We are trying to get it to the form; #z=r(costheta+isintheta)#, usually called #z=rcistheta#
Where;
The complex number is #z=a+bi#
The modulus (or absolute value) of the complex number #z# is #r=|z|=sqrt(a^2+b^2)#
And #tantheta=b/a#
Now to the question, first we find the modulus of #|z|# or #r#
#z=2-2i#
#|z|=r=sqrt(2^2+(-2^2))=sqrt(4+4)=sqrt8=2sqrt2#
Let's find #theta#
#tantheta=(-2)/2=-1#
We know that #tantheta=(opposite)/(adjacent)=sintheta/costheta#,
when #r=1# for the latter.
If you sketch a right angled triangle and put #-2# opposite #theta# and #2# adjacent to #theta#, the hypotenuse will be #-sqrt2#.
If you are familiar with the unit circle, you will know that the reference angle for the triangle is #45° or pi/4#.
If you are not familiar with the unit circle, using the triangle drawn,
we know this because #cos=(adjacent)/(hypoten use)=sqrt2/2# and
#sin=(opposite)/(hypoten use)=sqrt2/2# which are the values for #45°#
Here though the tangent of #45° or pi/4# is #1# and not #-1#.
How do we figure out what gives us a tangent of #-1#?
Easy! We know that #a=2# and #b=-2#.
If #a#, the abscissa (what we normally refer to as the #x#- coordinate) is positive, then it means we are looking at Quadrant #I# or #IV#.
If #b#, the ordinate (what we normally refer to as the #y#-coordinate) is negative, then it means we are looking at Quadrant #IV#.
Thus, we know that the complex number #z# lies in Quadrant #IV#.
If it lies in #IV#, then the angle that makes tangent #-1# is;
#theta=315° or 7pi/4# because that is the #45° or pi/4# angle in #IV#
Now, we have the modulus and #theta# so we can go ahead and substitute it into our trig form of #z#.
#z=2sqrt2[cos(7pi/4) +isin(7pi/4)]#
I hope this was well explained...
