Question

If the complex number z is to be satisfy |z| = 3, |z – {a(1 + i) – i}| £ 3 and |z + 2a – (a + 1)i| > 3 simultaneously for at least one z then range of ‘a’ is –

• A. $\left( \frac{1 - \sqrt{71}}{2},\frac{- 1 - 4\sqrt{11}}{5} \right)$Č$\left( \frac{- 1 + 4\sqrt{11}}{5},\frac{1 + \sqrt{71}}{2} \right)$
• B. $\left( \frac{+ 2 - \sqrt{41}}{3},\frac{- 2–\sqrt{41}}{5} \right)$Č$\left( \frac{- 2 + \sqrt{41}}{5},\frac{2 + \sqrt{41}}{3} \right)$
• C. $\left( \frac{1 - 2\sqrt{71}}{2},\frac{- 1 - 8\sqrt{11}}{5} \right)$Č$\left( \frac{- 1 + 8\sqrt{11}}{5},\frac{1 + 2\sqrt{71}}{2} \right)$
• D. None of these

Answer 1

Answer: (A)

$\left( \frac{1 - \sqrt{71}}{2},\frac{- 1 - 4\sqrt{11}}{5} \right)$Č$\left( \frac{- 1 + 4\sqrt{11}}{5},\frac{1 + \sqrt{71}}{2} \right)$

Answer 2

Sol. Equation |z| = 3 represents boundary of a circle and equation |z – {a (1 + i) – 0i| £ 3 represents the interior and the boundary of a circle and equation |z + 2a – (a + 1) i| > 3 represents the exterior of a circle. Then any point which satisfies all the three conditions will lie on first circle, on or inside the second circle and outside the third circle.

For the existence of such a point first two circles must cut or at least touch each other and first and third circle must not intersect each other. The arc ABC of first circle lying inside the second but outside the third circle, represents all such possible points

Let z = x + iy then equation of circles are

x² + y² = 9 … (1)

(x – a)² + (y – a + 1)² =9 … (2)

And (x – 2a)² + (y – a – 1)² = 9 … (3)

Circles (1) and (2) should cut or touch then distance between their centres £ sum of their radii.

Ž $\sqrt{(a - 0)^{2} + (a–1–0)^{2}}$£ 3 + 3

Ž a² + (a – 1)² £ 36

Ž 2a² – 2a – 35 £ 0 Ž a² – a – $\frac{35}{2}$ £ 0

Ž $\left( a - \frac { 1 + \sqrt { 71 } } { 2 } \right)$ $\left( a - \frac{1 - \sqrt{71}}{2} \right)$ £ 0

\ $\frac{1 - \sqrt{71}}{2}$ £ a £ $\frac{1 + \sqrt{71}}{2}$ … (4)

Again circles (1) and (3) should not cut or touch then distance between their centres > sum of their radii.

$\sqrt{( - 2a - 0)^{2} + (a + 1–0)^{2}}$> 3 + 3 or$\sqrt{5a^{2} + 2a + 1}$> 6

Ž 5a² + 2a + 1 > 36 or 5a² + 2a – 35 >0

Ž a² + $\frac{2a}{5}$ – 7 > 0

then $\left( a - \frac { - 1 - 4 \sqrt { 11 } } { 5 } \right)$ $\left( a - \frac{- 1 + 4\sqrt{11}}{5} \right)$ > 0

\ a Ī $\left( –\infty,\frac{- 1 - 4\sqrt{11}}{5} \right)$ Č $\left( \frac{- 1 + 4\sqrt{11}}{5},\infty \right)$ … (5)

Hence, the common values of a satisfying (4) and (5) are

aĪ $\left( \frac{1 - \sqrt{71}}{2},\frac{- 1 - 4\sqrt{11}}{5} \right)$Č $\left( \frac{- 1 + 4\sqrt{11}}{5},\frac{1 + \sqrt{71}}{2} \right)$.