Square Roots of Complex Numbers

A2 · P3 · 14 min

Every non-zero complex number has exactly two square roots, and they are negatives of each other, just as $9$ has square roots $3$ and $-3$ . The syllabus asks you to find them "in exact Cartesian form", for example the square roots of $5 + 12i$ , with full details of the working. The method is to write the root as $x + iy$ , square it, and equate real and imaginary parts. With square roots available you can then solve any quadratic equation, even one whose coefficients are complex.

Why there are two roots, and why they are negatives

If $w^2 = z$ then also $(-w)^2 = w^2 = z$ . So square roots always come as a pair $\pm w$ . There are no others: if $u^2 = w^2$ then $(u - w)(u + w) = 0$ , so $u = w$ or $u = -w$ .

Geometrically, the two roots are diametrically opposite points on a circle centred at the origin. In the polar form note you will see that squaring doubles the argument, so a square root halves it, and the two square roots have arguments that differ by $\pi$ .

x^2 + y^2 = 2 (0, 2) (1, 1) (-1, -1) (0, 0) -> (0, 2) (-1, -1) -- (1, 1)

The square roots of $2i$ are $1 + i$ and $-1 - i$ . They lie on the circle of radius $\sqrt{2} = \sqrt{|2i|}$ , opposite each other, at half the argument of $2i$ (and half plus $\pi$ ).

The method

Suppose you want the square roots of $a + ib$ . Let a root be $x + iy$ , with $x$ and $y$ real. Then

(x + iy)^2 = x^2 + 2ixy + i^2y^2 = (x^2 - y^2) + 2xyi.

Setting this equal to $a + ib$ and equating parts gives two real equations.

Equations for a square root

(x + iy)^2 = a + ib \quad\Longleftrightarrow\quad x^2 - y^2 = a \quad\text{and}\quad 2xy = b.

Square roots of a + ib in exact form

Let $(x + iy)^2 = a + ib$ with $x, y$ real. Expand and write $x^2 - y^2 + 2xyi$ .
Equate parts: $x^2 - y^2 = a$ and $2xy = b$ .
From the second, $y = \dfrac{b}{2x}$ . Substitute into the first to get $x^2 - \dfrac{b^2}{4x^2} = a$ .
Multiply by $x^2$ : a quadratic in $x^2$ , namely $x^4 - ax^2 - \dfrac{b^2}{4} = 0$ . Factorise or use the formula.
Reject the negative value of $x^2$ (since $x$ is real, $x^2 \ge 0$ ). Take $x = \pm\sqrt{\ }$ of the positive one.
Find each $y$ from $y = \dfrac{b}{2x}$ , and write both roots.

The quadratic in $x^2$ always has one positive and one negative root (their product is $-\tfrac{b^2}{4} < 0$ when $b \ne 0$ ), so there is never a choice to make in step 5.

The modulus shortcut

There is a third equation that makes the algebra quicker. Taking the modulus of both sides of $(x + iy)^2 = a + ib$ gives $|x + iy|^2 = |a + ib|$ , that is

x^2 + y^2 = \sqrt{a^2 + b^2}.

Combined with $x^2 - y^2 = a$ , adding and subtracting gives $x^2$ and $y^2$ directly. The signs then come from $2xy = b$ : if $b > 0$ , $x$ and $y$ have the same sign; if $b < 0$ , opposite signs.

Tip

The shortcut is a fine method and a good check, but if a question says "showing all your working", the examiner expects to see $x^2 - y^2 = a$ and $2xy = b$ written down. Write those two equations first whichever route you then take.

Special cases

Square root of a negative real number. For $k > 0$ , the square roots of $-k$ are $\pm i\sqrt{k}$ . No working needed: $\sqrt{-16} = \pm 4i$ .
Square root of a purely imaginary number. For $bi$ , the equations are $x^2 - y^2 = 0$ and $2xy = b$ , so $x = \pm y$ . For $2i$ : $x = y$ and $2x^2 = 2$ , giving $\pm(1 + i)$ .

Worked examples

The syllabus example: square roots of 5 + 12i

Find the square roots of $5 + 12i$ , giving your answers in exact Cartesian form.

Solution

Let $(x + iy)^2 = 5 + 12i$ , so $x^2 - y^2 + 2xyi = 5 + 12i$ .

Equating parts:

x^2 - y^2 = 5, \qquad 2xy = 12.

From the second, $y = \dfrac{6}{x}$ . Substitute:

x^2 - \frac{36}{x^2} = 5 \quad\Longrightarrow\quad x^4 - 5x^2 - 36 = 0 \quad\Longrightarrow\quad (x^2 - 9)(x^2 + 4) = 0.

