Let $z \in \mathbb{C}$ be such that $|z|<1$. If $w=\frac{5+3 z}{5(1-z)}$, then

(C) $5 \operatorname{Re}(w)>1$

(A) $5 \operatorname{Im}(w)<1$

(B) $4 \operatorname{Im}(w)>5$

(D) $5 \operatorname{Re}(w)>4$

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Complex Numbers

Let

z \in C

be such that

∣ z ∣ < 1

. If

w = \frac{5 + 3 z}{5 ( 1 - z )}

, then

Answer: (C)

Solve the defining relation for

z

: from

w = \frac{5 + 3 z}{5 ( 1 - z )}

we get

5 w (1 - z) = 5 + 3 z

, so

z = \frac{5 w - 5}{5 w + 3}

. The restriction

∣ z ∣ < 1

then reads

∣5 w - 5∣ < ∣5 w + 3∣

. Writing

w = u + i v

and squaring,

(5 u - 5)^{2} + (5 v)^{2} < (5 u + 3)^{2} + (5 v)^{2}

; the

v

terms cancel, leaving

(5 u - 5)^{2} < (5 u + 3)^{2}

. Expanding,

- 50 u + 25 < 30 u + 9

, so

80 u > 16

and

u > \frac{1}{5}

. Hence for every

z

inside the unit circle,

Re (w) > \frac{1}{5}

, that is

5 Re (w) > 1

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option