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JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Complex Numbers

If

z = 2 + 3 i

, then

z^{5} + (\overline{z})^{5}

is equal to :

(A)244✓ Correct
(B)224
(C)245
(D)265

Answer: (A)

Solution

Using the binomial theorem and the fact that

z^{5} + \overset{z}{ˉ}^{5}

is real (sum of conjugates):

z^{5} + (\overset{z}{ˉ})^{5} = (2 + 3 i)^{5} + (2 - 3 i)^{5}

Expanding using binomial theorem and keeping only real terms:

= 2 ((0 5) 2^{5} + (2 5) 2^{3} (3 i)^{2} + (4 5) 2^{1} (3 i)^{4})

= 2 (32 + 10 \times 8 \times (- 9) + 5 \times 2 \times 81)

= 2 (32 - 720 + 810) = 2 (122) = 244

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option

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