Complex Numbers — JEE Main · 2024 · 05 Apr 2024 · Shift 1 Question with Solution

Consider the following two statements: Statement I: For any two non-zero complex numbers

z_{1}, z_{2}

(∣ z_{1} ∣ + ∣ z_{2} ∣) \frac{z _{1}}{∣ z _{1} ∣} + \frac{z _{2}}{∣ z _{2} ∣} \leq 2 (∣ z_{1} ∣ + ∣ z_{2} ∣)

; Statement II: if

x, y, z

are three distinct complex numbers and

a, b, c

are three positive real numbers such that

\frac{a}{∣ y - z ∣} = \frac{b}{∣ z - x ∣} = \frac{c}{∣ x - y ∣}

, then

\frac{a ^{2}}{y - z} + \frac{b ^{2}}{z - x} + \frac{c ^{2}}{x - y} = 1

. Between the above two statements,

Answer: (B)

Solution

Statement I:

\frac{z _{1}}{∣ z _{1} ∣} + \frac{z _{2}}{∣ z _{2} ∣} \leq 2

(sum of two unit vectors), so

(∣ z_{1} ∣ + ∣ z_{2} ∣) \frac{z _{1}}{∣ z _{1} ∣} + \frac{z _{2}}{∣ z _{2} ∣} \leq 2 (∣ z_{1} ∣ + ∣ z_{2} ∣)

— true.
Statement II: with

a^{2} = ∣ y - z ∣^{2} = (y - z) \overline{(y - z)}

etc. and the common ratio,

\frac{a ^{2}}{y - z} + \frac{b ^{2}}{z - x} + \frac{c ^{2}}{x - y} = \overline{(y - z)} + \overline{(z - x)} + \overline{(x - y)} = 0

, not

1

— false.
So I correct, II incorrect.

JEE Main · 2024 · 05 Apr 2024 · Shift 1 · Single correct option