Answer
$z=6(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Work Step by Step
See p. 604,
A complex number $z=a+bi$ has the polar (or trigonometric) form
$z=r(\cos\theta+i\sin\theta)$
where $r=|z|=\sqrt{a^{2}+b^{2}}$ and $\tan\theta=b/a$.
The number $r$ is the modulus of $z$, and $\theta$ is an argument of $z$.
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$z=3+3i\sqrt{3}$
$r=|z|=\sqrt{(3)^{2}+(3\sqrt{3})^{2}}=\sqrt{9+9\cdot 3}=\sqrt{36}=6$
$\displaystyle \tan\theta=\frac{3\sqrt{3}}{3}=\sqrt{3}$,
In quadrant I, $\displaystyle \tan\frac{\pi}{3}=\sqrt{3}$.
By symmetry (on the unit circle),
$\theta$ can also be $\displaystyle \frac{4\pi}{3}$ (quadrant III).
$3+3i\sqrt{3}$
lies in quadrant I, so we select the appropriate argument:
$z=6(\displaystyle \cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$