Question

Express the complex number in trigonometric form.
-3 + 3 square root of three i

(i already found the answer i just need someone to verify if it is right)

Answer 1

\(\bf -3+3\sqrt{3}\ i\quad ?\)

Answer 2

\(\bf \begin{array}{ccllll}
-3&+3\sqrt{3}\ i\\
\uparrow &\uparrow \\
{\color{brown}{ a}}&{\color{brown}{ b}}\\
\end{array}\qquad
\begin{array}{llll}
r=\sqrt{{\color{brown}{ a}}^2+{\color{brown}{ b}}^2}\\
r=\sqrt{(-3)^2+(3\sqrt{3})^2}\\
r=\sqrt{9+3^2\cdot 3}\\
r=\sqrt{36}\to 6
\end{array}
\\ \quad \\
\theta=tan^{-1}\left(\cfrac{\cancel{ 3 }\sqrt{3}}{-\cancel{ 3 }}\right)\implies \theta=tan^{-1}(-\sqrt{3})
\\ \quad \\
thus\implies r[cos(\theta)+i\ sin(\theta)]\)

Answer 3

6(cos(2pi/3) + i sin(2pi/3))

Answer 4

is that right?^

Answer 5

hmmm well for \(\bf tan^{-1}(-\sqrt{3})\) I get -60 degrees or \(\bf -\cfrac{\pi}{3}\to \cfrac{5\pi}{3}\)