Polar Form from Cartesian Form

Question

What is the polar form of $\tfrac{1}{\sqrt{2}} + i\tfrac{1}{\sqrt{2}}$? I know that polar form looks like $r(\cos \theta + i \sin \theta)$ and I believe $r=1$ in this case upon solving it, but I don't understand how you find $\theta$.

Answer 1

If you compare $$\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$$ with the form $$r\left( \cos\theta + i \sin \theta \right) = r\cos\theta + i r\sin \theta$$ it is clear that you're looking for $r$ and $\theta$ such that: $$r\cos\theta = \frac{1}{\sqrt{2}} \quad\mbox{ and } \quad r\sin \theta = \frac{1}{\sqrt{2}}$$ Squaring and adding both equations confirms that you are right about $r$ being equal to $1$ (*), so this reduces the problem to finding $\theta$ such that: $$\cos\theta = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \quad\mbox{ and } \quad \sin \theta = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ Do you know such a $\theta$?

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(*) You could get this directly since $a+bi = r\cos\theta + i r\sin \theta$ implies that $r^2=a^2+b^2$; immediately giving $r=1$ in your case.

Answer 2

Shortcut: In the first quadrant take $\arctan(y/x)$