If (sqrt{3}-i)^{n}=2^{n}, n in N,then the lea | vedclass

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(D) Given the equation $(\sqrt{3}-i)^{n}=2^{n}$.First,express the complex number $z = \sqrt{3}-i$ in polar form.The modulus is $|z| = \sqrt{(\sqrt{3})

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Given the equation $(\sqrt{3}-i)^{n}=2^{n}$.First,express the complex number $z = \sqrt{3}-i$ in polar form.The modulus is $|z| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2$.The argument $	heta$ is given by $	an 	heta = \frac{-1}{\sqrt{3}}$,which implies $	heta = -\frac{\pi}{6}$ (since it lies in the fourth quadrant).Thus,$z = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})) = 2e^{-i\pi/6}$.Substituting this into the equation: $(2e^{-i\pi/6})^n = 2^n$.$2^n e^{-in\pi/6} = 2^n$.Dividing by $2^n$,we get $e^{-in\pi/6} = 1$.This implies $-\frac{n\pi}{6} = 2k\pi$ for some integer $k$.$n = -12k$.Since $n \in N$ (natural numbers),the smallest positive value for $n$ occurs when $k = -1$,giving $n = 12$.

If $(\sqrt{3}-i)^{n}=2^{n}, n \in N$,then the least possible value of $n$ is

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