For a complex number a+bi we can repesent this in trigonometric form, as z=r(costheta+isintheta)
r=sqrt(a^2+b^2)
theta=arctan(b/a)
So, for z_1=-8-9i, we can find that r_1=sqrt((-8)^2+(-9)^2)=sqrt(145)
theta_1=arctan((-9)/(-8))
However, -8-9i is in quadrant 3, but theta is in quadrant 1, so we need to add pi
theta_1=arctan((-9)/(-8))+pi~~3.9857
z_1~~sqrt(145)(cos(3.9857)+isin(3.9857))
For z_2=3-6i, r_2=sqrt(3^2+(-6)^2)=sqrt(45)=3sqrt(5)
theta_2=arctan(-3/6)=arctan(-2)
3-6i is in quadrant 4, while theta is in quadrant 4, however, we need theta to be positive, while being in the range thetain[0,2pi). So, we must add 2pi, even though this is going all the way around the circle (in terms of the 4 quadrants), we will effectively arrive at the same place.
theta_2=arctan(-3/6)=arctan(-2)+2pi~~5.1760
z_2~~3sqrt(5)(cos(5.1760)+isin(5.1760))
We now have z_1 and z_2 in trig form.
z_1+z_2=r_1costheta_1+ir_1sintheta_1+r_2costheta_2+ir_2sintheta_2
color(white)(z_1+z_2)=(r_1costheta_1+r_2costheta_2)+i(r_1sintheta_1+r_2sintheta_2)
color(white)(z_1+z_2)~~(sqrt(145)cos3.9857+3sqrt(5)cos5.1760)+i(sqrt(145)sin3.9857+3sqrt(5)sin5.1760)
color(white)(z_1+z_2)~~-5.000639287 +i(-14.99973664)
color(white)(z_1+z_2)~~-5.000639287-14.99973664i
color(white)(z_1+z_2)~~-5-15i
If we use theta_1=arctan(9/8)+pi and theta_2=arctan(-2)+2pi we get:
z_1+z_2=(sqrt(145)cos(arctan(9/8)+pi)+3sqrt(5)cos(arctan(-2)+2pi))+i(sqrt(145)sin(arctan(9/8)+pi)+3sqrt(5)sin(arctan(-2)+2pi))
color(white)(z_1+z_2)=-5+i(-15)
color(white)(z_1+z_2)=-5-15i
