9709 P3 §3.9 — Square Roots of Complex Numbers

💡 Lead In: What Number Squared Gives $5 + 12\mathrm{i}$?

The Scenario

For real numbers, finding a square root is straightforward: $\sqrt{9} = 3$ or $-3$, because $3^2 = 9$ and $(-3)^2 = 9$. But what about complex numbers? Imagine a locked box labelled with the complex number $5 + 12\mathrm{i}$. Inside, there are two "hidden" numbers — when you square each one, you get $5 + 12\mathrm{i}$.

The question is: what are these two hidden numbers?

The Core Question

If $z = a + b\mathrm{i}$ is a square root of $5 + 12\mathrm{i}$, then $(a + b\mathrm{i})^2 = 5 + 12\mathrm{i}$. How can we find the real numbers $a$ and $b$ that satisfy this equation? And why should there be two answers, not just one?

💡 Think About It (click to reveal)

Remember that for real numbers, every positive number has two square roots: $\sqrt{9} = \pm 3$. For complex numbers, the same property holds — every non-zero complex number has exactly two square roots, which are negatives of each other. If $w$ is one square root, then $-w$ is the other!

Today's method: set $(a + b\mathrm{i})^2 = z$, expand, equate real and imaginary parts, and solve the resulting simultaneous equations for $a$ and $b$.

📖 Key Concepts: Square Roots of Complex Numbers

Every non-zero complex number $z$ has exactly two square roots, which are negatives of each other. To find them, we set $(a + b\mathrm{i})^2 = z$, expand, and solve for $a$ and $b$.

Method — Finding square roots of $z = x + y\mathrm{i}$:

Let $(a + b\mathrm{i})^2 = x + y\mathrm{i}$, where $a, b$ are real.

Step 1: Expand: $a^2 - b^2 + 2ab\mathrm{i} = x + y\mathrm{i}$

Step 2: Equate real and imaginary parts:

$a^2 - b^2 = x \quad$ (real part)

$2ab = y \quad$ (imaginary part)

Step 3: Use the identity $|z|^2 = a^2 + b^2 = \sqrt{x^2 + y^2}$

Step 4: Solve for $a^2$ and $b^2$ from $a^2 - b^2 = x$ and $a^2 + b^2 = |z|$.

Step 5: Determine the signs of $a$ and $b$ using $2ab = y$.

$$\boxed{\;(a + b\mathrm{i})^2 = z \;\Longrightarrow\; a^2 - b^2 = \operatorname{Re}(z),\; 2ab = \operatorname{Im}(z)\;}$$

Key property — Square roots come in opposite pairs:

If $w$ is a square root of $z$, then $-w$ is also a square root. This is because $(-w)^2 = w^2 = z$. The two square roots are symmetric about the origin on the Argand diagram.

Since $w$ and $-w$ are separated by an angle of $\pi$ (180°), the two square roots always lie on opposite sides of the origin.

Group · Ch13 Activity 1: Root Race — Competing to Find Square Roots

Pairs compete to be the first to correctly find the square roots of assigned complex numbers. Each pair works on a different problem, then checks against the answer key.

Think-First (60 seconds of silence): Pose this question: "If $(a + b\mathrm{i})^2 = 3 + 4\mathrm{i}$, what two equations can you write down for $a$ and $b$?" Wait silently for 60 seconds, then take 2-3 responses. (Answer: $a^2 - b^2 = 3$ and $2ab = 4$.)

Round 1 — Pair Race: Each pair is assigned one of these complex numbers. Find its two square roots. The first pair to finish with the correct answer wins!

(A) $z = 3 + 4\mathrm{i}$ (B) $z = -5 - 12\mathrm{i}$ (C) $z = 21 + 20\mathrm{i}$ (D) $z = -9 + 40\mathrm{i}$

Round 2 — Peer Check: Swap your answer with another pair. Check their working: did they correctly equate $a^2 - b^2$ and $2ab$? Did they use $a^2 + b^2 = |z|$? Were the signs of $a$ and $b$ determined correctly from $2ab = y$?

👁 Answer Key

(A) $3 + 4\mathrm{i}$: $a^2 - b^2 = 3$, $2ab = 4$. $|z| = 5$, so $a^2 + b^2 = 5$. Solving: $a^2 = 4$, $b^2 = 1$. Since $ab = 2 > 0$, $a$ and $b$ have same sign. Roots: $\pm(2 + \mathrm{i})$.

