FIND THE INDICATED ROOTS OF THE COMPLEX NUMBER IN POLAR FORM
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Find all the complex roots. Write roots in polar form with Īø 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 2kš+7š6+i sin 2kš+7š6Taking square root on both sides,z12=25cos 2kš+7š6+i sin 2kš+7š612Using De Moivre's theorem, bringing the power insidez12=2512cos 12kš+7š6+i sin 12kš+7š6=5cos š6(12k+7)+i sin š6(12k+7)Put k=0 and 1 When k=0=5cos š6(12(0)+7)+i sin š6(12(0)+7)=5cos 7š6+i sin 7š6---(1)When k=1=5cos š6(12(1)+7)+i sin š6(12(1)+7)=5cos 19š6+i sin 19š6---(2)So, the roots are5cos 7š6+i sin 7š6 and 5cos 19š6+i sin 19š6
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 2kš+7š6+i sin 2kš+7š6Taking cube root on both sides,z13=8cos 2kš+7š6+i sin 2kš+7š613Using De Moivre's theorem, bringing the power insidez13=813cos 12kš+7š6+i sin 12kš+7š6=2cos š6(12k+7)+i sin š6(12k+7)Put k=0,1 and 2 When k=0=2cos š6(12(0)+7)+i sin š6(12(0)+7)=2cos 7š6+i sin 7š6---(1)When k=1=2cos š6(12(1)+7)+i sin š6(12(1)+7)=2cos 19š6+i sin 19š6---(2)When k=2=2cos š6(12(2)+7)+i sin š6(12(2)+7)=2cos 31š6+i sin 31š6---(3)So, the roots are2cos 7š6+i sin 7š6,2cos 19š6+i sin 19š6 and 2cos 31š6+i sin 31š6
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 2kš+17š10+i sin 2kš+17š10Taking cube root on both sides,z13=27cos 2kš+17š10+i sin 2kš+17š1013Using De Moivre's theorem, bringing the power insidez13=2713cos 20kš+17š10+i sin 20kš+17š10=3cos š10(20k+17)+i sin š10(20k+17)Put k=0,1 and 2 When k=0=3cos š10(20(0)+17)+i sin š10(20(0)+17)=3cos 17š10+i sin 17š10---(1)When k=1=3cos š10(20(1)+17)+i sin š10(20(1)+17)=3cos 37š10+i sin 37š10---(2)When k=2=3cos š10(20(2)+17)+i sin š10(20(2)+17)=3cos 57š10+i sin 57š10---(3)So, the roots are3cos 17š10+i sin 17š10,3cos 37š10+i sin 37š10 and 3cos 57š10+i sin 57š10
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 4š3+i sin 4š3
Solution:
Let z=81cos 4š3+i sin 4š3
The given number is in polar form.
z=81cos 2kš+4š3+i sin 2kš+4š3Taking fourth root on both sides,z14=81cos 2kš+4š3+i sin 2kš+4š314Using De Moivre's theorem, bringing the power insidez14=8114cos 6kš+4š3+i sin 6kš+4š3=3cos š3(6k+4)+i sin š3(6k+4)Put k=0,1,2 and 3When k=0=3cos š3(6(0)+4)+i sin š3(6(0)+4)=3cos 4š3+i sin 4š3---(1)When k=1=3cos š3(6(1)+4)+i sin š3(6(1)+4)=3cos 10š3+i sin 10š3---(2)When k=2=3cos š3(6(2)+4)+i sin š3(6(2)+4)=3cos 16š3+i sin 16š3---(3)When k=3=3cos š3(6(3)+4)+i sin š3(6(3)+4)=3cos 22š3+i sin 22š3---(4)So, the roots are3cos 4š3+i sin 4š3,3cos 10š3+i sin 10š3,3cos 16š3+i sin 16š3 and 3cos 22š3+i sin 22š3
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