Question

Express the complex number in trigonometric form.

-3 + 3 square root of threei

Answer 1

\[-3+i3\sqrt3 \]

Answer 2

\[r = a^2 + b^2\]
\[\theta =\tan^{-1}(b/a)\]

Answer 3

\[r^2 = 3^2 + (3\sqrt3)^2 = 36\]

Answer 4

sorry, I made a typo when I was writing the formula for r, it should be:
\[r =\sqrt{a^2 + b^2}\]

Answer 5

Anyway, we have that r = 6

Answer 6

We also get that theta = -60

Answer 7

This, however, is not correct, and highlights a flaw of the arctan function.

Answer 8

If we just plug in the values we got, we get this:
\[6(\cos(-60) + isin(-60)) = 3 - i3\sqrt3\]

Answer 9

Which is actually our answer times -1

Answer 10

In order to rectify this, we add 180 degrees to theta.

Answer 11

In the end, we have:
\[-3 + i3\sqrt3 = 6(\cos(120)+i \sin(120))\]

Answer 12

How did you get that theta=-60?