1
$\begingroup$

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 $5\pi/12$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.