Respuesta :
[tex]\bf \begin{array}{llll}
3&-&3i\\
\uparrow &&\uparrow \\
a&&b
\end{array}\qquad
\begin{cases}
r=\sqrt{a^2+b^2}\\
\theta=tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\\\\
-----------------------------\\\\
r=\sqrt{3^2+3^2}\implies r=\sqrt{18}\implies r=3\sqrt{2}
\\\\\\
tan(\theta)=\cfrac{3}{3}\implies tan(\theta)=1\implies \measuredangle \theta=tan^{-1}(1)
\\\\\\
\measuredangle \theta=\frac{\pi }{4}\impliedby \textit{reference angle}[/tex]
now, b is -3, and a is 3
or in rectangular if you wish, y = -3 and x = 3.... on what quadrant is "y" negative and "x" positive? well, the 4th quadrant
so, using our reference angle, that'd be [tex]\bf \cfrac{7\pi }{4}\quad or \quad -\cfrac{\pi }{4}[/tex]
so, let's use the first one
[tex]\bf (3,3i)\implies \begin{array}{llll} r[cos(\theta)+i\ sin(\theta)]\\\\ 3\sqrt{2}\left[ cos\left( \frac{7\pi }{4} \right)+i\ sin\left( \frac{7\pi }{4} \right) \right] \end{array}[/tex]
now, b is -3, and a is 3
or in rectangular if you wish, y = -3 and x = 3.... on what quadrant is "y" negative and "x" positive? well, the 4th quadrant
so, using our reference angle, that'd be [tex]\bf \cfrac{7\pi }{4}\quad or \quad -\cfrac{\pi }{4}[/tex]
so, let's use the first one
[tex]\bf (3,3i)\implies \begin{array}{llll} r[cos(\theta)+i\ sin(\theta)]\\\\ 3\sqrt{2}\left[ cos\left( \frac{7\pi }{4} \right)+i\ sin\left( \frac{7\pi }{4} \right) \right] \end{array}[/tex]
