How do I convert $\frac{1+ \sqrt{3i} }{1- \sqrt{3i} }$ to polar form? I came across it in this question but I don't know much about complex numbers and have no idea how to figure out $\theta$.
2026-03-31 10:37:56.1774953476
Convert $\frac{1+ \sqrt{3i} }{1- \sqrt{3i} }$ to polar form
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Hint: $\arg(xz)=\arg(x)+\arg(z)$ which (setting $z=\frac{y}x$ and rearranging) yields: $$\arg\left(\frac{y}x\right)=\arg(y)-\arg(x)$$ So it is only necessary to calculate the argument of $1+\sqrt{3}i$ and its conjugate (which has argument equal to the negative of that of $1+\sqrt{3}i$), then take their difference.