Let
#z_1=4+5i#
#z_2=-3+7i#
We need to find
#z=z_1z_2#
#r_1=sqrt(4^2+5^2)=sqrt(16+25)=sqrt41#
#r1=sqrt41#
#theta_1=tan^-1(5/4)#
#z_1=r_1cistheta_1#
#z_1=sqrt41cis(tan^-1(5/4))#
#r_2=sqrt((-3)^2+7^2)=sqrt(9+49)=sqrt58#
#r_2=sqrt58#
#theta_2=tan^-1(7/-3)#
#z_2=r_2cistheta_2#
#z_2=sqrt58cis(tan^-1(7/-3))#
#z=z_1z_2#
#z=(sqrt41cis(tan^-1(5/4)))(sqrt58cis(tan^-1(7/-3)))#
#z=sqrt41sqrt58cis(tan^-1(5/4))cis(tan^-1(7/-3))#
#sqrt41sqrt58=sqrt(41xx58)=sqrt2378#
By De-Moivre's Theorem
#cis(tan^-1(5/4))cis(tan^-1(7/-3))=cis(tan^-1(5/4)+tan^-1(7/-3))#
#tan^-1(5/4)+tan^-1(7/-3)=tan^-1((5/4+7/-3)/(1-5/4xx7/-3))#
#5/4+7/-3=(5xx-3+7xx4)/(4xx-3)=(-15+28)/-12=13/-12#
#z=rcistheta#
#r=sqrt2378#
#theta=tan^-1(13/-12)#
#z=sqrt2378cis(tan^-1(13/-12))#
#r=48.765#
#tan^-1(13/-12)#
adjacent side is negative,
the angle lies in the second quadrant
#tan^-1(13/-12)=pi-tan^-1(13/12)#
#tan^-1(13/12)=0.825#
expressed in radians
#tan^-1(13/12)=pi-0.825#
#theta=2.317#
#cos0.825=-0.679#
since the complex number is in 2nd quadrant
#sin0.825=0.735#
#r=48.765#
#x=rcostheta=48.765xx(-0.679)#
#x=-33.111#
#y=rsintheta=48.765xx0.735#
#y=35.842#
#z=x+iy#
#z=(-33.111)+i(35.842)#
Thus,
#(4+5i)(-3+7i)=48.765cis2.317#
#(4+5i)(-3+7i)=-33.111+35.842i#
