Let's start by solving this in the form given and see what we get:
Multiply the conjugate of the denominator:
#( -3i+2) / (-1-i)*(-1+i)/(-1+i)=(3i-2-3i^2+2i)/(1-i^2)=(5i+1)/2=5/2i+1/2#
#1/2+5/2i#
We can convert this to trigonometric form by finding #r#:
#r=sqrt(a^2+b^2)#
#theta=arctan(b/a)#
#r=sqrt((1/2)^2+(5/2)^2)#
#r= sqrt26/2#
#theta=arctan(5)#
#thetaapprox78.69^@#
So we can say the solution in trig form is
#z=sqrt26/2(cos(78.69^@)+isin(78.69^@))#
Let's double check by solving in trigonometric form:
#z_1= 2-3i#
#r= sqrt((2)^2+(-3)^2)#
#r= sqrt(13)#
#theta= arctan(-3/2)approx-56.31^@approx303.69^@#
#z_2=-1-i#
#r= sqrt((-1)^2+(-1)^2)#
#r=sqrt2#
#theta= arctan(-1/-1)= 45^@#, however this in the third quadrant so#+180^@= 225^@#
#z_1/z_2= (sqrt13(cos(303.69^@)+isin(303.69^@)))/(sqrt2(cos(225^@)+isin(225^@))#
WHEN DIVIDING IN TRIGONOMETRIC FORM:
Divide the r by the other r
And subtract the angles
#sqrt13/sqrt2*sqrt2/sqrt2= sqrt26/2#
#303.69^@-225^@= 78.69^@#
So same solution:
#z= sqrt26/2(cos(78.69^@)+isin(78.69^@))#
