Answer
$$z= 2\left ( \cos \pi +i \sin \pi \right )$$
Work Step by Step
We multiply the complex numbers as follows:
$$z_1z_2=(1+i)(-1+i)=-1+i-i+i^2=-1-1=-2$$
Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi $.
So, for the complex number $z=-2$ we have$$r=\sqrt{(-2)^2+0^2}=\sqrt{4}=2, \\ \cos \theta =\frac{a}{r}=\frac{-2}{2}=-1.$$ So$$\theta =\pi $$Thus, the complex number has the polar form$$z= 2\left ( \cos \pi +i \sin \pi \right ).$$
