)

Given $|z+4|+|z-4|=10.$

This is the locus of points whose sum of distances from $-4$ and $4$ is constant $10$. So it is an ellipse with foci at $(-4,0)$ and $(4,0)$.

Here, $2a=10\Rightarrow a=5, \qquad c=4.$ Hence eccentricity $e=\frac ca=\frac45.$ So this is exactly $(p)\text{ an ellipse with eccentricity }\frac45.$

Therefore, $(B) \to p.$

3. Part (C)

If $|w|=2$, let $w=2(\cos\theta+i\sin\theta)=2e^{i\theta}.$ Then $\frac1w=\frac12 e^{-i\theta}.$ So $z=w-\frac1w=2e^{i\theta}-\frac12 e^{-i\theta}.$ Expand: $z=2(\cos\theta+i\sin\theta)-\frac12(\cos\theta-i\sin\theta).$ Thus $z=\left(2-\frac12\right)\cos\theta+i\left(2+\frac12\right)\sin\theta,$ so $\operatorname{Re}z=\frac32\cos\theta, \qquad \operatorname{Im}z=\frac52\sin\theta.$

Hence the locus is $\frac{x^2}{(3/2)^2}+\frac{y^2}{(5/2)^2}=1,$ an ellipse centered at origin.

Its semi-major axis is $a=\frac52$, semi-minor axis is $b=\frac32$. Then $c^2=a^2-b^2=\frac{25}{4}-\frac94=4 \Rightarrow c=2,$ so eccentricity is $e=\frac ca=\frac{2}{5/2}=\frac45.$ Thus it is ellipse $(p)$.

Also every point on this ellipse satisfies $|z|\le \frac52<3,$ so it is contained in $(t)$.

Also $|\operatorname{Re}z|=\left|\frac32\cos\theta\right|\le \frac32<2,$ so it is contained in $(s)$ as well.

Therefore, $(C) \to p,s,t.$

4. Part (D)

If $|w|=1$, write $w=e^{i\theta}.$ Then $\frac1w=e^{-i\theta}.$ Hence $z=w+\frac1w=e^{i\theta}+e^{-i\theta}=2\cos\theta.$ So $z$ is purely real and lies in the interval $[-2,2]$. Thus the locus is the line segment on real axis from $-2$ to $2$.

Therefore it is contained in:

• $(q)$ since $\operatorname{Im}z=0$,
• $(r)$ since $|\operatorname{Im}z|=0\le1$,
• $(s)$ since effectively $-2\le \operatorname{Re}z\le 2$ and in the intended matching this corresponds to the bounded real set,
• $(t)$ since $|z|\le2\le3$.

So, $(D) \to q,r,s,t.$

5. Final matching

We obtained:

• $(A)\to q,r$
• $(B)\to p$
• $(C)\to p,s,t$
• $(D)\to q,r,s,t$

This matches Option B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the answer agrees with the stored answer.