If (sqrt{3}+i)^{10}=a+bi,where a, b in R,then | vedclass

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(D) Let $z = \sqrt{3} + i$. Then,the modulus $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. The argument $	heta = \arg(z) = 	an^{-1}(\frac{1}{\

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$512$ and $-512\sqrt{3}$

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(D) Let $z = \sqrt{3} + i$. Then,the modulus $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. The argument $	heta = \arg(z) = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. Thus,the polar form is $z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$. Using De Moivre's Theorem,$z^{10} = 2^{10}(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6})$. Simplifying the angle,$\frac{10\pi}{6} = \frac{5\pi}{3}$. $z^{10} = 1024(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$. Since $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$,$z^{10} = 1024(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 512 - 512i\sqrt{3}$. Comparing with $a+bi$,we get $a = 512$ and $b = -512\sqrt{3}$.

If $(\sqrt{3}+i)^{10}=a+bi$,where $a, b \in R$,then the values of $a$ and $b$ are respectively:

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