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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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  • $\begingroup$ 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 $0 \le \theta < 2 \pi$ $\endgroup$ – neofoxmulder Feb 15 '14 at 10:05
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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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  • $\begingroup$ check your values for sin 60º , cos 60º. :) $\endgroup$ – neofoxmulder Feb 15 '14 at 10:24
  • $\begingroup$ 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. $\endgroup$ – DonAntonio Feb 15 '14 at 10:42
  • $\begingroup$ Maybe so but $ sin \frac{ \pi}{3} \ne \frac{1}{2}$. You switched the values , inadvertantly i presume. :) $\endgroup$ – neofoxmulder Feb 15 '14 at 10:46
  • $\begingroup$ Oh, I see what you meant, @neofoxmulder. Thanks, already fixed. $\endgroup$ – DonAntonio Feb 15 '14 at 10:48
  • $\begingroup$ 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 :) $\endgroup$ – neofoxmulder Feb 15 '14 at 10:54
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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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