Information that will be useful:
#color(red)("~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~")#
#color(white)("XX ")color(blue)(a+bi harr r * [cos(theta)+i * sin(theta)])#
#color(white)("XXX")color(blue)("where " r=sqrt(a^2+b^2))#
#color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})#
#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#color(red)("~~~ Trigonometric Division ~~~")#
#color(white)("XX ")color(blue)((cos(theta)+i *
sin(theta))/(cos(phi)+i * sin(phi)))#
#color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta)cos(phi)-cos(theta)sin(phi))]#
#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
==========================================================
If #A=-i-2#
#color(white)("XXX")r_A = sqrt((-1)^2+(-2)^) = sqrt(5)#
and (since #(-1,-2)# is in Quadrant III)
#color(white)("XXX")theta_A = "arctan"((-2)/(-1))+pi = arctan(2)+pi#
#color(white)("XXXXX")~~4.248741371# (with the aid of a caculator)
If #B=2i-5#
#color(white)("XXX")r_B=sqrt(2^2+(-5)^2)=sqrt(29)#
and
#color(white)("XXX")theta_B="arctan"(-5/2)#
#color(white)("XXXXX")~~-1.19028995# (again with calculator)
So #A/B=(-i-2)/(2i-5)=(sqrt(5) * [cos(4.248...)+i * sin(4.248...)])/(sqrt(29) * [cos(-1.190...)+ i * sin(-1.190...)])#
#color(white)("XXX")~~[cos(4.249) * cos(-1.190) + sin(4.249) * sin(-1.190)] + i[sin(4.248) * cos(1.190) - cos(4.248) * sin(-1.190)]#
(and, once more using a calculator)
#color(white)("XXX")~~0.2758620689 - i (0.3103448276)#
