MCQ
If $(\sqrt{8}+i)^{50}=3^{49}(a+i b)$, then $a^2+b^2$ is
- A3
- B8
- ✓9
- D$\sqrt{8}$
✓
Answer
Correct option: C.
9
(C)
$(\sqrt{8}+ i )^{50}=3^{49}( a + ib )$
Taking modulus and squaring on both sides,
we get
$(8+1)^{50}=3^{98}\left(a^2+b^2\right)$
$\Rightarrow 9^{50}=3^{98}\left(a^2+b^2\right)$
$\Rightarrow 3^{100}=3^{98}\left( a ^2+ b ^2\right)$
$\Rightarrow\left( a ^2+ b ^2\right)=9$
$(\sqrt{8}+ i )^{50}=3^{49}( a + ib )$
Taking modulus and squaring on both sides,
we get
$(8+1)^{50}=3^{98}\left(a^2+b^2\right)$
$\Rightarrow 9^{50}=3^{98}\left(a^2+b^2\right)$
$\Rightarrow 3^{100}=3^{98}\left( a ^2+ b ^2\right)$
$\Rightarrow\left( a ^2+ b ^2\right)=9$
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