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 $5\pi/12$.

  • $\begingroup$ 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.) $\endgroup$ – Clive Newstead May 8 at 19:19
  • $\begingroup$ $\arctan(2\sqrt 3)$ is not equal to $\pi/12$. Check it with your calculator. $\endgroup$ – TonyK May 8 at 19:20
  • $\begingroup$ sorry, I meant $5\pi /12$ $\endgroup$ – Mollie Passacantando May 8 at 19:21
  • 3
    $\begingroup$ 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 $\arctan \left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )=\arctan \left ( 2 + \sqrt{3} \right )$. $\endgroup$ – Ian May 8 at 19:22
  • $\begingroup$ $\arctan(2\sqrt 3)$ is not equal to $5\pi/12$ either. $\endgroup$ – 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.


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