If left(frac{sqrt{3}+i}{sqrt{3}-i}right)^m=1 | vedclass

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(C) Given the equation $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^m=1$. First,simplify the base: $\frac{\sqrt{3}+i}{\sqrt{3}-i} = \frac{(\sqrt{3}+i)(

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$2028$

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(C) Given the equation $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^m=1$. First,simplify the base: $\frac{\sqrt{3}+i}{\sqrt{3}-i} = \frac{(\sqrt{3}+i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)} = \frac{3-1+2\sqrt{3}i}{3+1} = \frac{2+2\sqrt{3}i}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. This is equal to $\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) = e^{i\pi/3}$. So,the equation becomes $(e^{i\pi/3})^m = 1$,which implies $e^{im\pi/3} = 1$. This holds if $\frac{m\pi}{3} = 2n\pi$ for some integer $n$,meaning $m = 6n$. We are given $2022 Checking multiples of $6$: $2022/6 = 337$ and $2028/6 = 338$. The only multiple of $6$ in the range $(2022, 2029)$ is $2028$. Thus,$m = 2028$.

If $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^m=1$ and $2022 < m < 2029$,then $m=$

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