How to find roots of any complex number?

Table Of Content

Brush Up Basics

Let z = (a + i b) be any complex number. The nth root of complex number z is given by z1/n where n → θ (i.e. set of rational numbers).

Steps to Convert

Step 1

Convert the given complex number, into polar form.

Complex no. in Polar Form

Amplitude and Argument

Step 2

Add 2kπ to the argument of the complex number converted into polar form.

Add 2kπ to the argument

Step 3

Raise index 1/n to the power of z to calculate the nth root of complex number.

Raise index 1/n to the power of z

Step 4

Apply De Moivre’s Theorem

De Moivre’s Theorem

Step 5

Obtain n distinct values. After applying Moivre’s Theorem in step (4) we obtain  which has n distinct values. These values can be obtained by putting k = 0, 1, 2… n – 1 (i.e. one less than the number in the denominator of the given index in lowest form).

Therefore n roots of complex number for different values of k can be obtained as follows:

Roots of Complex No. for different values of K

Example to clear it all

Find the cube root of iota i.e. (i1/3)

Let z = i (iota) and n = 3

Step 1

To convert iota into polar form, z can be expressed as

Z = 0 + 0.i

We need to calculate the value of amplitude r and argument θ. Now,

Value of Amplitude and Argument

Step 2

Add 2kπ to the argument.

All 2kπ to the Argument

Step 3

Take cube root of both sides.

Take Cube Root

Step 4

Apply De Moivre’s Theorem

Apply De Moivre’s Theorem

Step 5

Put k = 0, 1, and 2 to obtain three distinct values.

Put Values of K

Thus, three values of cube root of iota (i) are

Values of iota(i)

It is interesting to note that sum of all roots is zero.

Find the nth root of unity. (1)1/n, Explained here.

Observation to give you insight

(z)1/n has only n distinct values which can be found out by putting k = 0, 1, 2, ….. n-1, n.

Put Values of K here

When we put k = n, the value comes out to be identical with that corresponding to k = 0.

Equations

Which is same value corresponding to k = 0.

When we put k = n + 1, the value comes out to be identical with that corresponding to k = 1. Thus value of each root repeats cyclically when k exceeds n – 1. Hence (z)1/n have only n distinct values.