Square roots of Complex numbers
Let z = a +i b, and let the square root of z be the complex number x +i y.
Then (a +i b) = x +i y
Squaring both sides, a +i b = (x +i y)² = (x² -y²) +i (2 x y).
Equating real and imaginary parts, we get x² -y² = a, 2 x y = b.
Solving these two equations for real x, y yields x + i y.
There are two square roots of a complex number, of the form ± (x +i y).
Illustrative Examples
Example
Find square roots of
- z = -3 +4 i. Verify your result.
- z = i
- z =
Solution
- Let (-3 +4 i) = x +i y.
Squaring both sides, we get -3 +4 i = (x +i y)² = (x² -y²) +2 x y i.
Equating real and imaginary parts, we get x² -y² = -3, 2 x y = 4
=> y = 2; x² - = -3
=> x4 - 4 = -3 x² => x4 +3 x² -4 = 0
=> (x² +4)(x² -1) = 0 => x² = -4 or x² = 1.
Now x being real, x² = -4 is ruled out, so x² = 1 => x = ±1.
When x = 1, y = 2/x = 2;
when x = -1, y = 2/x = -2
So we get (-3 + 4i) = ± (1 +2 i)
Let us verify our result:
[± (1 +2 i)]² = (1 +2 i)² = 1 +4 i² +4 i = 1 -4 +4 i = -3 +4 i
- Let i = x +i y
=> i = (x² -y²) +2 i x y
=> x² -y² = 0; 2xy = 1
=> y = 1; x² -
= 0
=> x² = 1/(4x²) => x4 = 1/4
=> x² = ±1/2
Now x being real, x² = -1/2 is ruled out
x² = 1/2 => x = ± 1/2
When
When x = -1/2, y = 1/(2x) = -1/2
Verification: .
- Given z = , which is a real number.
So square roots of z are ±
Exercise
- Find square roots of
(i) -i
(ii) 1
(iii) -1
(iv) e
(v) -8
- Find square roots of
(i) -1 +22 i
(ii) 5 -12 i
(iii) -8 i
(iv) 3 -4 i
(v) 3 +4 i
- If (a +b i) = ± (x +i y) prove that
(a -b i) = ± (x -i y).
Answers
1. (i) ± (1-i)/2
(ii) ±1
(iii) ±i
(iv) ±e
(v) ±22 i
2. (i) ±(1 +2 i)
(ii) ±(3 -2 i)
(iii) ±2 (1 -i)
(iv) ±(2 -i)
(v) ±(2 +i)
3. [Hint.(a +b i) = ±(x +i y) =>
a +i b = (x² + y²) +2 i x y
=>
a -i b = x² +y² -2 i x y = (x -i y)²]