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Problem 1 :
Solution:
Let z=32cos 5š3+i sin 5š3
The given number is in polar form.
z=32cos 2kš+5š3+i sin 2kš+5š3Taking fifth root on both sides,z15=32cos 2kš+5š3+i sin 2kš+5š315Using De Moivre's theorem, bringing the power insidez15=3215cos 6kš+5š3+i sin 6kš+5š3=2cos š3(6k+5)+i sin š3(6k+5)Put k=0,1,2,3 and 4When k=0=2cos š3(6(0)+5)+i sin š3(6(0)+5)=2cos 5š3+i sin 5š3---(1)When k=1=2cos š3(6(1)+5)+i sin š3(6(1)+5)=2cos 11š3+i sin 11š3---(2)When k=2=2cos š3(6(2)+5)+i sin š3(6(2)+5)=2cos 17š3+i sin 17š3---(3)When k=3=2cos š3(6(3)+5)+i sin š3(6(3)+5)=2cos 23š3+i sin 23š3---(4)When k=4=2cos š3(6(4)+5)+i sin š3(6(4)+5)=2cos 29š3+i sin 29š3---(5)So, the roots are2cos 5š3+i sin 5š3,2cos 11š3+i sin 11š3,2cos 17š3+i sin 17š3, 2cos 23š3+i sin 23š3 and 2cos 29š3+i sin 29š3