FINDING ROOTS OF A COMPLEX NUMBER

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To find the nth root of a complex number, we follow the steps given below.

Step 1 :

Write the given complex number from rectangular form to polar form.

Step 2 :

Add 2kĻ€ to the argument

Step 3 :

Apply De' Moivre's theorem (bring the power to inside)

Step 4 :

Put k = 0, 1, 2, ............ up to n - 1

Solve each equation in the complex number system. Express solutions in polar and rectangular form.

Problem 1 :

x5 - 32i = 0

Solution:

x5 - 32i = 0

x5 = 32i

x=532i

z = 0 + 32i

z = r(cos Īø + i sin Īø)

Finding the r :

r = āˆš(02 + 322)

r = āˆš1024

r = 32

Finding the Ī± :

Ī± = tan-1(32/0)

Ī± = tan-1(āˆž)

Ī± = Ļ€/2

Īø = Ļ€/2

z=32cos šœ‹2 + i sin šœ‹2z15=32cos 2kšœ‹+šœ‹2+i sin 2kšœ‹+šœ‹215Using De Moivre's theorem, bringing the power insidez15=3215cos 4kšœ‹+šœ‹2+i sin 4kšœ‹+šœ‹2=2cos šœ‹2(4k+1)+i sin šœ‹2(4k+1)Put k=0,1,2,3 and 4When k=0=2cos šœ‹2(4(0)+1)+i sin šœ‹2(4(0)+1)=2cos šœ‹2+i sin šœ‹2---(1)When k=1=2cos šœ‹2(4(1)+1)+i sin šœ‹2(4(1)+1)=2cos 5šœ‹2+i sin 5šœ‹2---(2)When k=2=2cos šœ‹2(4(2)+1)+i sin šœ‹2(4(2)+1)=2cos 9šœ‹2+i sin 9šœ‹2---(3)When k=3=2cos šœ‹2(4(3)+1)+i sin šœ‹2(4(3)+1)=2cos 13šœ‹2+i sin 13šœ‹2---(4)When k=4=2cos šœ‹2(4(4)+1)+i sin šœ‹2(4(4)+1)=2cos 17šœ‹2+i sin 17šœ‹2---(5)

Problem 2 :

x3 - (1 + iāˆš3) = 0

Solution:

x3 - (1 + iāˆš3) = 0

x3 = (1 + iāˆš3)

x=31+i3

z = 1 + iāˆš3

z = r(cos Īø + i sin Īø)

Finding the r :

r = āˆš((1)2 + āˆš32)

r = āˆš(1 + 3)

r = āˆš4

r = 2

Finding the Ī± :

Ī± = tan-1(āˆš3/1)

Ī± = tan-1(āˆš3)

Ī± = Ļ€/3

Īø = Ļ€/3 

z=2cos šœ‹3 + i sin šœ‹3z13=2cos 2kšœ‹+šœ‹3+i sin 2kšœ‹+šœ‹313Using De Moivre's theorem, bringing the power insidez13=213cos 6kšœ‹+šœ‹3+i sin 6kšœ‹+šœ‹3=32cos šœ‹3(6k+1)+i sin šœ‹3(6k+1)Put k=0,1 and 2When k=0=32cos šœ‹3(6(0)+1)+i sin šœ‹3(6(0)+1)=32cos šœ‹3+i sin šœ‹3---(1)When k=1=32cos šœ‹3(6(1)+1)+i sin šœ‹3(6(1)+1)=32cos 7šœ‹3+i sin 7šœ‹3---(2)When k=2=32cos šœ‹3(6(2)+1)+i sin šœ‹3(6(2)+1)=32cos 13šœ‹3+i sin 13šœ‹3---(3)

Problem 3 :

x3 - (1 - iāˆš3) = 0

Solution:

x3 - (1 - iāˆš3) = 0

x3 = (1 - iāˆš3)

x=31-i3

z = 1 - iāˆš3

z = r(cos Īø + i sin Īø)

Finding the r :

r = āˆš((1)2 + āˆš32)

r = āˆš(1 + 3)

r = āˆš4

r = 2

Finding the Ī± :

Ī± = tan-1(-āˆš3/1)

Ī± = tan-1(-āˆš3)

Ī± = -Ļ€/3

Īø = -Ļ€/3 

z=2cos -šœ‹3 + i sin-šœ‹3z13=2cos 2kšœ‹-šœ‹3+i sin 2kšœ‹-šœ‹313Using De Moivre's theorem, bringing the power insidez13=213cos 6kšœ‹-šœ‹3+i sin 6kšœ‹-šœ‹3=32cos šœ‹3(6k-1)+i sin šœ‹3(6k-1)Put k=0,1 and When k=0=32cos šœ‹3(6(0)-1)+i sin šœ‹3(6(0)-1)=32cos -šœ‹3+i sin -šœ‹3---(1)When k=1=32cos šœ‹3(6(1)-1)+i sin šœ‹3(6(1)-1)=32cos 5šœ‹3+i sin 5šœ‹3---(2)When k=2=32cos šœ‹3(6(2)-1)+i sin šœ‹3(6(2)-1)=32cos 11šœ‹3+i sin 11šœ‹3---(3)

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