Answer
$z=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
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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$z=2i(1+i)=2i+2i^{2}=-2+2i$
lies in Q.II.
$r=|z|=\sqrt{(-2)^{2}+(2)^{2}}=\sqrt{8}=2\sqrt{2}$
$\tan\theta =\displaystyle \frac{2}{-2}=-1$
In Q.I,$\quad \displaystyle \tan(\frac{\pi}{4}) =1,$
In Q.II,$\quad \displaystyle \tan(\frac{3\pi}{4}) =-1,$
So,
$z=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
