# FIND THE NTH ROOT OF A COMPLEX NUMBER

Finding the nth roots of a complex number.

Let z  =  r(cos θ + i sin θ) and n be a positive integer.

Then z has n distinct nth roots given by,

zk  =  n√r[cos ((θ + 2πk)/n) + i sin ((θ + 2πk)/n)]

(where k  =  0, 1, 2, 3, … , n -1)

Find the nth root of a complex number for the specified value of n.

Example 1 :

1 + i,   n  =  4

Solution :

Given, standard form of z  =  1 + i

The polar form of the complex number z is

1 + i  =  r(cos θ + i sin θ) ---(1)

 Finding r :r  =  √[(1)2 + (1)2]r  = √2 Finding the α :α  =  tan-1(1/1)α  =  π/4

Since the complex number 1 + i is positive, z lies in the second quadrant.

So, the principal value θ  =  π/4

By applying the value of r and θ in equation (1), we get

1 + i  =  √2(cos π/4 + i sin π/4)

So, the polar form is

√2(cos π/4 + i sin π/4)

Then,

Given, n  =  4

Using the nth formula :

zk  =  n√r[cos ((θ + 2πk)/n) + i sin ((θ + 2πk)/n)]

For k  =  0, 1, 2, and 3 we obtain the roots.

Here n  =  4, r  =  √2, and θ  =  π/4

If k  =  0

z0  4√(√2)[cos (π/4 + 2π(0))/4) + i sin (π/4 + 2π(0))/4)]

By m√(n√a)  =  mn√a, we get

z0  8√2[cos π/16 + i sin π/16]

If k  =  1

z1  4√(√2)[cos (π/4 + 2π(1))/4) + i sin (π/4 + 2π(1))/4)]

z1  8√2[cos 9π/16 + i sin 9π/16]

If k  =  2

z2  4√(√2)[cos (π/4 + 2π(2))/4) + i sin (π/4 + 2π(2))/4)]

z2  8√2[cos 17π/16 + i sin 17π/16]

If k  =  3

z3  4√(√2)[cos (π/4 + 2π(3))/4) + i sin (π/4 + 2π(3))/4)]

z3  8√2[cos 25π/16 + i sin 25π/16]

Example 2 :

1 - i,   n  =  6

Solution :

Given, standard form of z  =  1 - i

The polar form of the complex number z is

1 - i  =  r(cos θ + i sin θ) ---(1)

 Finding r :r  =  √[(1)2 + (1)2]r  = √2 Finding the α :α  =  tan-1(1/1)α  =  π/4

Since the complex number 1 - i is positive and negative, z lies in the second quadrant.

So, the principal value θ  =  -π/4

By applying the value of r and θ in equation (1), we get

1 - i  =  √2[cos (-π/4) + i sin (-π/4)]

So, the polar form is

√2[cos (-π/4) + i sin (-π/4)]

Then,

Given, n  =  6

Using the nth formula :

zk  =  n√r[cos ((θ + 2πk)/n) + i sin ((θ + 2πk)/n)]

For k  =  0, 1, 2, 3, 4 and 5 we obtain the roots.

Here n  =  6, r  =  √2, and θ  =  -π/4

If k  =  0

z0  6√(√2)[cos (-π/4 + 2π(0))/6) + i sin (-π/4 + 2π(0))/6)]

By m√(n√a)  =  mn√a, we get

z0  12√2[cos π/24 + i sin π/24]

If k  =  1

z1  6√(√2)[cos (-π/4 + 2π(1))/6) + i sin (-π/4 + 2π(1))/6)]

z1  =  8√2[cos 7π/24 + i sin 7π/24]

If k  =  2

z2  6√(√2)[cos (-π/4 + 2π(2))/6) + i sin (-π/4 + 2π(2))/6)]

z2  12√2[cos 15π/24 + i sin 15π/24]

z2  =  12√2(cos 5π/8 + i sin 5π/8)

