Foundations path · F09

Before this lesson: Powers, radicals and logarithms, Equations and quadratic tools

Your goal: Interpret complex roots before the advanced algebra branches.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Complex Numbers and Conjugates: the key idea

The complex number i satisfies i²=−1. Every complex number has the form a+bi with real a,b. Addition combines real and imaginary parts separately; multiplication uses ordinary distribution followed by i²=−1. The conjugate of a+bi is a−bi, and their product is a²+b². This makes division possible when the denominator is nonzero. A real-coefficient polynomial takes conjugate values at conjugate inputs, so its nonreal roots occur in conjugate pairs, with multiplicity.

A worked example

Solve x²−4x+13=0 over the complex numbers.

Complete the square: (x−2)²=−9. Thus x−2=±3i and x=2±3i. The two roots are conjugates, consistent with the real coefficients.

Your turn: change one thing

Simplify 1/(1+i).

Try this on paper before opening the explanation.

Compare your reasoning

Multiply by the conjugate: 1/(1+i)=(1−i)/[(1+i)(1−i)]=(1−i)/2.

Pause and check

A trap to avoid: State the domain, keep exact values and explain why each step is allowed. A correct answer still needs a reason.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Prove that if z is a root of a polynomial with real coefficients, then its conjugate is also a root.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Conjugation preserves sums and products, and leaves each real coefficient unchanged. Consequently P(conjugate(z))=conjugate(P(z)). If P(z)=0, the right side is 0. Hence conjugate(z) is a root. A real root equals its own conjugate.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. What is i²?

Foundation

  1. i
  2. −i
  3. −1
  4. 1
Hint

This is the defining relation.

Answer and reasoning

−1. The complex unit is defined by i²=−1.

2. What is the conjugate of 3−2i?

Foundation

  1. −3−2i
  2. −3+2i
  3. 2−3i
  4. 3+2i
Hint

Reverse the sign of the imaginary part.

Answer and reasoning

3+2i. Conjugation preserves the real part 3 and changes −2i to +2i.

3. Evaluate (2+i)(2−i).

Core

  1. 4
  2. 6
  3. 5
  4. 3
Hint

Multiply conjugates.

Answer and reasoning

5. The product is 2²−i²=4+1=5.

4. What is i⁷?

Core

  1. −1
  2. −i
  3. i
  4. 1
Hint

The powers repeat every four.

Answer and reasoning

−i. i⁷=i⁴i³=1·(−i)=−i.

5. A real-coefficient polynomial has root 1+4i. Which root is forced?

Stretch

  1. −1+4i
  2. 4+i
  3. −1−4i
  4. 1−4i
Hint

Use conjugation of the polynomial equation.

Answer and reasoning

1−4i. Conjugating a zero of a real-coefficient polynomial produces another zero.

6. How many real roots does x²+1=0 have?

Stretch

  1. Infinitely many
  2. None
  3. One
  4. Two
Hint

A real square is nonnegative.

Answer and reasoning

None. For real x, x²+1≥1, so it cannot vanish. Its complex roots are i and −i.

7. What is i²?

Foundation

  1. i
  2. 0
  3. −1
  4. 1
Hint

This defines the imaginary unit.

Answer and reasoning

−1. The number i satisfies i²=−1.

8. Compute (2+i)(2−i).

Core

  1. 5
  2. 3
  3. 4−i
  4. 4+i
Hint

Use difference of squares.

Answer and reasoning

5. 4−i²=4+1=5.

9. Why do nonreal roots of real-coefficient polynomials occur in conjugate pairs?

Stretch

  1. Every coefficient is positive
  2. Every root is imaginary
  3. Conjugation commutes with addition and multiplication
  4. All polynomials are quadratics
Hint

Conjugate the equation P(z)=0.

Answer and reasoning

Conjugation commutes with addition and multiplication. Real coefficients stay fixed, so P(conjugate z)=conjugate P(z)=0.

Choose your next step

Continue to Trigonometry foundations. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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