Since $x$ is real, $x^2 = 9$ (reject $x^2 = -4$ ). So $x = 3$ , $y = 2$ or $x = -3$ , $y = -2$ .

The square roots are $3 + 2i$ and $-3 - 2i$ .

Check: $(3 + 2i)^2 = 9 + 12i + 4i^2 = 5 + 12i$ .

Negative imaginary part

Find the square roots of $-8 - 6i$ .

Solution

$(x + iy)^2 = -8 - 6i$ gives

x^2 - y^2 = -8, \qquad 2xy = -6 \;\Rightarrow\; y = -\frac{3}{x}.

x^2 - \frac{9}{x^2} = -8 \quad\Longrightarrow\quad x^4 + 8x^2 - 9 = 0 \quad\Longrightarrow\quad (x^2 - 1)(x^2 + 9) = 0.

So $x^2 = 1$ , $x = \pm 1$ . When $x = 1$ , $y = -3$ ; when $x = -1$ , $y = 3$ .

The square roots are $1 - 3i$ and $-1 + 3i$ . (Because $b < 0$ , $x$ and $y$ have opposite signs.)

Using the modulus shortcut

Find the square roots of $-7 + 24i$ .

Solution

$(x + iy)^2 = -7 + 24i$ gives $x^2 - y^2 = -7$ and $2xy = 24$ .

Modulus: $x^2 + y^2 = |-7 + 24i| = \sqrt{49 + 576} = 25$ .

Adding $x^2 - y^2 = -7$ and $x^2 + y^2 = 25$ : $2x^2 = 18$ , $x = \pm 3$ . Subtracting: $2y^2 = 32$ , $y = \pm 4$ .

Since $2xy = 24 > 0$ , $x$ and $y$ have the same sign. The square roots are $3 + 4i$ and $-3 - 4i$ .

Surds in the answer

Find the square roots of $1 + 2\sqrt{2}\,i$ in exact form.

Solution

$x^2 - y^2 = 1$ and $2xy = 2\sqrt{2}$ , so $y = \dfrac{\sqrt{2}}{x}$ .

x^2 - \frac{2}{x^2} = 1 \quad\Longrightarrow\quad x^4 - x^2 - 2 = 0 \quad\Longrightarrow\quad (x^2 - 2)(x^2 + 1) = 0.

So $x^2 = 2$ , $x = \pm\sqrt{2}$ , and $y = \dfrac{\sqrt{2}}{\pm\sqrt{2}} = \pm 1$ (same sign).

The square roots are $\sqrt{2} + i$ and $-\sqrt{2} - i$ .

Check: $(\sqrt{2} + i)^2 = 2 + 2\sqrt{2}\,i + i^2 = 1 + 2\sqrt{2}\,i$ .

Quadratic equations with complex coefficients

The quadratic formula $z = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is just completing the square, so it works for complex $a$ , $b$ , $c$ too. The only new step is that the discriminant may be a complex number, and you need its square roots.

Solving a quadratic with complex coefficients

Identify $a$ , $b$ , $c$ (they may be complex).
Compute the discriminant $\Delta = b^2 - 4ac$ , showing the expansion.
Find the square roots $\pm w$ of $\Delta$ by the method above.
$z = \dfrac{-b \pm w}{2a}$ . Simplify each root to $x + iy$ (dividing by a complex $2a$ if necessary).
Check: the roots should add to $-\dfrac{b}{a}$ and multiply to $\dfrac{c}{a}$ .

The roots of a quadratic with complex coefficients need not be conjugates. The conjugate-pair rule only applies when the coefficients are real.

Exam style: a quadratic with complex coefficients

(a) Find the square roots of $-8 - 6i$ .

(b) Hence solve the equation $z^2 - (3 + i)z + (4 + 3i) = 0$ , giving your answers in the form $x + iy$ .

Solution

(a) From the earlier example, the square roots are $\pm(1 - 3i)$ .

(b) Here $a = 1$ , $b = -(3 + i)$ , $c = 4 + 3i$ . The discriminant is

(3 + i)^2 - 4(4 + 3i) = (9 + 6i + i^2) - 16 - 12i = 8 + 6i - 16 - 12i = -8 - 6i.

Its square roots are $\pm(1 - 3i)$ , so

z = \frac{(3 + i) \pm (1 - 3i)}{2}.

With $+$ : $z = \dfrac{4 - 2i}{2} = 2 - i$ . With $-$ : $z = \dfrac{2 + 4i}{2} = 1 + 2i$ .

Check: sum $3 + i$ (which is $-b$ ) and product $(2 - i)(1 + 2i) = 2 + 4i - i - 2i^2 = 4 + 3i$ (which is $c$ ).