If k  =  3

z3  6√(√2)[cos (-π/4 + 2π(3))/6) + i sin (-π/4 + 2π(3))/6)]

z3  12√2[cos 23π/24 + i sin 23π/24]

If k  =  4

z4  =  6√(√2)[cos (-π/4 + 2π(4))/6) + i sin (-π/4 + 2π(4))/6)]

z4  =  12√2[cos 31π/24 + i sin 31π/24]

If k  =  5

z5  =  6√(√2)[cos (-π/4 + 2π(5))/6) + i sin (-π/4 + 2π(5))/6)]

z5  =  12√2[cos 39π/24 + i sin 39π/24]

Example 3 :

2 + 2i,   n  =  3

Solution :

Given, standard form of z  =  2 + 2i

The polar form of the complex number z is

2 + 2i  =  r(cos θ + i sin θ) ---(1)

 Finding r :r  =  √[(2)2 + (2)2]r  = √8 Finding the α :α  =  tan-1(2/2)α  =  π/4

Since the complex number 2 + 2i is positive, z lies in the second quadrant.

So, the principal value θ  =  π/4

By applying the value of r and θ in equation (1), we get

2 + 2i  =  √8(cos π/4 + i sin π/4)

So, the polar form is

√8(cos π/4 + i sin π/4)

Then,

Given, n  =  3

For k  =  0, 1, and 2 we obtain the roots.

Here n  =  3, r  =  √8, and θ  =  π/4

If k  =  0

z0  3√(√8)[cos (π/4 + 2π(0))/3) + i sin (π/4 + 2π(0))/3)]

z0  6√8[cos π/12 + i sin π/12]

If k  =  1

z1  3√(√8)[cos (π/4 + 2π(1))/3) + i sin (π/4 + 2π(1))/3)]

z1  6√8[cos 9π/12 + i sin 9π/12]

z1  6√8[cos 3π/4 + i sin 3π/4]

If k  =  2

z2  3√(√8)[cos (π/4 + 2π(2))/3) + i sin (π/4 + 2π(2))/3)]

z2  6√8[cos 17π/12 + i sin 17π/12]

Example 4 :

-2 + 2i,   n  =  4

Solution :

Given, standard form of z  =  -2 + 2i

The polar form of the complex number z is

-2 + 2i  =  r(cos θ + i sin θ) ---(1)

 Finding r :r  =  √[(2)2 + (2)2]r  = √8 Finding the α :α  =  tan-1(2/2)α  =  π/4

Since the complex number -2 + 2i is negative and positive, z lies in the second quadrant.

So, the principal value θ  =  π - π/4

θ  =  3π/4

By applying the value of r and θ in equation (1), we get

-2 + 2i  =  √8(cos 3π/4 + i sin 3π/4)

So, the polar form is

√8(cos 3π/4 + i sin 3π/4)

Then,

Given, n  =  4

For k  =  0, 1, 2 and 3 we obtain the roots.

Here n  =  4, r  =  √8, and θ  =  3π/4

If k  =  0

z0  4√(√8)[cos (3π/4 + 2π(0))/4) + i sin (3π/4 + 2π(0))/4)]

z0  8√8[cos 3π/16 + i sin 3π/16]

If k  =  1

z1  4√(√8)[cos (3π/4 + 2π(1))/4) + i sin (3π/4 + 2π(1))/4)]

z1  8√8[cos 11π/16 + i sin 11π/16]

If k  =  2

z2  4√(√8)[cos (3π/4 + 2π(2))/4) + i sin (3π/4 + 2π(2))/4)]

z2  8√8[cos 19π/16 + i sin 19π/16]

If k  =  3

z3  =  4√(√8)[cos (3π/4 + 2π(3))/4) + i sin (3π/4 + 2π(3))/4)]

z3  =  8√8[cos 27π/16 + i sin 27π/16] Apart from the stuff given above if you need any other stuff in math, please use our google custom search here.

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