Solutions to Problems on Complex Numbers

# Thomas' Calculus: Early Transcendentals

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Mat104 Solutions to Problems on Complex Numbers from Old Exams (1) Solve z 5 = 6 i . Let z = r (cos θ + i sin θ ). Then z 5 = r 5 (cos 5 θ + i sin 5 θ ). This has modulus r 5 and argument 5 θ . We want this to match the complex number 6 i which has modulus 6 and infinitely many possible arguments, although all are of the form π/ 2 , π/ 2 ± 2 π, π/ 2 ± 4 π, π/ 2 ± 6 π, π/ 2 ± 8 π, π/ 2 ± 10 π, . . . . (We will see that we don’t really lose anything if we drop the ± in our list of possible arguments for 6 i .) So we choose r 5 = 6 and 5 θ = π 2 or π 2 + 2 π or π 2 + 4 π or π 2 + 6 π or π 2 + 8 π or π 2 + 10 π or . . . In other words, in order to have z 5 = 6 i we should take z of the form r (cos θ + i sin θ ) where r = 5 6 and θ = π 10 or π 10 + 2 π 5 or π 10 + 4 π 5 or π 10 + 6 π 5 or π 10 + 8 π 5 or π 10 + 10 π 5 or . . . Notice that there are only really 5 choices for theta. The sixth choice θ = π/ 10 + 10 π/ 5 = π/ 10+2 π gives the same complex number as the first choice, where we simply take θ = π/ 10. So there are exactly 5 solutions to z 5 = 6 i corresponding to r = 5 6 and θ = π/ 10 , π/ 10 + 2 π/ 5 , π/ 10 + 4 π/ 5 , π/ 10 + 6 π/ 5 and π/ 10 + 8 π/ 5. If we sketch these complex numbers we would see that they all lie on the circle of radius 5 6 1 . 43 and they are separated from each other by an angle of 2 π/ 5. (2) Find the real part of (cos 0 . 7 + i sin 0 . 7) 53 . This is the same as ( e 0 . 7 i ) 53 = e 53 · 0 . 7 i = e 37 . 1 i = cos(37 . 1) + i sin(37 . 1) . So the real part is simply cos(37 . 1). (3) Find all complex numbers z in rectangular form such that ( z - 1) 4 = - 1. Solve w 4 = - 1 first and then z = w + 1. The complex number - 1 has modulus 1 and argument of the form ± π, ± 3 π, ± 5 π, ± 7 π, . . . . If w = r (cos θ + i sin θ ) then w 4 = r 4 (cos 4 θ + i sin 4 θ ). So r 4 = 1 and 4 θ = ± π, ± 3 π, ± 5 π, ± 7 π, ± 9 π, . . .

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