Exam-hard: a quartic that is a quadratic in z squared

Solve the equation $z^4 + 6z^2 + 25 = 0$ , giving all four roots in the form $x + iy$ .

Solution

Let $u = z^2$ : $u^2 + 6u + 25 = 0$ , so

u = \frac{-6 \pm \sqrt{36 - 100}}{2} = \frac{-6 \pm 8i}{2} = -3 \pm 4i.

Now find $z$ with $z^2 = -3 + 4i$ . Let $(x + iy)^2 = -3 + 4i$ : $x^2 - y^2 = -3$ , $2xy = 4$ , and $x^2 + y^2 = |-3 + 4i| = 5$ . So $x^2 = 1$ , $y^2 = 4$ , with the same sign: $z = \pm(1 + 2i)$ .

For $z^2 = -3 - 4i$ : $x^2 - y^2 = -3$ , $2xy = -4$ , $x^2 + y^2 = 5$ , so $x^2 = 1$ , $y^2 = 4$ with opposite signs: $z = \pm(1 - 2i)$ .

The four roots are $1 + 2i$ , $-1 - 2i$ , $1 - 2i$ , $-1 + 2i$ , which can be written $\pm 1 \pm 2i$ .

Notice that the quartic has real coefficients and its roots form two conjugate pairs, $1 \pm 2i$ and $-1 \pm 2i$ , as they must.

Common mistakes

Splitting the square root

$\sqrt{5 + 12i} \ne \sqrt{5} + \sqrt{12}\,i$ . Square roots do not distribute over addition, even for real numbers: $\sqrt{9 + 16} = 5$ , not $3 + 4$ .

Pairing the signs wrongly

From $x = \pm 3$ and $y = \pm 2$ there are four combinations, but only two are square roots. Use $2xy = b$ to pair them: $y = \dfrac{b}{2x}$ . Writing $3 - 2i$ as a square root of $5 + 12i$ is wrong; it squares to $5 - 12i$ .

Keeping the negative value of x squared

$x^2 = -4$ gives no real $x$ . Students sometimes write $x = \pm 2i$ and produce nonsense. $x$ and $y$ are real by definition, so discard the negative root.

Giving one root

"Find the square roots" means both. Give the pair, for example $\pm(3 + 2i)$ or " $3 + 2i$ and $-3 - 2i$ ".

Assuming conjugate roots

In $z^2 - (3 + i)z + (4 + 3i) = 0$ the roots are $2 - i$ and $1 + 2i$ , which are not conjugates. The coefficients are complex, so the conjugate-pair rule does not apply.

Exam technique

Exam tip

The syllabus says "full details of the working should be shown". The marks are typically: equate parts to get both equations; eliminate to get a quadratic in $x^2$ (or $y^2$ ); solve; state both roots. Show each stage.
Using a calculator's complex mode to get the roots and writing them down without working will not earn the method marks.
Give the roots in exact form: surds, not decimals.
Many questions come in two parts: "(a) find the square roots of $w$ ; (b) hence solve the quadratic". The "hence" tells you the discriminant of (b) is $w$ (or a simple multiple of it). If your discriminant does not match, recheck the expansion.
Always check one root by squaring it. It is quick, and an error here spoils every later part.

Summary

Every non-zero complex number has two square roots, $\pm w$ ; on an Argand diagram they are opposite points on a circle about the origin.
Let $(x + iy)^2 = a + ib$ : then $x^2 - y^2 = a$ and $2xy = b$ .
Eliminate $y$ to get a quadratic in $x^2$ ; reject the negative value of $x^2$ .
Shortcut: $x^2 + y^2 = \sqrt{a^2 + b^2}$ ; the sign of $b$ tells you whether $x$ and $y$ have the same or opposite signs.
Quadratics with complex coefficients: use the formula, finding the square roots of the complex discriminant.
Roots of a quadratic with complex coefficients are not necessarily conjugates.
Quartics like $z^4 + pz^2 + q = 0$ : solve for $z^2$ first, then take square roots.

Practice

Question

Find the square roots of $3 + 4i$ .
Find the square roots of (a) $-16$ ; (b) $2i$ ; (c) $-2i$ .
Find the square roots of $21 - 20i$ .
Find the square roots of $-5 - 12i$ .
Solve $z^2 - 2z + (1 - 4i) = 0$ , giving your answers in exact form.
(a) Find the square roots of $-15 - 8i$ . (b) Hence solve $z^2 - (3 - 2i)z + (5 - i) = 0$ .
Solve $z^4 + 4 = 0$ , giving all four roots in the form $x + iy$ .
Solve $z^2 - (5 + i)z + (8 + i) = 0$ .
(a) Find the square roots of $1 - 4\sqrt{3}\,i$ in exact form. (b) Hence find the four roots of $z^4 - 2z^2 + 49 = 0$ .