(B) $-5 - 12\mathrm{i}$: $a^2 - b^2 = -5$, $2ab = -12$. $|z| = 13$, so $a^2 + b^2 = 13$. Solving: $a^2 = 4$, $b^2 = 9$. Since $ab = -6 < 0$, $a$ and $b$ have opposite signs. Roots: $\pm(2 - 3\mathrm{i})$.

(C) $21 + 20\mathrm{i}$: $a^2 - b^2 = 21$, $2ab = 20$. $|z| = 29$, so $a^2 + b^2 = 29$. Solving: $a^2 = 25$, $b^2 = 4$. Since $ab = 10 > 0$, $a$ and $b$ have same sign. Roots: $\pm(5 + 2\mathrm{i})$.

(D) $-9 + 40\mathrm{i}$: $a^2 - b^2 = -9$, $2ab = 40$. $|z| = 41$, so $a^2 + b^2 = 41$. Solving: $a^2 = 16$, $b^2 = 25$. Since $ab = 20 > 0$, $a$ and $b$ have same sign. Roots: $\pm(4 + 5\mathrm{i})$.

Individual · Ch16 Activity 2: Find the Square Roots — Independent Practice

Work through these problems independently. Use the five-step method: expand, equate, compute $|z|$, solve for $a^2$ and $b^2$, determine signs.

Problem 1: Find the two square roots of $z = -40 - 42\mathrm{i}$.

Problem 2: Find the two square roots of $z = -7 - 24\mathrm{i}$.

Problem 3: Find the two square roots of $z = -16 + 30\mathrm{i}$.

👁 Check Your Answers

Problem 1 ($-40 - 42\mathrm{i}$): $a^2 - b^2 = -40$, $2ab = -42$. $|z| = 58$, so $a^2 + b^2 = 58$. Solving: $a^2 = 9$, $b^2 = 49$. $ab = -21 < 0$, so $a$ and $b$ have opposite signs. Roots: $\pm(3 - 7\mathrm{i})$.

Problem 2 ($-7 - 24\mathrm{i}$): $a^2 - b^2 = -7$, $2ab = -24$. $|z| = 25$, so $a^2 + b^2 = 25$. Solving: $a^2 = 9$, $b^2 = 16$. $ab = -12 < 0$, so $a$ and $b$ have opposite signs. Roots: $\pm(3 - 4\mathrm{i})$.

Problem 3 ($-16 + 30\mathrm{i}$): $a^2 - b^2 = -16$, $2ab = 30$. $|z| = 34$, so $a^2 + b^2 = 34$. Solving: $a^2 = 9$, $b^2 = 25$. $ab = 15 > 0$, so $a$ and $b$ have same sign. Roots: $\pm(3 + 5\mathrm{i})$.

Group · Guided Worked Example: Square Roots of $5 + 12\mathrm{i}$ Exercise 11B Q10 (pattern)

Find the two square roots of $z = 5 + 12\mathrm{i}$, giving your answers in the form $a + b\mathrm{i}$ where $a$ and $b$ are real.

👁 Reveal Solution

Step 1: Set up the equation

Let $a$ and $b$ be real numbers such that $(a + b\mathrm{i})^2 = 5 + 12\mathrm{i}$.

Expand: $$(a + b\mathrm{i})^2 = a^2 + 2ab\mathrm{i} + b^2\mathrm{i}^2 = (a^2 - b^2) + 2ab\mathrm{i}$$

Step 2: Equate real and imaginary parts

$$a^2 - b^2 = 5 \quad \text{(1)}$$

$$2ab = 12 \quad \text{(2)}$$

From (2): $ab = 6$.

Step 3: Use the modulus to find a second equation

We know $|z| = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$.

Since $|(a + b\mathrm{i})^2| = |a + b\mathrm{i}|^2$, we have $a^2 + b^2 = |z| = 13$.

$$a^2 + b^2 = 13 \quad \text{(3)}$$

Step 4: Solve for $a^2$ and $b^2$

Add (1) and (3): $2a^2 = 18 \;\Longrightarrow\; a^2 = 9 \;\Longrightarrow\; a = \pm 3$

Subtract (1) from (3): $2b^2 = 8 \;\Longrightarrow\; b^2 = 4 \;\Longrightarrow\; b = \pm 2$

Step 5: Determine the signs

From $ab = 6$ (from equation 2), we know $a$ and $b$ must have the same sign (since their product is positive).

So the possible pairs are: $a = 3, b = 2$ or $a = -3, b = -2$.

