4-1.Complex numbers
medium

If ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$ then ${a^2} + {b^2}$ is

A

$3$

B

$8$

C

$9$

D

$\sqrt 8 $

Solution

(c) ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$
Taking modulus and squaring on both sides, we get
${(8 + 1)^{50}} = {3^{98}}({a^2} + {b^2})$
${9^{50}} = {3^{98}}({a^2} + {b^2})$
${3^{100}} = {3^{98}}({a^2} + {b^2})$
==> $({a^2} + {b^2}) = 9$.

Standard 11
Mathematics

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