Question

Ok so, I need the trig form of 3 + √3i. I know the whole proccess up until the point where you find theta. I don't understand how to do that. Help?

Answer 1

This is the theta you are looking for, correct?
This is the drawn figure of 3 + sqrt(3) i

Answer 2

In this setting, you are looking at a right triangle with certain side lengths. It is very closely related to finding the tangent of theta equal to sqrt(3)/3:
\( \displaystyle \tan \theta = \frac{\sqrt{3}}{3} \)

Answer 3

If you know the unit circle and its angles well, this is suspiciously close to one of those angles:
\( \tan \theta = \dfrac{\sqrt{3}}{3} \)

\( \tan \theta = \dfrac{1}{\sqrt{3}} \)

\( \tan \theta = \dfrac{1/2}{\sqrt{3}/2} \) <--- opposite 1/2, adjacent sqrt(3)/2

Answer 4

Okay so that means that the theta for this problem, based on the unit circle would be what?

Answer 5

pi/6, or 30 degrees.

Or, if you did not know the unit circle, a calculator would also tell you the same (although perhaps not with pi in radians)

Answer 6

Just have to input tan^-1 (sqrt(3)/ 3), which is usually 2nd Tan

Answer 7

okay, so can this only be used for this problems, or can I use it (tan^-1) for other problems. Like if I wanted to find the theta of 5 + 2i, what would be the way to do that?

Answer 8

tan^(-1) is a more general application. The unit circle is strictly for problems with 'nice enough angles' which are usually easily spotted.
For 5 + 2i, you would have tan theta = 2/5, or tan^(-1) (2/5).

The approach mainly changes when you have zero for the real part, or tan^(-1) (A/ 0)

Answer 9

oh okay. Thanks so much! Like I said I understood everything for the process except for that step and it was very frustrating, so thank you for helping me understand!

Answer 10

Yep! Happy to help! :)