Final Answer

$$\boxed{\text{Square roots: } 3 + 2\mathrm{i} \text{ and } -3 - 2\mathrm{i}}$$

Verification: $(3 + 2\mathrm{i})^2 = 9 + 12\mathrm{i} + 4\mathrm{i}^2 = 9 + 12\mathrm{i} - 4 = 5 + 12\mathrm{i}$ ✓

$(-3 - 2\mathrm{i})^2 = 9 + 12\mathrm{i} + 4\mathrm{i}^2 = 5 + 12\mathrm{i}$ ✓

Notice that the two roots are negatives of each other: $-(3 + 2\mathrm{i}) = -3 - 2\mathrm{i}$. This is always true for square roots!

✏️ Concept Quiz

Score: 0 / 4

1. To find the square roots of $z = 7 - 24\mathrm{i}$, we set $(a + b\mathrm{i})^2 = 7 - 24\mathrm{i}$. Which pair of equations must $a$ and $b$ satisfy?

$a^2 + b^2 = 7$, $2ab = -24$

$a^2 - b^2 = 7$, $2ab = -24$

$a^2 - b^2 = 7$, $ab = -24$

$a^2 + b^2 = 7$, $ab = -12$

Expanding $(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i}$. Equating real parts: $a^2 - b^2 = 7$. Equating imaginary parts: $2ab = -24$, so $ab = -12$. The equation $a^2 + b^2 = |z|$ is a separate step we use to solve, not from direct equating.

2. For $z = -5 + 12\mathrm{i}$, what is $a^2 + b^2$ equal to?

$5$

$7$

$13$

$17$

$|z| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. Since $|(a + b\mathrm{i})^2| = |a + b\mathrm{i}|^2$, we have $a^2 + b^2 = |z| = 13$. This is a key step: $a^2 + b^2$ always equals the modulus of $z$ (not $|z|^2$).

3. If $a^2 = 9$ and $b^2 = 4$ and $2ab = -12$, what are the correct signs of $a$ and $b$?

$a$ positive, $b$ positive

$a$ negative, $b$ negative

$a$ positive, $b$ negative OR $a$ negative, $b$ positive

$a$ positive, $b$ positive OR $a$ negative, $b$ negative

Since $2ab = -12$, we have $ab = -6$, which is negative. So $a$ and $b$ must have opposite signs. The possibilities are $a = 3, b = -2$ or $a = -3, b = 2$, giving the two square roots $\pm(3 - 2\mathrm{i})$.

4. If $w = 2 + 5\mathrm{i}$ is one square root of $z$, what is the other square root?

$2 - 5\mathrm{i}$

$-2 + 5\mathrm{i}$

$-2 - 5\mathrm{i}$

$-5 + 2\mathrm{i}$

The two square roots of any complex number are negatives of each other: if $w$ is one root, then $-w$ is the other. Since $(-w)^2 = w^2 = z$. Here $w = 2 + 5\mathrm{i}$, so the other root is $-w = -2 - 5\mathrm{i}$. The conjugate $2 - 5\mathrm{i}$ is only relevant if you need to find a real quadratic factor.

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✎ More Practice

Exercise 11B Q2(i) Standard

Find real numbers $a$ and $b$ with $a > 0$ such that $(a + b\mathrm{i})^2 = 21 + 20\mathrm{i}$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = 21 + 20\mathrm{i}$

Equating: $a^2 - b^2 = 21$, $2ab = 20$ so $ab = 10$.

$|21 + 20\mathrm{i}| = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29$

So $a^2 + b^2 = 29$.

Solving with $a^2 - b^2 = 21$: $a^2 = 25$, $b^2 = 4$.

$a = 5$, $b = 2$ (since $a > 0$ and $ab = 10 > 0$).

$$\boxed{a = 5,\; b = 2}$$

Exercise 11B Q2(ii) Standard

Find real numbers $a$ and $b$ with $a > 0$ such that $(a + b\mathrm{i})^2 = -40 - 42\mathrm{i}$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = -40 - 42\mathrm{i}$

Equating: $a^2 - b^2 = -40$, $2ab = -42$ so $ab = -21$.

$|-40 - 42\mathrm{i}| = \sqrt{(-40)^2 + (-42)^2} = \sqrt{1600 + 1764} = \sqrt{3364} = 58$

So $a^2 + b^2 = 58$.

Solving with $a^2 - b^2 = -40$: $a^2 = 9$, $b^2 = 49$.

