The Square Root of a Complex Number

Introduction

According to linear equation with one and two variables and quadratic equation with one variable,

x^2 + 1 = 0

x^2 = -1

The square of every real number is positive. Thus, we have to use complex numbers to extend the real number system to a larger system. Complex numbers are the sum of a real number and an imaginary number. For example, if Z is a complex number, then

Z = a + ib

,here, Z = complex number

a = real part

i = iota (√-1 ) ( imaginary)

ib = imaginary number

Iota (i) is useful to find the square root of complex numbers.

We have the value of i^2 = -1, an example to show its use: √-9  =  √i29  = +3i

i is the solution of equation, x^2 + 1 = 0  

Square Root of Complex Number 

The formula to find the square root of a complex number a + ib is given by 

(a+ib ) = ±(x + iy),

Here, x and y are real numbers

Two complex numbers z1 = a + ib and z2 = c + id will be equal if a = c and b = d.

The square roots of the complex number and imaginary numbers can be found with the help of the above-given formula.  

The square root of a complex number calculator

Take the square root of complex number a + ib to be x + iy,

 (a+ib ) = x + iy

Taking square on both sides of the equation,

[ (a+ib ) ]^2 = (x + iy)^2

⇒ a + ib = x^2 + (iy)^2 + i^2xy

⇒ a + ib = x^2 – y^2 + i^2xy     [i^2 = -1]

Compare real and imaginary parts of the above equation,

a = x^2 – y^2, b = 2xy    ……(1)

(x^2 + y^2)^2 = (x^2 – y^2)^2 + 4x^2y^2 ……..(2)

(x^2 + y^2)^2 = a^2 + b^2 ……from (1) and (2)

x^2 + y^2 = (a^2 +b^2 )   [ x^2 + y^2 is always positive as sum of squares of non-zero real numbers is always greater than zero]

 We have x^2 + y^2 =  (a^2 +b^2 )   and  a = x^2 – y^2

Solve these two values

x  = ± (a^2 + b^2) + a^2 and y = ± (a^2 + b^2) – a^2

2xy = b, so

x and y have the same sign if b > 0

x and y have opposite signs if b < 0

Square root of a complex number examples

Example 1 – Find square root of 8 – 6i

To determine the square root of 8 – 6i, determine its magnitude and compare it with a + ib.

8 – 6i = (x + iy)^2

8 – 6i = x^2 + (iy)^2 + i^2xy

8 – 6i = x^2 – y^2 + i^2xy [i^2 = -1]

So, x^2 – y^2 = 8 and 2xy =  -6

x^2 + y^2 = (x^2 – y^2)+ (2xy)^2 

x^2 + y^2 =  8+ 2

x^2 + y^2 = 10

Similarly, x^2 – y^2 = 8

On adding and subtracting a and b

x = ± 3 and y = ± 1

When b is negative, xy is negative

Thus, both x and y have different sign

(8 – 6i) = ± (x – iy)

(8 – 6i) = ± (3 – i)

Answer: Hence, (8 – 6i)  = ± (3 – i)

Example 2 – Find the square root of complex number z = 3 + 4i

Solution:

z = 3 +4i

3 +4i = (x + iy)^2

3 + 4i = x^2 + (iy)^2 + i^2xy

3+4i = x^2 – y^2 + i^2xy [i^2 = -1]

So, x^2 – y^2 = 3 and 2xy = 4

x^2 + y^2 = (x^2 – y^2)+ (2xy)^2

x^2 + y^2 = 32 + 42

x^2 + y^2 = 5

Similarly, x^2 – y^2 = 3, xy = 2

x = 2 and y = 1

Answer : √ (3 + 4i) =  ±(2 + i)

Square root of complex number in polar form

The nth root theorem can be used for complex numbers to find the square root of a complex number in polar form

According to the nth root theorem:

For a complex number  z = r (cosθ + i sinθ),

the nth root is given by z1/n = r1/n [cos [(θ + 2kπ)/n] + i sin [(θ + 2kπ)/n]], where k = 0, 1, 2, 3, …, n-1

To obtain the periodic roots of the complex number, add 2kπ to θ.

On using the formula for nth root,

The formula for the square root of a complex number in polar form is

z1/2 = r1/2 [cos [(θ + 2kπ)/2] + i sin [(θ + 2kπ)/2]],

where k = 0, 1

The square root of an imaginary number

Examples of imaginary numbers are 5i, i√7 and -4i. They are present in the form of bi, where b is a non-zero real number.

(±5i)^2 = 5^2i^2

       = 25i^2

We know that i^2 = -1, so

25i^2 = 25 (-1)

       = -25

This means, (±5i)^ 2 = -25

or, square root of -25 is ±5i

Examples of square root of imaginary numbers

Find square root of -18

Solution: √-18 = i√18

     = i√(9×2)

     = i√9 √2

     = i x (±3) x √2

     = ±3i√2

Answer = √-18 = ± 3i√2

Conclusion

The square root of a negative number does not exist in the real number system. The square root of complex numbers helps find numerous roots in the polynomial equation. In a number of the form a + ib, where a and b are real numbers, a is called the natural part and b is the imaginary part of the complex number.