FIND THE CUBE ROOT OF A COMPLEX NUMBER

Example 1 :

2(cos 2π + i sin 2π)

Solution :

Given, z  =  2(cos 2π + i sin 2π)

Using the n^th roots formula :

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

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

Here n  =  3, r  =  2, and θ  =  2π

If k  =  0

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

z₀  =  ³√2[cos (2π/3) + i sin (2π/3)]

If k  =  1

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

z₁  =  ³√2[cos (8π/3) + i sin (8π/3)]

If k  =  2

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

z₂  =  ³√2[cos (10π/3) + i sin (10π/3)]

Example 2 :

2(cos π/4 + i sin π/4)

Solution :

Given, z  =  2(cos π/4 + i sin π/4)

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

Here n  =  3, r  =  2, and θ  =  π/4

If k  =  0

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

z₀  =  ³√2[cos (π/12) + i sin (π/12)]

If k  =  1

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

z₁  =  ³√2[cos (9π/12) + i sin (9π/12)]

If k  =  2

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

z₂  =  ³√2[cos (17π/12) + i sin (17π/12)]

Example 3 :

3(cos 4π/3 + i sin 4π/3)

Solution :

Given, z  =  3(cos 4π/3 + i sin 4π/3)

Here n  =  3, r  =  3, and θ  =  4π/3

If k  =  0

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

z₀  =  ³√3[cos (4π/9) + i sin (4π/9)]

If k  =  1

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

z₁  =  ³√3[cos (10π/9) + i sin (10π/9)]

If k  =  2

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

z₂  =  ³√2[cos (16π/9) + i sin (16π/9)]

Example 4 :

27(cos 11π/6 + i sin 11π/6)

Solution :

Given, z  =  27(cos 11π/6 + i sin 11π/6)

Here n  =  3, r  =  27, and θ  =  11π/6

If k  =  0

z₀  =  ³√27[cos ((11π/6 + 2π(0)/3) + i sin ((11π/6 + 2π(0)/3)]

z₀  =  3[cos (11π/18) + i sin (11π/18)]

If k  =  1

z₁  =  ³√27[cos ((11π/6 + 2π(1))/3) + i sin ((11π/6 + 2π(1))/3)]

z₁  =  3[cos (23π/18) + i sin (23π/18)]

If k  =  2

z₂  =  ³√27[cos ((11π/6 + 2π(2)/3) + i sin ((11π/6 + 2π(2)/3)]

z₂  =  3[cos (35π/18) + i sin (35π/18)]

Example 5 :

-2 + 2i

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)

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)

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

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

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

If k  =  0

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

By ^m√(ⁿ√a)  =  ^mn√a, we get

z₀  =  ⁶√8[cos 3π/12 + i sin 3π/12]

If k  =  1

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

z₁  =  ⁶√8[cos 11π/12 + i sin 11π/12]

If k  =  2

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

z₂  =  ⁶√8[cos 19π/12 + i sin 19π/12]

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