Question
If $n$ is odd, then the value of the expression $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{\prime}+1$ is(a) $\omega$(b) $\omega^{2}$(c) 0(d) $-1$
Step 1
We have $\frac{\sqrt{3}+i}{\sqrt{3}-i}$. We can multiply the numerator and the denominator by the conjugate of the denominator to get rid of the imaginary part in the denominator. This gives us $\frac{(\sqrt{3}+i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}$. Show more…
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