Square Roots of Complex Numbers

Complex numbers can also be square rooted.

Example 1

Find the square root of 3 +4i

(a+ib)² = 3 +4i

a² + 2iab - b² = 3 +4i

By equating real and imaginary parts

a² - b² = 3

2ab = 4

By solving simultaneously

a = 2 or -2

When a = 2, b =1

When a = -2, b = -1

So 2 + i and -2 - i are the square roots of 3 +4i

Example 2

Find the square root of i

(a +ib)² = i

a² +2iab -b² = i

By equating real and imaginary parts

a² -b² = 0 (1)

2ab =1

a = 1/2b (2)

Substitute (2) into (1)

∴ (1/2b)² - b² = 0

1/4b² - b² = 0

1 - 4b⁴ = 0

(1+2b²)(1-2b²) = 0

(1+2b²)(1-√2b)(1+√2b)=0

b = 1/√2 or -1/√2

When b = 1/√2, a = 1/√2

When b = -1/√2, a = -1/√2

Square roots of i are ±(1/√2 + i/√2)

With this, we are able to solve any quadratic equation even if the coefficients are complex.

Example

x² -(1+2i)x - (1 - i) =0

Δ = b² - 4ac

= (1+2i)² -4(-1+2i)

= 1+4i-4 +4 -4i

= 1 (in this case Δ is a real number, there are cases where Δ is complex and the above method would have to be used to find the roots to Δ)

x = (-b ± √Δ)/2a

= [(1+2i) ± 1]/2

= i or 1+i