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

  1. z = -3 +4 i. Verify your result.
  2. z = i
  3. z =

Solution

  1. 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
  2. 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: .
  3. Given z = , which is a real number.
    So square roots of z are ±

Exercise

  1. Find square roots of
    (i) -i
    (ii) 1
    (iii) -1
    (iv) e
    (v) -8
  2. Find square roots of
    (i) -1 +22 i
    (ii) 5 -12 i
    (iii) -8 i
    (iv) 3 -4 i
    (v) 3 +4 i
  3. 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]