FIND THE INDICATED ROOTS OF THE COMPLEX NUMBER IN POLAR FORM
Find all the complex roots. Write roots in polar form with θ in degrees.
Problem 1 :
The complex square roots of 9(cos 30° + i sin 30°)
Solution:
Let z = 9(cos 30° + i sin 30°)
The given number is in polar form.
Problem 2 :
The complex square roots of 25 (cos 210° + i sin 210°)
Solution:
Let z = 25(cos 210° + i sin 210°)
The given number is in polar form.
z=25cos 2k𝜋+7𝜋6+i sin 2k𝜋+7𝜋6Taking square root on both sides,z12=25cos 2k𝜋+7𝜋6+i sin 2k𝜋+7𝜋612Using De Moivre's theorem, bringing the power insidez12=2512cos 12k𝜋+7𝜋6+i sin 12k𝜋+7𝜋6=5cos 𝜋6(12k+7)+i sin 𝜋6(12k+7)Put k=0 and 1 When k=0=5cos 𝜋6(12(0)+7)+i sin 𝜋6(12(0)+7)=5cos 7𝜋6+i sin 7𝜋6---(1)When k=1=5cos 𝜋6(12(1)+7)+i sin 𝜋6(12(1)+7)=5cos 19𝜋6+i sin 19𝜋6---(2)So, the roots are5cos 7𝜋6+i sin 7𝜋6 and 5cos 19𝜋6+i sin 19𝜋6
Problem 3 :
The complex cube roots of 8(cos 210° + i sin 210°)
Solution:
Let z = 8(cos 210° + i sin 210°)
The given number is in polar form.
z=8cos 2k𝜋+7𝜋6+i sin 2k𝜋+7𝜋6Taking cube root on both sides,z13=8cos 2k𝜋+7𝜋6+i sin 2k𝜋+7𝜋613Using De Moivre's theorem, bringing the power insidez13=813cos 12k𝜋+7𝜋6+i sin 12k𝜋+7𝜋6=2cos 𝜋6(12k+7)+i sin 𝜋6(12k+7)Put k=0,1 and 2 When k=0=2cos 𝜋6(12(0)+7)+i sin 𝜋6(12(0)+7)=2cos 7𝜋6+i sin 7𝜋6---(1)When k=1=2cos 𝜋6(12(1)+7)+i sin 𝜋6(12(1)+7)=2cos 19𝜋6+i sin 19𝜋6---(2)When k=2=2cos 𝜋6(12(2)+7)+i sin 𝜋6(12(2)+7)=2cos 31𝜋6+i sin 31𝜋6---(3)So, the roots are2cos 7𝜋6+i sin 7𝜋6,2cos 19𝜋6+i sin 19𝜋6 and 2cos 31𝜋6+i sin 31𝜋6
Problem 4 :
The complex cube roots of 27(cos 306° + i sin 306°)
Solution:
Let z = 27(cos 306° + i sin 306°)
The given number is in polar form.
z=27cos 2k𝜋+17𝜋10+i sin 2k𝜋+17𝜋10Taking cube root on both sides,z13=27cos 2k𝜋+17𝜋10+i sin 2k𝜋+17𝜋1013Using De Moivre's theorem, bringing the power insidez13=2713cos 20k𝜋+17𝜋10+i sin 20k𝜋+17𝜋10=3cos 𝜋10(20k+17)+i sin 𝜋10(20k+17)Put k=0,1 and 2 When k=0=3cos 𝜋10(20(0)+17)+i sin 𝜋10(20(0)+17)=3cos 17𝜋10+i sin 17𝜋10---(1)When k=1=3cos 𝜋10(20(1)+17)+i sin 𝜋10(20(1)+17)=3cos 37𝜋10+i sin 37𝜋10---(2)When k=2=3cos 𝜋10(20(2)+17)+i sin 𝜋10(20(2)+17)=3cos 57𝜋10+i sin 57𝜋10---(3)So, the roots are3cos 17𝜋10+i sin 17𝜋10,3cos 37𝜋10+i sin 37𝜋10 and 3cos 57𝜋10+i sin 57𝜋10
Find all the complex roots. Write roots in rectangular form. If necessary, round to the nearest tenth.
Problem 5 :
The complex fourth roots of 81cos 4𝜋3+i sin 4𝜋3
Solution:
Let z=81cos 4𝜋3+i sin 4𝜋3
The given number is in polar form.
z=81cos 2k𝜋+4𝜋3+i sin 2k𝜋+4𝜋3Taking fourth root on both sides,z14=81cos 2k𝜋+4𝜋3+i sin 2k𝜋+4𝜋314Using De Moivre's theorem, bringing the power insidez14=8114cos 6k𝜋+4𝜋3+i sin 6k𝜋+4𝜋3=3cos 𝜋3(6k+4)+i sin 𝜋3(6k+4)Put k=0,1,2 and 3When k=0=3cos 𝜋3(6(0)+4)+i sin 𝜋3(6(0)+4)=3cos 4𝜋3+i sin 4𝜋3---(1)When k=1=3cos 𝜋3(6(1)+4)+i sin 𝜋3(6(1)+4)=3cos 10𝜋3+i sin 10𝜋3---(2)When k=2=3cos 𝜋3(6(2)+4)+i sin 𝜋3(6(2)+4)=3cos 16𝜋3+i sin 16𝜋3---(3)When k=3=3cos 𝜋3(6(3)+4)+i sin 𝜋3(6(3)+4)=3cos 22𝜋3+i sin 22𝜋3---(4)So, the roots are3cos 4𝜋3+i sin 4𝜋3,3cos 10𝜋3+i sin 10𝜋3,3cos 16𝜋3+i sin 16𝜋3 and 3cos 22𝜋3+i sin 22𝜋3