# Finding argument of complex number without calculator

I am solving some exercises in the book I am reading. In this particular exercise I should find real and imaginary part of

$$\left ( {1 + i \sqrt{3}\over 2}\right )^n$$

My idea was to calculate the argument and the absolute value and then use polar representation.

But I think the idea is not to use a calculator. So I am stuck on $$\arctan \sqrt{3}$$

If I use a calculator to find this value I can easily solve the exercise.

How to calculate $\arctan \sqrt{3}$ without using a calculator? Is it possible?

Edit

If it's possible any general method is most appreciated since I am already stuck at the next exercise where I am trying to find the argument of $-3+i$.

• Remember, $\arctan\sqrt{3} = \theta \implies \tan \theta = \sqrt{3}$.. Then draw your exact triangles to find the value of $\theta$. Feb 13 '15 at 3:01
• $$z = 1 \angle{60}$$ $$z^n = 1 \angle{n 60}$$ $$tan(30)=\frac{\sqrt3}3$$ $$tan(60)={\sqrt3}$$ Feb 13 '15 at 3:07
• @Mattos What do you mean by draw the exact triangles? I am now trying to find the argument of $-3 +i$ and I drew a triangle with sides $1$ and $3$.
– Anna
Feb 13 '15 at 3:11
• there are at least three special angles $\pi/6, \pi/4,$ and $\pi/3$ whose $\sin$ and $\cos$ one is supposed to know. $\cos \pi/6 = \sin \pi/3 = \frac{\sqrt 3}2, \sin \pi/4 = \cos \pi/4 = \sqrt 2/2$ and $\cos \pi/3 = \sin \pi/ 6 = 1/2$
– abel
Feb 13 '15 at 3:13
• @Anna See abels answer above for the exact triangles. Also, to calculate the argument, just take $\tan \theta = \frac{b}{a} = \frac{1}{-3}$, where $a, b$ are just the coefficients in your equation $z = a + ib$.. Feb 13 '15 at 3:16

There's an even easier way. You can verify by direct computation that

$$z^6=\left({1+i\sqrt 3\over 2}\right)^6=1$$

and that no smaller power works. This along with the fact that both real and imaginary parts are positive, i.e. in the first quadrant tells you that $z=e^{i\pi\over 3}$, since this is the only $6^{th}$ root of $1$ in the first quadrant. (all others aside from $1$ have an argument at least $120^\circ={2\pi\over 3}$ which is outside the first quadrant.

Then you know that if $n=6k+r$ with $0\le r\le 5$ that the argument is $r$ times that of $z$, i.e. $\theta={\pi r\over 3}$.

• Thanks, is this applicable to any case? I'm already stuck at computing the argument of $-3 +i$.
– Anna
Feb 13 '15 at 3:10
• Remember, $z = a + ib$, so the principle argument is given by $\tan \theta = \frac{b}{a}$ (draw the graph if it confuses you). Feb 13 '15 at 3:12
• @Mattos Like I said: I already drew a triangle. I don't see what to do with it.
– Anna
Feb 13 '15 at 3:14
• @Anna that one doesn't work as simply since the normalized real and imaginary parts have square roots of $10$ in the denominator, which are not as easy to work with as the more familiar trig functions. That one you can just list as $\arctan-{1\over 3}$, that cannot be easily computed without a calculator. The point is no method can be applicable to any case, because only some of them have $n^{th}$ powers which are real numbers. And if it worked for every case we could always determine the arctan function everywhere, which is known to be impossible. Feb 13 '15 at 3:15
• Adam: Then is it likely that the author of the book assumes the reader will use a calculator?
– Anna
Feb 13 '15 at 3:17

$$(\frac{1+i\sqrt{3}}{2})^n$$ $$(\frac{1}{2} + \frac{i\sqrt{3}}{2})^n$$ $$Let \ \ z=\frac{1}{2} + \frac{i\sqrt{3}}{2}$$ $$r = |z| = \sqrt{ (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$$ $$\sin\theta = \frac{y}{r} \ \ , \ \ \cos\theta = \frac{x}{r}$$ $$\sin\theta = \frac{\sqrt{3}}{2} \ \ , \ \ \cos\theta = \frac{1}{2}$$ Here both $\sin\theta$ and $\cos\theta$ are in 1st quadrant so $$\theta = \frac{\pi}{3}$$ $$\operatorname{Arg} z = \frac{\pi}{3}$$