a+bia+bi in trig form is r(cos\theta+isin\thetar(cosθ+isinθ), where:
- r=sqrt(a^2+b^2)r=√a2+b2
- \theta=abs(arctan(b/a))θ=∣∣∣arctan(ba)∣∣∣
(5+3i)(3+2i)~=(r_1(cos\theta_1+isin\theta_1))(r_2(cos\theta_2+isin\theta_2))(5+3i)(3+2i)≅(r1(cosθ1+isinθ1))(r2(cosθ2+isinθ2))
=r_1r_2(cos(\theta_1+\theta_2)+isin(\theta_1+\theta_2))=r1r2(cos(θ1+θ2)+isin(θ1+θ2))
=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))))=√52+32√32+22(cos(∣∣∣arctan(35)∣∣∣+∣∣∣arctan(23)∣∣∣)+isin(∣∣∣arctan(35)∣∣∣+∣∣∣arctan(23)∣∣∣))
~~sqrt(34)sqrt(13)(cos(64.65)+isin(64.65))≈√34√13(cos(64.65)+isin(64.65))
=sqrt(442)(cos(64.65)+isin(64.65))=√442(cos(64.65)+isin(64.65))
~~21.02(cos(64.65)+isin(64.65))≈21.02(cos(64.65)+isin(64.65))
Given tantheta_1=3/5tanθ1=35 and tantheta_1=2/3tanθ1=23
tantheta=(3/5+2/3)/(1-3/5xx2/3)=(19/15)/(3/5)=19/9tanθ=35+231−35×23=191535=199
and in exact form we can write product as
sqrt442(cos(arctan(19/9))+isin((arctan(19/9)))√442(cos(arctan(199))+isin((arctan(199)))
