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Say you have a complex number like this: 2-2i

You want to convert it to polar form $(re^{i \theta }$)

I know how to do this using tangent (and then adjusting the resulting answer to make it correct), but I would like to know how to do it using cosine and sine. My complex analysis book says that the argument is fixed by sine and cosine. That is, $\theta$ is uniqely determined by the pair of equations:

$$ \cos \theta = \frac{x}{|z|} $$ $$ \sin \theta = \frac{y}{|z|} $$

It never says exactly how I should find $\theta$ using these two equations however. (in all examples it simply says: and then using, $\cos \theta$ and $\sin \theta$, we see that $\theta$ is...)

So, when converting to polar form, what is the standard procedure for finding the argument of a complex number using cosine and sine?

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So, going with your example,

$z = 2-2i$

$|z| = \sqrt{2^2 + (-2)^2} = \sqrt 8 = 2\sqrt 2$

$\cos \theta = \frac{2}{2\sqrt 2} = \frac{1}{\sqrt 2}$

$\sin\theta = \frac{-2}{2\sqrt 2} = -\frac{1}{\sqrt 2}$

The reference angle for $\theta$ in both cases is $\frac{\pi}{4}$ ($\arccos |\frac{1}{\sqrt 2}| = \arcsin |-\frac{1}{\sqrt 2}| = \frac{\pi}{4}$). Remember your special triangles, in this case, the $1,1,\sqrt 2$ right triangle.

Moreover, cosine is positive, so $\theta$ lies in either $1$st or $4$th quadrant.

Sine is negative, so $\theta$ lies in either $3$rd or $4$th quadrant.

Hence $\theta$ lies in the $4$th quadrant, giving $\theta = -\frac{\pi}{4}$

When your complex analysis book says the sine and cosine uniquely fix the argument, they mean that the equations for sine and cosine will always give the same reference angle and that by considering the signs, there can only be one quadrant that the argument can lie in. Your example serves to illustrate this nicely.

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