Rectangular to polar form using exact values. I'm in a first year math course at university, and we've been asked to convert a rectangular form complex number into polar form, using exact values only.
I have the modulus, that's all good. But I now have $\tan\theta=2-\sqrt{3}$ for the argument. I plugged this into my calculator and it spat out $15^{\circ}$ which I know can be represented as $\frac{\pi}{3}-\frac{\pi}{4}$ but I can't for the life of me figure out how we are supposed to be able to get from $\tan\theta=2-\sqrt{3}$ to $\theta=\frac{\pi}{3}-\frac{\pi}{4}$ without using a calculator!
Am I missing some elementary step? Many thanks.
 A: You can solve this equation exactly. Note that you can write
$$
\tan t =\frac{\sin t}{\cos t} = \frac{\frac{1}{2i}(e^{it}-e^{-it})}{\frac12(e^{it}+e^{-it})}=\frac 1i\frac{e^{2it}-1}{e^{2it}+1}
$$
Now suppose we are given $w$ (in your case,  $w=2-\sqrt 3$); to solve
$$w=\tan t$$ we use the expression above to write
$$w = \frac 1i\frac{e^{2it}-1}{e^{2it}+1}$$
$$iw(e^{2it}+1) = e^{2it}-1$$
$$e^{2it}(1-iw) =1+iw$$
$$e^{2it}=\frac{1+iw}{1-iw}$$
Using your particular value $w=2-\sqrt 3$, the expression $\frac{1+iw}{1-iw}$ simplifies after some tedious basic arithmetic to $\frac{\sqrt 3}{2}+\frac12 i  =e^{\pi i/6}$. Our equation, therefore, is
$$e^{2it} =e^{\pi i/6}$$
which means
$$2it = \frac{\pi}{6}i +2\pi ki$$
$$\boxed{t =\dfrac{\pi}{12} + \pi k}$$
for integral $k$, as desired.
A: Hint: Apply the identity $\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ as follows: 
$\tan 15^{\circ} = \tan(60^{\circ}-45^{\circ}) = \dfrac{\tan 60^{\circ} - \tan 45^{\circ}}{1 + \tan 60^{\circ} \tan 45^{\circ}}$. 
Can you continue from here?
Note that you could also use any pair $A,B$ such that $A-B = 15^{\circ}$, however, you want to stick to $A,B$ such that you know the values of $\tan A$ and $\tan B$. One such pair is $(A,B) = (60^{\circ},45^{\circ})$. Another pair you could have used is $(A,B) = (45^{\circ},30^{\circ})$.
