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If $x+i y=(-1+i \sqrt{3})^{2010}$, then $x$ is
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Verified Answer
The correct answer is:
$2^{2010}$
$x+i y=(-1+i \sqrt{3})^{2010}$
$\Rightarrow \quad x+i y=(2)^{2010}\left(\frac{-1+i \sqrt{3}}{2}\right)^{2010}$
$\Rightarrow \quad(x+i y)=(2)^{2010} \omega^{2010} \quad\left(\because \omega=\frac{-1+i \sqrt{3}}{2}\right)$
$\Rightarrow \quad(x+i y)=(2)^{2010}\left(\omega^{3}\right)^{670} \quad\left(\because \omega^{3}=1\right)$
$\Rightarrow \quad(x+i y)=(2)^{2010}(1)^{670}=2^{2010}+i \cdot 0$
On comparing real part
$\Rightarrow \quad \quad x=2010$
$\Rightarrow \quad x+i y=(2)^{2010}\left(\frac{-1+i \sqrt{3}}{2}\right)^{2010}$
$\Rightarrow \quad(x+i y)=(2)^{2010} \omega^{2010} \quad\left(\because \omega=\frac{-1+i \sqrt{3}}{2}\right)$
$\Rightarrow \quad(x+i y)=(2)^{2010}\left(\omega^{3}\right)^{670} \quad\left(\because \omega^{3}=1\right)$
$\Rightarrow \quad(x+i y)=(2)^{2010}(1)^{670}=2^{2010}+i \cdot 0$
On comparing real part
$\Rightarrow \quad \quad x=2010$
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