How do you divide −3−4i5+2i in trigonometric form?

1 Answer
1s2s2p
Apr 26, 2018

5√29(cos(0.540)+isin(0.540))≈0.79+0.48i

Explanation:

−3−4i5+2i=−3+4i5+2i

z=a+bi can be written as z=r(cosθ+isinθ), where

  • r=√a2+b2
  • θ=tan−1(ba)

For z1=3+4i:
r=√32+42=5
θ=tan−1(43)=≈0,927

For z2=5+2i:
r=√52+22=√29
θ=tan−1(25)=≈0.381

For z1z2:
z1z2=r1r2(cos(θ1−θ2)+isin(θ1−θ2))

z1z2=5√29(cos(0.921−0.381)+isin(0.921−0.381))

z1z2=5√29(cos(0.540)+isin(0.540))=0.79+0.48i

Proof:
−3+4i5+2i⋅5−2i5−2i=−15+20i−6i+825+4=23+14i29=0.79+0.48i

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