How do you add (7+4i) and (2+6i) in trigonometric form?

1 Answer
1s2s2p
Jan 29, 2018

9+10i

Explanation:

If we have a complex number such that z=a+bi, then z=r(cosθ+isinθ), where:

  • r=√a2+b2
  • θ=tan−1(ba)

For z1=7+4i :
r=√72+42=√49+16=√65
θ=tan−1(47)≈29.7
=√65(cos(29.7)+isin(29.7))

For z2=2+6i :
r=√22+62=√4+36=√40
θ=tan−1(62)≈71.6
=√40(cos(71.6)+isin(71.6))

For z1+z2 :
z1+z2=√65cos(29.7)+i√65sin(29.7)+√40cos(71.6)+i√40sin(71.6)

z1+z2=√65cos(29.7)+√40cos(71.6)+i(√65sin(29.7)+√40sin(71.6))

z1+z2=8.999470954+9.995734316i#

z1+z2≈9+10i

Proof:
7+4i+2+6i=(7+2)+i(4+6)=9+10i

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