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I am trying to find the argument of $2+i$, $\tan \theta=\frac{1}{2}$, but not getting anything. How would I find this?


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What's keeping you from taking the next step after having $\displaystyle \tan (\theta )=\dfrac{1}{2}?$ – Git Gud Apr 21 '13 at 20:20
@GitGud I was looking for an exact value, not a decimal. – Alti Apr 21 '13 at 20:22
You don't know what $\displaystyle \arctan \left(\frac{1}{2}\right)$ is, is that it? – Git Gud Apr 21 '13 at 20:23
Yeah, I tried using the unit circle but it's not working. Unless I am overlooking the obvious – Alti Apr 21 '13 at 20:24
You can use Tangant, sine, or cosine tables and then come up with general forms. Also you can use arc-tan to refer to the solution? – user25004 Apr 21 '13 at 20:25
up vote 4 down vote accepted

You're almost done. Take $\theta=\tan^{-1}\frac12$ to get the principal argument, since $2+i$ lies to the right of the imaginary axis.

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Can I get an exact value? – Alti Apr 21 '13 at 20:23
$\theta=\tan^{-1}\frac12$ is an exact value. It isn't a nice fraction of $\pi$, if that's what you're looking for. – Cameron Buie Apr 21 '13 at 20:25
Oh okay, that's what I thought, but wanted to make sure I wasn't overlooking anything. – Alti Apr 21 '13 at 20:28
Don't be scared of the arctangent! Imagine someone saying "I have found that $x = \sqrt(3)$, but I don't know where to go from here!" ... – The Chaz 2.0 Apr 21 '13 at 20:56

Hint: $z=x+iy \to tan\phi=\frac{y}{x}$

enter image description here

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The OP already figured out that much. – Cameron Buie Apr 21 '13 at 20:27

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