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I have the complex roots $x_1=1+3i, x_2=1-3i$ and i want to express them in trig form

Since the trig form is $p(cosθ + sinθ)$ and $p=\sqrt{a^2 + b^2}, tanθ=\frac{b}{a}$

we have $p=\sqrt{1^2 + 3^2}=\sqrt{10}$

and $tanθ = \frac{3}{1}=3$

Issue is, that the $tan^{-1}(3)$ isnt a "nice" $\frac{π}{n}$ number and i have the impression that this exercise is supposed to give you something of the latter.

So am i making a mistake here? Or there is any way to make the final answer "prettier"? My trig skills arent the best

Than1
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    You can't make it better. $\arctan 3$ is not expressible in the form $p\pi/q$. – Trebor Sep 21 '21 at 09:11
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    I see two problems with the arctan function : [1] it's range is $(-\pi/2, \pi/2)$, where you need $\theta$ to be in a half-open interval of width $(2\pi)$, such as $(-\pi, \pi].$ [2] Since you are looking for an expression of form $p[\cos(\theta) + i\sin(\theta)]$ and since specifying the two values $\cos(\theta), \sin(\theta)$ is always enough to uniquely identify $\theta$, within a modulus of $(2\pi)$, I recommend specifying $\theta$ by simultaneously specifying $\cos(\theta)$ and $\sin(\theta).$ – user2661923 Sep 21 '21 at 09:39
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    In your stated problem, I would then specify $x_1 = (1 + 3i)$ as $\sqrt{10}[\cos(\theta) + i\sin(\theta)]$, where $\cos(\theta) = \frac{1}{\sqrt{10}}$ and $\sin(\theta) = \frac{3}{\sqrt{10}}.$ – user2661923 Sep 21 '21 at 09:41
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    Note that expressing (for example) $(-1 - 3i)$ which is in the 3rd quadrant, precludes attempting to directly use either the arcsine or arccosine function. The arcsine function allows 1st or 4th quadrant angles only, while the arccosine function allows 1st or 2nd quadrant angles only. Neither function gives you 3rd quadrant angles. Indirectly, you could say that since the number is in the 3rd quadrant, then its argument is $(\theta + \pi)$ [or $(\theta - \pi)$], where $\theta = \text{arccos}\left(\frac{1}{\sqrt{10}}\right)$ or $\theta = \text{arcsin}\left(\frac{3}{\sqrt{10}}\right)$. – user2661923 Sep 21 '21 at 09:50
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    It might be worth double checking that you shouldn't have $1 \pm \sqrt 3 i$. That would come out more nicely. – Blitzer Sep 21 '21 at 10:36

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