Answers

$x^2 - y^2 = 3$ , $2xy = 4$ , so $y = \dfrac{2}{x}$ and $x^4 - 3x^2 - 4 = 0$ , $(x^2 - 4)(x^2 + 1) = 0$ . So $x^2 = 4$ , $x = \pm 2$ , $y = \pm 1$ : roots $\pm(2 + i)$ .
(a) $\pm 4i$ . (b) $x^2 - y^2 = 0$ , $2xy = 2$ : $x = y$ , $x^2 = 1$ : $\pm(1 + i)$ . (c) $x^2 - y^2 = 0$ , $2xy = -2$ : $x = -y$ , $x^2 = 1$ : $\pm(1 - i)$ .
$x^2 - y^2 = 21$ , $2xy = -20$ , $x^2 + y^2 = \sqrt{441 + 400} = 29$ . So $x^2 = 25$ , $y^2 = 4$ , opposite signs: $\pm(5 - 2i)$ .
$x^2 - y^2 = -5$ , $2xy = -12$ , $x^2 + y^2 = 13$ . So $x^2 = 4$ , $y^2 = 9$ , opposite signs: $\pm(2 - 3i)$ .
Discriminant $4 - 4(1 - 4i) = 16i$ . Square roots of $16i$ : $x^2 - y^2 = 0$ , $2xy = 16$ , so $x = y$ and $x^2 = 8$ : $\pm(2\sqrt{2} + 2\sqrt{2}\,i)$ . Then $z = \dfrac{2 \pm (2\sqrt{2} + 2\sqrt{2}\,i)}{2} = 1 \pm (\sqrt{2} + \sqrt{2}\,i)$ , i.e. $z = (1 + \sqrt{2}) + \sqrt{2}\,i$ or $z = (1 - \sqrt{2}) - \sqrt{2}\,i$ .
(a) $x^2 - y^2 = -15$ , $2xy = -8$ , $x^2 + y^2 = 17$ . So $x^2 = 1$ , $y^2 = 16$ , opposite signs: $\pm(1 - 4i)$ . (b) Discriminant $(3 - 2i)^2 - 4(5 - i) = (5 - 12i) - 20 + 4i = -15 - 8i$ . So $z = \dfrac{(3 - 2i) \pm (1 - 4i)}{2}$ : $z = \dfrac{4 - 6i}{2} = 2 - 3i$ or $z = \dfrac{2 + 2i}{2} = 1 + i$ .
$z^2 = \pm 2i$ . Square roots of $2i$ are $\pm(1 + i)$ ; of $-2i$ are $\pm(1 - i)$ . Roots: $1 + i$ , $-1 - i$ , $1 - i$ , $-1 + i$ .
Discriminant $(5 + i)^2 - 4(8 + i) = 24 + 10i - 32 - 4i = -8 + 6i$ . Square roots: $x^2 - y^2 = -8$ , $2xy = 6$ , $x^2 + y^2 = 10$ , so $x^2 = 1$ , $y^2 = 9$ , same sign: $\pm(1 + 3i)$ . Then $z = \dfrac{(5 + i) \pm (1 + 3i)}{2}$ : $z = 3 + 2i$ or $z = 2 - i$ . Check: product $(3 + 2i)(2 - i) = 6 - 3i + 4i - 2i^2 = 8 + i$ .
(a) $x^2 - y^2 = 1$ , $2xy = -4\sqrt{3}$ , so $y = -\dfrac{2\sqrt{3}}{x}$ and $x^2 - \dfrac{12}{x^2} = 1$ , giving $x^4 - x^2 - 12 = 0$ , $(x^2 - 4)(x^2 + 3) = 0$ . So $x = \pm 2$ and $y = \mp\sqrt{3}$ : roots $\pm\left(2 - \sqrt{3}\,i\right)$ . (b) With $u = z^2$ : $u^2 - 2u + 49 = 0$ , $u = \dfrac{2 \pm \sqrt{4 - 196}}{2} = 1 \pm \dfrac{\sqrt{-192}}{2} = 1 \pm 4\sqrt{3}\,i$ . From (a), $z^2 = 1 - 4\sqrt{3}\,i$ gives $z = \pm(2 - \sqrt{3}\,i)$ . Taking conjugates, $z^2 = 1 + 4\sqrt{3}\,i$ gives $z = \pm(2 + \sqrt{3}\,i)$ . The four roots are $\pm 2 \pm \sqrt{3}\,i$ .

Why there are two roots, and why they are negatives

The method

The modulus shortcut

Special cases

Worked examples

Quadratic equations with complex coefficients

Common mistakes

Exam technique

Summary

Practice

Console