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Question

If $$(\sqrt {3} + i)^{100} = 2^{99} (a + ib)$$, then show that $$a^{2} + b^{2} = 4$$


Solution

Since $$\omega = \dfrac{-1+i\sqrt 3}{3}=\dfrac{i^2+i\sqrt 3}{2}$$
$$\Rightarrow \sqrt 3+i=\dfrac{2\omega }{i}=-2\omega i$$

Thus $$(\sqrt 3+i)^{100}=(-2\omega i)^{100}=2^{100}\cdot \omega^{100}i^{100}$$

$$ = 2^{100}\omega (1)=2^{100}\left(\dfrac{-1+i\sqrt 3}{2}\right)=2^{99}(-1+i\sqrt 3)$$

So $$a=-1$$ and $$b =\sqrt 3$$

Hence $$a^2+b^2=1+3=4$$

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