How to find the $\arctan(2\sqrt{3})$ by hand?

Question

I'm trying to find the polar form of the complex number $zw$ where $z = 1 + i$. and $w = \sqrt{3} + i$.

I multiplied foiled the complex numbers, grouped the real and imaginary terms together to get a modulus of $\sqrt{8}$ and an angle of $\theta = \arctan(2\sqrt{3})$. I dont know how to find this, i do know that $\arctan(\sqrt{3})$ is $\pi/3$ but i dont know how to incorporate the multiplied 2. The answer is given as $5\pi/12$.

Answer 1

Hint: For the solution of the problem you do not need to evaluate this $\arctan$. Recall that the argument of the product is the sum of arguments of the factors and the latter are very easy to evaluate.

$$\frac\pi4+\frac\pi6=\frac {5\pi}{12}.$$

Answer 2

You should recognize that the argument of $z=1+i$ is $\frac\pi4.$ You should also recognize that the argument of $w = \sqrt3 + i$ is $\arctan(1/\sqrt3),$ which you should realize is $\frac\pi2 - \frac\pi3 = \frac\pi6.$

So $zw$ is the product of a number whose argument is $\frac\pi4$ and a number whose argument is $\frac\pi6.$ The argument of $zw$ is therefore

$$\frac\pi4 + \frac\pi6.$$

I will let you finish from there!

In my opinion, any attempt to evaluate $\arctan(2+\sqrt3)$ in this context by any other method than the above is a waste of time.