#a+bi# in trig form is #r(cos\theta+isin\theta#), where:
- #r=sqrt(a^2+b^2)#
- #\theta=abs(arctan(b/a))#
#(5+3i)(3+2i)~=(r_1(cos\theta_1+isin\theta_1))(r_2(cos\theta_2+isin\theta_2))#
#=r_1r_2(cos(\theta_1+\theta_2)+isin(\theta_1+\theta_2))#
#=sqrt(5^2+3^2)sqrt(3^2+2^2)(cos(abs(arctan(3/5))+abs(arctan(2/3)))+isin(abs(arctan(3/5))+abs(arctan(2/3))))#
#~~sqrt(34)sqrt(13)(cos(64.65)+isin(64.65))#
#=sqrt(442)(cos(64.65)+isin(64.65))#
#~~21.02(cos(64.65)+isin(64.65))#
Given #tantheta_1=3/5# and #tantheta_1=2/3#
#tantheta=(3/5+2/3)/(1-3/5xx2/3)=(19/15)/(3/5)=19/9#
and in exact form we can write product as
#sqrt442(cos(arctan(19/9))+isin((arctan(19/9)))#
