IIT JEE Advanced 2010 Paper 2

(A)→(Q) $\left| \frac{z}{|z|} - i \right|$ = $\left| \frac{z}{|z|} + i \right|$ z ≠ 0 Where $\frac{z}{|z|}$ is unimodular complex number and lies on perpendicular bisector of i and −i. ⇒ $\frac{z}{|z|}$ = ± 1 ⇒ z = ± 1 |z| ⇒ a is real number ⇒Im(z) = 0. (B)→(P) We have |z + 4| + |z - 4| = 10 Where z lies on an ellipse whose focus are (4, 0) and (−4, 0) and length of major axis is 10. Therefore, 2ae = 8 and 2a = 10 ⇒ e = $\frac{4}{5}$ Therefore, Re(z) ≤ 5. (C)→(P), (T) We have |w| 2 ⇒ w = 2 (cos θ + i sin θ) x + iy = 2 (cos θ + i sin θ) - $\frac{1}{2}$ (cos θ - i sin θ) = $\frac{3}{2}\cos\theta + i\frac{5}{2}\sin\theta$ ⇒ $\frac{x^2}{\left(\frac{3}{2}\right)^2} + \frac{y^2}{\left(\frac{5}{2}\right)^2}$ = 1 Thus,we have $e^2$ = 1 - $\frac{\frac{9}{4}}{\frac{25}{4}}$ = 1 - $\frac{9}{25}$ = $\frac{16}{25}$ ⇒ e = $\frac{4}{5}$ (D)→(Q), (T) We have |w| = 1 ⇒ x + iy = cos + i sin θ + cos θ - i sin θ x + iy = 2 cos θ Hence, |Re (z)| ≤ 1, and Im (z) = 0.