How do you divide 2i−7−5i−8 in trigonometric form?

1 Answer
1s2s2p
Apr 25, 2018

0.51−0.58i

Explanation:

We have z=−7+2i−8−5i=7−2i8+5i

For z=a+bi, z=r(cosθ+isinθ), where:

  • r=√a2+b2
  • θ=tan−1(ba)

For 7−2i:

r=√72+22=√53
θ=tan−1(−27)≈−0.28c, however 7−2i is in quadrant 4 and so must add 2π to it to make it positive, also 2π would be going around a circle back.

θ=tan−1(−27)+2π≈6c

For 8+5i:
r=√82+52=√89
θ=tan−1(58)≈0.56c

When we have z1z1 in trig form, we do r1r1(cos(θ1−θ2)+isin(θ1−θ2)

z1z2=√53√89(cos(6−0.56)+isin(6−0.56))=√471789(cos(5.44)+isin(5.44))=0.51−0.58i

Proof:
7−2i8+5i⋅8−5i8−5i=56−51i−1064+25=46−51i89=0.52−0.57

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