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JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Complex Numbers

Let

z_{1}

and

z_{2}

be two complex number such that

z_{1} + z_{2} = 5

and

z_{1}^{3} + z_{2}^{3} = 20 + 15 i

. Then

z_{1}^{4} + z_{2}^{4}

equals-

(A) $303$
(B)75✓ Correct
(C) $1515$
(D) $253$

Answer: (B)

Solution

Given

z_{1} + z_{2} = 5

and

z_{1}^{3} + z_{2}^{3} = 20 + 15 i

.

Using

z_{1}^{3} + z_{2}^{3} = (z_{1} + z_{2})^{3} - 3 z_{1} z_{2} (z_{1} + z_{2})

:

20 + 15 i = 125 - 15 z_{1} z_{2}

15 z_{1} z_{2} = 105 - 15 i \Rightarrow z_{1} z_{2} = 7 - i

.

Now

(z_{1} + z_{2})^{2} = z_{1}^{2} + z_{2}^{2} + 2 z_{1} z_{2}

, so

z_{1}^{2} + z_{2}^{2} = 25 - 2 (7 - i) = 11 + 2 i

.

Then

z_{1}^{4} + z_{2}^{4} = (z_{1}^{2} + z_{2}^{2})^{2} - 2 (z_{1} z_{2})^{2}

.

(11 + 2 i)^{2} = 121 + 44 i - 4 = 117 + 44 i

(7 - i)^{2} = 49 - 14 i - 1 = 48 - 14 i

z_{1}^{4} + z_{2}^{4} = 117 + 44 i - 2 (48 - 14 i) = 117 + 44 i - 96 + 28 i = 21 + 72 i

∣ z_{1}^{4} + z_{2}^{4} ∣ = 2 1^{2} + 7 2^{2} = 441 + 5184 = 5625 = 75

.

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option

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