FINDING ROOTS OF A COMPLEX NUMBER

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 2kπ 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 θ + i sin θ)

Finding the r :

r = √(02 + 322)

r = √1024

r = 32

Finding the α :

α = tan-1(32/0)

α = tan-1(∞)

α = π/2

θ = π/2

z=32cos 𝜋2 + i sin 𝜋2z15=32cos 2k𝜋+𝜋2+i sin 2k𝜋+𝜋215Using De Moivre's theorem, bringing the power insidez15=3215cos 4k𝜋+𝜋2+i sin 4k𝜋+𝜋2=2cos 𝜋2(4k+1)+i sin 𝜋2(4k+1)Put k=0,1,2,3 and 4When k=0=2cos 𝜋2(4(0)+1)+i sin 𝜋2(4(0)+1)=2cos 𝜋2+i sin 𝜋2---(1)When k=1=2cos 𝜋2(4(1)+1)+i sin 𝜋2(4(1)+1)=2cos 5𝜋2+i sin 5𝜋2---(2)When k=2=2cos 𝜋2(4(2)+1)+i sin 𝜋2(4(2)+1)=2cos 9𝜋2+i sin 9𝜋2---(3)When k=3=2cos 𝜋2(4(3)+1)+i sin 𝜋2(4(3)+1)=2cos 13𝜋2+i sin 13𝜋2---(4)When k=4=2cos 𝜋2(4(4)+1)+i sin 𝜋2(4(4)+1)=2cos 17𝜋2+i sin 17𝜋2---(5)

Problem 2 :

x3 - (1 + i√3) = 0

Solution:

x3 - (1 + i√3) = 0

x3 = (1 + i√3)

x=31+i3

z = 1 + i√3

z = r(cos θ + i sin θ)

Finding the r :

r = √((1)2 + √32)

r = √(1 + 3)

r = √4

r = 2

Finding the α :

α = tan-1(√3/1)

α = tan-1(√3)

α = π/3

θ = π/3 

z=2cos 𝜋3 + i sin 𝜋3z13=2cos 2k𝜋+𝜋3+i sin 2k𝜋+𝜋313Using De Moivre's theorem, bringing the power insidez13=213cos 6k𝜋+𝜋3+i sin 6k𝜋+𝜋3=32cos 𝜋3(6k+1)+i sin 𝜋3(6k+1)Put k=0,1 and 2When k=0=32cos 𝜋3(6(0)+1)+i sin 𝜋3(6(0)+1)=32cos 𝜋3+i sin 𝜋3---(1)When k=1=32cos 𝜋3(6(1)+1)+i sin 𝜋3(6(1)+1)=32cos 7𝜋3+i sin 7𝜋3---(2)When k=2=32cos 𝜋3(6(2)+1)+i sin 𝜋3(6(2)+1)=32cos 13𝜋3+i sin 13𝜋3---(3)

Problem 3 :

x3 - (1 - i√3) = 0

Solution:

x3 - (1 - i√3) = 0

x3 = (1 - i√3)

x=31-i3

z = 1 - i√3

z = r(cos θ + i sin θ)

Finding the r :

r = √((1)2 + √32)

r = √(1 + 3)

r = √4

r = 2

Finding the α :

α = tan-1(-√3/1)

α = tan-1(-√3)

α = -π/3

θ = -π/3 

z=2cos -𝜋3 + i sin-𝜋3z13=2cos 2k𝜋-𝜋3+i sin 2k𝜋-𝜋313Using De Moivre's theorem, bringing the power insidez13=213cos 6k𝜋-𝜋3+i sin 6k𝜋-𝜋3=32cos 𝜋3(6k-1)+i sin 𝜋3(6k-1)Put k=0,1 and When k=0=32cos 𝜋3(6(0)-1)+i sin 𝜋3(6(0)-1)=32cos -𝜋3+i sin -𝜋3---(1)When k=1=32cos 𝜋3(6(1)-1)+i sin 𝜋3(6(1)-1)=32cos 5𝜋3+i sin 5𝜋3---(2)When k=2=32cos 𝜋3(6(2)-1)+i sin 𝜋3(6(2)-1)=32cos 11𝜋3+i sin 11𝜋3---(3)

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