$a = 3$, $b = -7$ (since $a > 0$ and $ab = -21 < 0$).

$$\boxed{a = 3,\; b = -7}$$

Exercise 11B Q2(iii) Standard

Find real numbers $a$ and $b$ with $a > 0$ such that $(a + b\mathrm{i})^2 = -5 - 12\mathrm{i}$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = -5 - 12\mathrm{i}$

Equating: $a^2 - b^2 = -5$, $2ab = -12$ so $ab = -6$.

$|-5 - 12\mathrm{i}| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

So $a^2 + b^2 = 13$.

Solving with $a^2 - b^2 = -5$: $a^2 = 4$, $b^2 = 9$.

$a = 2$, $b = -3$ (since $a > 0$ and $ab = -6 < 0$).

$$\boxed{a = 2,\; b = -3}$$

Exercise 11B Q2(iv) Standard

Find real numbers $a$ and $b$ with $a > 0$ such that $(a + b\mathrm{i})^2 = -9 + 40\mathrm{i}$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = -9 + 40\mathrm{i}$

Equating: $a^2 - b^2 = -9$, $2ab = 40$ so $ab = 20$.

$|-9 + 40\mathrm{i}| = \sqrt{(-9)^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41$

So $a^2 + b^2 = 41$.

Solving with $a^2 - b^2 = -9$: $a^2 = 16$, $b^2 = 25$.

$a = 4$, $b = 5$ (since $a > 0$ and $ab = 20 > 0$).

$$\boxed{a = 4,\; b = 5}$$

Exercise 11B Q10(ii) Medium

Find the two square roots of $3 + 4\mathrm{i}$.

💡 Hint

Use the method: $(a + b\mathrm{i})^2 = 3 + 4\mathrm{i}$. Compute $|3 + 4\mathrm{i}| = 5$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = 3 + 4\mathrm{i}$

Equating: $a^2 - b^2 = 3$, $2ab = 4$ so $ab = 2$.

$|3 + 4\mathrm{i}| = 5$, so $a^2 + b^2 = 5$.

Solving: $a^2 = 4$, $b^2 = 1$. $a = \pm 2$, $b = \pm 1$.

$ab = 2 > 0$, so $a$ and $b$ have same sign.

$$\boxed{\text{Square roots: } 2 + \mathrm{i} \text{ and } -2 - \mathrm{i}}$$

Exercise 11B Q10(iii) Medium

Find the two square roots of $-16 + 30\mathrm{i}$.

💡 Hint

Compute $|{-16 + 30\mathrm{i}}| = \sqrt{256 + 900} = \sqrt{1156} = 34$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = -16 + 30\mathrm{i}$

Equating: $a^2 - b^2 = -16$, $2ab = 30$ so $ab = 15$.

$|{-16 + 30\mathrm{i}}| = 34$, so $a^2 + b^2 = 34$.

Solving: $a^2 = 9$, $b^2 = 25$. $a = \pm 3$, $b = \pm 5$.

$ab = 15 > 0$, so $a$ and $b$ have same sign.

$$\boxed{\text{Square roots: } 3 + 5\mathrm{i} \text{ and } -3 - 5\mathrm{i}}$$

Exercise 11B Q10(iv) Medium

Find the two square roots of $-7 - 24\mathrm{i}$.

💡 Hint

Compute $|{-7 - 24\mathrm{i}}| = \sqrt{49 + 576} = \sqrt{625} = 25$.

👁 Show Solution

Solution

$(a + b\mathrm{i})^2 = a^2 - b^2 + 2ab\mathrm{i} = -7 - 24\mathrm{i}$

Equating: $a^2 - b^2 = -7$, $2ab = -24$ so $ab = -12$.

$|{-7 - 24\mathrm{i}}| = 25$, so $a^2 + b^2 = 25$.

Solving: $a^2 = 9$, $b^2 = 16$. $a = \pm 3$, $b = \pm 4$.

$ab = -12 < 0$, so $a$ and $b$ have opposite signs.

$$\boxed{\text{Square roots: } 3 - 4\mathrm{i} \text{ and } -3 + 4\mathrm{i}}$$

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Exit Ticket

Check the learning objectives you feel confident about:

Set up $(a + b\mathrm{i})^2 = z$ and equate real and imaginary parts to form equations for $a$ and $b$. Use $a^2 + b^2 = |z|$ together with $a^2 - b^2 = \operatorname{Re}(z)$ to solve for $a^2$ and $b^2$. Determine the correct signs of $a$ and $b$ using $2ab = \operatorname{Im}(z)$. State both square roots of a complex number as $\pm(a + b\mathrm{i})$ and verify by squaring.