JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Complex Numbers

(A) $|z-i|z||=|z+i|z||$ implies $z/|z|$ (a unimodular complex number) is equidistant from $i$ and $-i$, so it lies on the real axis. Thus $z$ is real, i.e., $\mathrm{Im}(z)=0$. Also, $|\mathrm{Im}(z)|\le 1$ is satisfied. Matches (q) and (r).

(B) $|z+4|+|z-4|=10$ is an ellipse with foci at $(\pm 4,0)$ and major axis length 10. So $2a=10$, $2ae=8$, giving $e=4/5$. Matches (p).

(C) If $|w|=2$, write $w=2e^{i\theta }$. Then $z=2e^{i\theta }-\frac{1}{2}{e}^{-i\theta }=\frac{3}{2}\cos \theta +i\frac{5}{2}\sin \theta$, which is an ellipse $\frac{x^2}{(3/2{)}^2}+\frac{y^2}{(5/2{)}^2}=1$ with $e=4/5$. Also, $|\mathrm{Re}(z)|\le 3/2<2$ and $|z|\le 5/2<3$. Matches (p), (s), and (t).

(D) If $|w|=1$, write $w=e^{i\theta }$. Then $z=e^{i\theta }+e^{-i\theta }=2\cos \theta$, which is real with $|z|\le 2$. So $\mathrm{Im}(z)=0$, $|\mathrm{Im}(z)|\le 1$, $|\mathrm{Re}(z)|\le 2$, and $|z|\le 3$. Matches (q), (r), (s), and (t).