# Determine polar form of complex number without a calculator

Determine the polar form of $\mathcal z_1 = 2 + \mathcal i \sqrt 3$.

This is how far I have gotten:

$\mathcal r = \sqrt{2^2 + (\sqrt{3^2})}= \sqrt7$

Therefore: $\mathcal cos\theta = \frac{x}{y} =\frac {2}{\sqrt7}$ and $\mathcal sin\theta = \frac {y}{r} = \frac{\sqrt3}{\sqrt7}$.

I don't know how to go any further... I am not allowed to use a calculator for this module so I can't just go about finding $\theta$ by using the $tan\theta = \frac{\sqrt3}{2}$ and then using the inverse function because I would need a calculator for this.

I understand that I need to use the unit circle, but there is no co-ordinate for $(\frac{2}{\sqrt7},\frac{\sqrt3}{\sqrt7})$. Is there some kind of calculation to do which allows me to find the radians without a calculator?

And how do I continue from this?

There is no such an angle except using the inverse trigonometri functions.

Otherwise, you can approximate it since it looks to be quite close to $\frac \pi 4$ and using a truncated Taylor expansion around this value, you would get $$\tan^{-1}(x)=1+2 \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ and then, ignoring the higher order terms, you could solve $$1+2 \left(x-\frac{\pi }{4}\right)=\frac{\sqrt{3}}2\implies x=\frac{\sqrt{3}+\pi-2 }{4}\approx 0.7184$$ while the exact value would be $0.7137$.

You could also use very nice approximations for $\sin(x)$ or $\cos(x)$ (have alook here).

Edit

Sooner or later, you will learn that, better than with Taylor series, functions can be locally approximated using Padé approximants. Using the simplest around $x=a$, we have

$$\tan(x)=\frac{1+(x-a)}{1-(x-a)}$$ Using it for your problem, we just need to solve $$\frac{1+(x-\frac \pi 4)}{1-(x-\frac \pi 4)}=\frac{\sqrt{3}}2\implies x=\frac \pi 4+4 \sqrt{3}-7\approx 0.7136$$ which is much better.

It's just $$2+\sqrt3i=\sqrt7\left(\cos\arccos\frac{2}{\sqrt7}+i\sin\arccos\frac{2}{\sqrt7}\right)$$

• What does arccos mean? @Michael_Rozenberg Sep 8, 2017 at 10:21
• Common notation for inverse cos Sep 8, 2017 at 10:33

You have all the ingredients. $tan\theta=\frac{\sqrt{3}}{2}$ can't be calculated exactly without a calculator. Although you can make a guess since it is $\frac{\sqrt{3}}{2} = 0.87$ roughly you can say $\theta$ should be between $30-45$.

Or use the series of $tan(x)$ in the list here.