First I will rewrite the expressions in the form of a+bi
3+i7−3i
For a complex number z=a+bi, z=r(cosθ+isinθ), where:
Let's call 3+i z1 and 7−3i z2.
For z1:
z1=r1(cosθ1+isinθ1)
r1=√32+12=√9+1=√10
θ1=tan−1(13)=0.32c
z1=√10(cos(0.32)+isin(0.32))
For z2:
z2=r2(cosθ2+isinθ2)
r2=√72+(−3)2=√58
θ2=tan−1(−37)=−0.40c
However, since 7−3i is in quadrant 4, we need to get a positive angle equivalent (the negative angle goes clockwise around the circle, and we need an anticlockwise angle).
To get a positive angle equivalent, we add 2π, tan−1(−37)+2π=5.88c
z2=√58(cos(5.88)+isin(5.88))
For z1z2:
z1z2=r1r2(cos(θ1−θ2)+isin(θ1−θ2))
z1z2=√10√58(cos[tan−1(13)−(tan−1(−37)+2π)]+isin[tan−1(13)−(tan−1(−37)+2π)])
z1z2=√14529(cos[tan−1(13)−tan−1(−37)−2π]+isin[tan−1(13)−tan−1(−37)−2π])
z1z2=√14529(cos(−5.56)+isin(−5.56))
z1z2=√14529cos(−5.56)+i√14529sin(−5.56)
z1z2=0.311+0.275i
Proof:
3+i7−3i⋅7+3i7+3i=(3+i)(7+3i)(7−3i)(7+3i)=21+7i+9i+3i249+21i−21i−9i2=21+16i+3i249−9i2
i2=−1
=21+16i−349+9=18+16i58=929+829i≈0.310+0.275i
