# How to calculate $\theta$ when we know $\tan \theta$.

Hej

I'm having difficulties calculating the angle given the tangent.

Example:

In a homework assignement I'm to express a complex variable $z = \sqrt{3} -i$ in polar form. I know how to solve this except for when I get to calculating the angle $\theta$.

I know that $\tan \theta = -\frac{1}{\sqrt{3}}$ but I do not know how to continue and compute the angle from that.

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If you plot the point you see it is in the 4th quadrant so you can find the reference angle $\theta_R$ by pretending it is in the 1st quadrant and then do $2 \pi - \theta_R$ to get an angle $0 \le \theta < 2 \pi$ – neofoxmulder Feb 15 '14 at 10:05

You shouldn't use the tangent for this kind of problems; compute $$|z|=\sqrt{z\bar{z}}=\sqrt{(\sqrt{3}-i)(\sqrt{3}+i)}= \sqrt{3+1}=2$$ Then you have $z=|z|u$, where $$u=\frac{\sqrt{3}}{2}-i\frac{1}{2}$$ and you need an angle $\theta$ such that $$\cos\theta=\frac{\sqrt{3}}{2},\quad\sin\theta=-\frac{1}{2}.$$ Since the sine is negative and the cosine is positive, you see that you can take $$\theta=-\frac{\pi}{6}$$ (the pair of values is well known). If you need an angle in the interval $[0,2\pi)$, just take $$-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}.$$

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$$\tan\theta=\frac{-1}{\sqrt3}=\pm\frac{\frac12}{\frac{\sqrt3}2}\implies\theta=\arctan\frac{-1}{\sqrt3}=\ldots$$

Remember though that

$$\sin\frac\pi6=\frac12\;,\;\;\cos\frac\pi6=\frac{\sqrt3}2$$

and since the minus sign belongs to the imaginary part (i.e., to the sine), it must be that

$$\theta=-\frac\pi6+2k\pi\;,\;\;k\in\Bbb Z$$

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check your values for sin 60º , cos 60º. :) – neofoxmulder Feb 15 '14 at 10:24
If you're addressing my answer @neofoxmulder then I can't understand what you say: in the first quadrant both sine and cosine are positive and thus $\;60^\circ \sim \frac\pi3\;rads.$ is irrelevant. – DonAntonio Feb 15 '14 at 10:42
Maybe so but $sin \frac{ \pi}{3} \ne \frac{1}{2}$. You switched the values , inadvertantly i presume. :) – neofoxmulder Feb 15 '14 at 10:46
Oh, I see what you meant, @neofoxmulder. Thanks, already fixed. – DonAntonio Feb 15 '14 at 10:48
No problem. I'm just an amateur newbie so I didn't want to edit your post because of my respect for your 85.5k :) – neofoxmulder Feb 15 '14 at 10:54

You can find the reference angle by disregarding the sign , you still need to figure out which quadrant you are in (which is easy) so you can add or subtract the reference angle accordingly

$$\theta_R = tan^{-1} \frac{1}{ \sqrt{3}}$$

$$\theta_R = \frac{ \pi}{6}$$

We are in the 4th quadrant so ,

$$\theta = 2 \pi - \frac{ \pi}{6} = \frac{ 11 \pi}{6}$$

Now , if you need more working angles just add integer multiples of $2 \pi$

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