You can simplify this as #-2(1+sqrt(3)i)#. This is not needed, but I like to work with easier terms. Next, you remember that #z=r(cos(theta)+isin(theta))#. #r# is called the modulus of #z# and it is defined as #sqrt(x^2+y^2)=|z|#. Yes, that's the same symbol for the absolute value, but in complex numbers, it defines the modulus. (This has to do with metric spaces)
Hence, #r=2#
Now, to find #theta# which is also called the argument of z, you use #tan^-1(sqrt(3)/1)=pi/3#
The proof of this lies in basic trigonometry, but it is more easily seen from Euler's Formula and unit vectors.
Now, not forgetting the #-2# at the beginning, we get:
#-2(1+sqrt(3)i)=-4(cos(pi/3)+isin(pi/3))#
