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 $\theta$.
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2Is the i inside or outside the square root sign? – MasB Dec 10 '14 at 01:16
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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.
Milo Brandt
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Ohhh... that's much better than what I was trying which was just getting the original fraction into $a + bi$. – Elliot Gorokhovsky Dec 10 '14 at 01:08
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@Rene Doing that actually wouldn't be too bad; if you multiply through by the denominator, you're just trying to solve: $$(a+bi)(1-\sqrt{3}i)=1+\sqrt{3}i$$ and after you separate into real and imaginary parts, you'd get two linear equations: $$a-\sqrt{3}b=1$$ $$\sqrt{3}a+b=\sqrt{3}$$ which is easy enough. – Milo Brandt Dec 10 '14 at 01:14