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.

z=9cos 2kšœ‹+šœ‹6+i sin 2kšœ‹+šœ‹6Taking square root on both sides,z12=9cos 2kšœ‹+šœ‹6+i sin 2kšœ‹+šœ‹612Using De Moivre's theorem, bringing the power insidez12=912cos 12kšœ‹+šœ‹6+i sin 12kšœ‹+šœ‹6=3cos šœ‹6(12k+1)+i sin šœ‹6(12k+1)Put k=0 and 1 When k=0=3cos šœ‹6(12(0)+1)+i sin šœ‹6(12(0)+1)=3cos šœ‹6+i sin šœ‹6---(1)When k=1=3cos šœ‹6(12(1)+1)+i sin šœ‹6(12(1)+1)=3cos 13šœ‹6+i sin 13šœ‹6---(2)So, the roots arecos šœ‹6+i sin šœ‹6 and 3cos 13šœ‹6+i sin 13šœ‹6

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œ‹+7šœ‹6+i sin 2kšœ‹+7šœ‹6Taking square root on both sides,z12=25cos 2kšœ‹+7šœ‹6+i sin 2kšœ‹+7šœ‹612Using De Moivre's theorem, bringing the power insidez12=2512cos 12kšœ‹+7šœ‹6+i sin 12kšœ‹+7šœ‹6=5cos šœ‹6(12k+7)+i sin šœ‹6(12k+7)Put k=0 and 1 When k=0=5cos šœ‹6(12(0)+7)+i sin šœ‹6(12(0)+7)=5cos 7šœ‹6+i sin 7šœ‹6---(1)When k=1=5cos šœ‹6(12(1)+7)+i sin šœ‹6(12(1)+7)=5cos 19šœ‹6+i sin 19šœ‹6---(2)So, the roots are5cos 7šœ‹6+i sin 7šœ‹6 and 5cos 19šœ‹6+i sin 19šœ‹6

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œ‹+7šœ‹6+i sin 2kšœ‹+7šœ‹6Taking cube root on both sides,z13=8cos 2kšœ‹+7šœ‹6+i sin 2kšœ‹+7šœ‹613Using De Moivre's theorem, bringing the power insidez13=813cos 12kšœ‹+7šœ‹6+i sin 12kšœ‹+7šœ‹6=2cos šœ‹6(12k+7)+i sin šœ‹6(12k+7)Put k=0,1 and 2 When k=0=2cos šœ‹6(12(0)+7)+i sin šœ‹6(12(0)+7)=2cos 7šœ‹6+i sin 7šœ‹6---(1)When k=1=2cos šœ‹6(12(1)+7)+i sin šœ‹6(12(1)+7)=2cos 19šœ‹6+i sin 19šœ‹6---(2)When k=2=2cos šœ‹6(12(2)+7)+i sin šœ‹6(12(2)+7)=2cos 31šœ‹6+i sin 31šœ‹6---(3)So, the roots are2cos 7šœ‹6+i sin 7šœ‹6,2cos 19šœ‹6+i sin 19šœ‹6 and 2cos 31šœ‹6+i sin 31šœ‹6

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œ‹+17šœ‹10+i sin 2kšœ‹+17šœ‹10Taking cube root on both sides,z13=27cos 2kšœ‹+17šœ‹10+i sin 2kšœ‹+17šœ‹1013Using De Moivre's theorem, bringing the power insidez13=2713cos 20kšœ‹+17šœ‹10+i sin 20kšœ‹+17šœ‹10=3cos šœ‹10(20k+17)+i sin šœ‹10(20k+17)Put k=0,1 and 2 When k=0=3cos šœ‹10(20(0)+17)+i sin šœ‹10(20(0)+17)=3cos 17šœ‹10+i sin 17šœ‹10---(1)When k=1=3cos šœ‹10(20(1)+17)+i sin šœ‹10(20(1)+17)=3cos 37šœ‹10+i sin 37šœ‹10---(2)When k=2=3cos šœ‹10(20(2)+17)+i sin šœ‹10(20(2)+17)=3cos 57šœ‹10+i sin 57šœ‹10---(3)So, the roots are3cos 17šœ‹10+i sin 17šœ‹10,3cos 37šœ‹10+i sin 37šœ‹10 and 3cos 57šœ‹10+i sin 57šœ‹10

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 4šœ‹3+i sin 4šœ‹3

Solution:

Let z=81cos 4šœ‹3+i sin 4šœ‹3

The given number is in polar form.

z=81cos 2kšœ‹+4šœ‹3+i sin 2kšœ‹+4šœ‹3Taking fourth root on both sides,z14=81cos 2kšœ‹+4šœ‹3+i sin 2kšœ‹+4šœ‹314Using De Moivre's theorem, bringing the power insidez14=8114cos 6kšœ‹+4šœ‹3+i sin 6kšœ‹+4šœ‹3=3cos šœ‹3(6k+4)+i sin šœ‹3(6k+4)Put k=0,1,2 and 3When k=0=3cos šœ‹3(6(0)+4)+i sin šœ‹3(6(0)+4)=3cos 4šœ‹3+i sin 4šœ‹3---(1)When k=1=3cos šœ‹3(6(1)+4)+i sin šœ‹3(6(1)+4)=3cos 10šœ‹3+i sin 10šœ‹3---(2)When k=2=3cos šœ‹3(6(2)+4)+i sin šœ‹3(6(2)+4)=3cos 16šœ‹3+i sin 16šœ‹3---(3)When k=3=3cos šœ‹3(6(3)+4)+i sin šœ‹3(6(3)+4)=3cos 22šœ‹3+i sin 22šœ‹3---(4)So, the roots are3cos 4šœ‹3+i sin 4šœ‹3,3cos 10šœ‹3+i sin 10šœ‹3,3cos 16šœ‹3+i sin 16šœ‹3 and 3cos 22šœ‹3+i sin 22šœ‹3

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