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

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 $$5\pi/12$$.

• LaTeX tip: use $\{\text{braces}\}$ instead of $(\text{parentheses})$ as arguments to commands, i.e. \sqrt{3} instead of \sqrt(3). (I fixed this in your post.) – Clive Newstead May 8 at 19:19
• $\arctan(2\sqrt 3)$ is not equal to $\pi/12$. Check it with your calculator. – TonyK May 8 at 19:20
• sorry, I meant $5\pi /12$ – Mollie Passacantando May 8 at 19:21
• Your end goal is to find the argument of $(1+i)(\sqrt{3}+i)$? Well, that number is $(\sqrt{3}-1)+i(1+\sqrt{3})$, which is in the first quadrant, so its argument is $\arctan \left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )=\arctan \left ( 2 + \sqrt{3} \right )$. – Ian May 8 at 19:22
• $\arctan(2\sqrt 3)$ is not equal to $5\pi/12$ either. – TonyK May 8 at 19:23

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.

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

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.