Question

If z satisfies |z + 1| < |z – 2|, then w = 3z + 2 + i satisfies –

• A. |w + 1| < |w – 8|
• B. |w + 1| < |w – 7|
• C. w +$\bar{\omega}$> 7
• D. |w + 5| < |w – 4|

Answer 1

Answer: (A)

|w + 1| < |w – 8|

Answer 2

Sol. We have |z + 1| < |z – 2|

Ž (z + 1) ($\bar{z}$ + 1) < (z – 2) ($\bar{z}$– 2)

Ž z$\bar{z}$ + z + $\bar{z}$ + 1 < z$\bar{z}$ – 2z – 2$\bar{z}$ + 4

Ž 3(z + $\bar{z}$) < 3 Ž z + $\bar{z}$ < 1. … (1)

Now w + $\bar{\omega}$ = (3z + 2 + i) + (3$\bar{z}$ + 2 – i)

= 3 (z +$\bar{z}$) + 4 < 3(1) + 4 = 7

\ w + $\bar{\omega}$ < 7 … (2)

(1) |w + 1| < |w – 8|

If (w + 1) ($\bar{\omega}$ + 1) < (w – 8) ($\bar{\omega}$ – 8)

If $\omega\bar{\omega}$ + w + $\bar{\omega}$ + 1 <$\omega\bar{\omega}$ – 8w – $8\bar{\omega}$ + 64

If 9(w + $\bar{\omega}$) < 63 if w + $\bar{\omega}$ < 7, which is true.