# What mistake did I make finding $\arg (\sqrt{3}-i)$

What mistake did I make finding $\arg (\sqrt{3}-i)$?

I figured it will be in the 4th quadrant and look like:

So $\arg{z} = - \arctan{\frac{\sqrt{3}}{1}} = -\frac{\pi}{3}$

The right answer is supposed to be $-\frac{\pi}{6}$

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Tangent is opposite over adjacent, which is $y$ coordinate over $x$ coordinate. You have it the other way around.