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Q.
If $\left(\sqrt{8}+i\right)^{50} =3^{49} \left(a+ib\right)$, then $a^{2}+b^{2}$ is
Complex Numbers and Quadratic Equations
Solution:
Writing $\sqrt{8}+i$ in polar form,
$\sqrt{8}+i=r\left(cos\,\theta+i\,sin\,\theta\right)$, where $r=\sqrt{\left(\sqrt{8}\right)^{2}+1^{2}}=3$
and $tan\, \theta=\frac{1 /3}{\sqrt{8}/3} =\frac{1}{\sqrt{8}}$
$\therefore \, \theta=tan^{-1} \frac{1}{2\sqrt{2}}$
Now, $\left(\sqrt{8}+i\right)^{50} =3^{50}\left(cos\,\theta+i\,sin\theta\right)^{50}$
$=3^{50}\left(cos\,50\theta+isin\,50\theta\right)$
But given that $\left(\sqrt{8}+i\right)^{50}=3^{49}\left(a+ib\right)$
$\therefore \, a=3 \, cos \, 50\theta, b=3 \, sin \,50\theta$
$\therefore \, a^{2}+b^{2}=9\left(cos^{2} 50\theta+sin^{2}\,50\theta\right)=9$