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4-1.Complex numbers
medium
જો ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$ તો ${a^2} + {b^2}$ = . . .
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