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Q.
If $(\sqrt3+i)^{100}=2^{99}(a+ib)$, then $a^2+b^2$ is equal to
Complex Numbers and Quadratic Equations
Solution:
Since $\left(\sqrt{3}+i\right)^{100}=2^{99}\left(a+ib\right)$
$\therefore \left(\sqrt{3}-i\right)^{100}=2^{99}\left(a-ib\right)$
Multiplying, we get
$\left(\sqrt{3}+i\right)^{100} \left(\sqrt{3}-i\right)^{100}=2^{198}\left(a^{2}+b^{2}\right)$
$\therefore a^{2}+b^{2}=\frac{\left(3+1\right)^{100}}{2^{198}}=\frac{2^{200}}{2^{198}}$
$=2^{2}=4$.