HOW TO FIND CUBE ROOT OF A GIVEN NUMBER

Problem 1 :

The complex fifth roots of 32cos 5šœ‹3+i sin 5šœ‹3

Solution:

Let z=32cos 5šœ‹3+i sin 5šœ‹3

The given number is in polar form.

z=32cos 2kšœ‹+5šœ‹3+i sin 2kšœ‹+5šœ‹3Taking fifth root on both sides,z15=32cos 2kšœ‹+5šœ‹3+i sin 2kšœ‹+5šœ‹315Using De Moivre's theorem, bringing the power insidez15=3215cos 6kšœ‹+5šœ‹3+i sin 6kšœ‹+5šœ‹3=2cos šœ‹3(6k+5)+i sin šœ‹3(6k+5)Put k=0,1,2,3 and 4When k=0=2cos šœ‹3(6(0)+5)+i sin šœ‹3(6(0)+5)=2cos 5šœ‹3+i sin 5šœ‹3---(1)When k=1=2cos šœ‹3(6(1)+5)+i sin šœ‹3(6(1)+5)=2cos 11šœ‹3+i sin 11šœ‹3---(2)When k=2=2cos šœ‹3(6(2)+5)+i sin šœ‹3(6(2)+5)=2cos 17šœ‹3+i sin 17šœ‹3---(3)When k=3=2cos šœ‹3(6(3)+5)+i sin šœ‹3(6(3)+5)=2cos 23šœ‹3+i sin 23šœ‹3---(4)When k=4=2cos šœ‹3(6(4)+5)+i sin šœ‹3(6(4)+5)=2cos 29šœ‹3+i sin 29šœ‹3---(5)So, the roots are2cos 5šœ‹3+i sin 5šœ‹3,2cos 11šœ‹3+i sin 11šœ‹3,2cos 17šœ‹3+i sin 17šœ‹3, 2cos 23šœ‹3+i sin 23šœ‹3 and 2cos 29šœ‹3+i sin 29šœ‹3

Problem 2 :

The complex fifth roots of 32.

Solution:

nth roots of z = r cis Īø

z1n=r1n cisšœƒ+2kšœ‹n for k=0,1,2,..n-1z15=3215 cis0Ā°+2kšœ‹5=2 cis0Ā°+2kšœ‹5Put k=0,1,2,3 and 4When k=0=2 (cis 0Ā°)=2---(1)When k=1=2cis 2(1)šœ‹5=2cis 2šœ‹5---(2)When k=2=2cis 2(2)šœ‹5=2cis 4šœ‹5---(3)When k=3=2cis 2(3)šœ‹5=2cis 6šœ‹5---(4)When k=4=2cis 2(4)šœ‹5=2cis 8šœ‹5---(5)So, roots are2,2cis 2šœ‹5,2cis 4šœ‹5,2cis 6šœ‹5 and 2cis 8šœ‹5

Problem 3 :

The complex sixth roots of 64.

Solution:

z1n=r1n cisšœƒ+2kšœ‹n for k=0,1,2,..n-1z16=6416 cis0Ā°+2kšœ‹6=2 cis0Ā°+2kšœ‹6Put k=0,1,2,3,4 and 5When k=0=2 (cis 0Ā°)=2---(1)When k=1=2cis 2(1)šœ‹6=2cis šœ‹3---(2)When k=2=2cis 2(2)šœ‹6=2cis 2šœ‹3---(3)When k=3=2cis 2(3)šœ‹6=2(cis šœ‹)---(4)When k=4=2cis 2(4)šœ‹6=2cis 4šœ‹3---(5)When k=5=2cis 2(5)šœ‹6=2cis 5šœ‹3---(5)So, roots are2,2cis šœ‹3,2cis 2šœ‹3,2(cis šœ‹),2cis 4šœ‹3 and 2cis 5šœ‹3

Problem 4 :

The complex cube roots of 1.

Solution:

z3 = 1

Let z = x + iy

(x + iy)3 = 1

z = r(cis Īø) 

= cis 0 

= cis(0 + 2kĻ€) 

= cis(2kĻ€) 

z=31=113=(cis 2kšœ‹)13z=cis 2kšœ‹3Put k=0,1 and 2When k=0z=cis 0=1When k=1z=cis 2(1)šœ‹3=cis 2šœ‹3=-12+i32When k=2z=cis 2(2)šœ‹3=cis 4šœ‹3=-12-i32So, roots are1,-12+i32 and -12-i32

Problem 5 :

The complex cube roots of i.

Solution:

r=1šœƒ=šœ‹2z=31 cis šœ‹2+2kšœ‹3Put k=0,1,and 2When k=0=31cis šœ‹6=32+12i---(1)When k=1=31cis šœ‹2+2(1)šœ‹3=31cis 5šœ‹6=-32+12i---(2)When k=2=31cis šœ‹2+2(2)šœ‹3=31cis 3šœ‹2=-i---(3)So, roots are 32+12i,-32+12i and -i

Problem 6 :

The complex fourth roots of 1 + i

Solution: 

r=12+12=2tan šœƒ=yx=11šœƒ=šœ‹41+i=2 cis šœ‹4z=214cis šœ‹4+2kšœ‹14=214 cis šœ‹4+2kšœ‹4Put k=0,1,2 and 3When k=0=82cis šœ‹16When k=1=82cis šœ‹4+2(1)šœ‹4=82cis 9šœ‹16=-82 cos7šœ‹16+82i sin7šœ‹16When k=2=82cis šœ‹4+2(2)šœ‹4=82cis 17šœ‹16=-82 cosšœ‹16-82i sinšœ‹16When k=3=82cis šœ‹4+2(3)šœ‹4=82cis 25šœ‹16=82 cos7šœ‹16-82i sin7šœ‹16So, roots are 82cis šœ‹16,-82 cos7šœ‹16+82i sin7šœ‹16,-82 cosšœ‹16-82i sinšœ‹16 and 82 cos7šœ‹16-82i sin7šœ‹16

Problem 7 :

The complex fifth roots of -1 + i

Solution: 

r=(-1)2+12=2tanšœƒ=yx=1-1=-1šœƒ=3šœ‹4z=215 cis 3šœ‹4+2kšœ‹5Put k=0,1,2,3 and 4When k=0=102cis 3šœ‹4+2(0)šœ‹5=102cis 3šœ‹20---(1)When k=1=102cis 3šœ‹4+2(1)šœ‹5=102cis 11šœ‹20---(2)When k=2=102cis 3šœ‹4+2(2)šœ‹5=102cis 19šœ‹20---(3)When k=3=102cis 3šœ‹4+2(3)šœ‹5=102cis 27šœ‹20---(4)When k=4=102cis 3šœ‹4+2(4)šœ‹5=102cis 7šœ‹4---(5)So, roots are 102cis 3šœ‹20,102cis 11šœ‹20,102cis 19šœ‹20,102cis 27šœ‹20 and 102cis 7šœ‹4

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