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#
