MathsISC Class 11 · Complex Numbers

Algebra of Complex Numbers: Powers of i, Conjugate, Modulus and Division

Q: Is every real number complex?

Yes: a real a is the complex number a + 0i.

Q: What does the conjugate represent?

The reflection of the point in the real axis of the Argand plane.

Updated 17 October 2026 · Fact-checked

A complex number is z = a + ib with i² = −1. Complex numbers add and multiply like binomials with i² = −1; division is done by multiplying by the conjugate. The conjugate of z is z̄ = a − ib and the modulus is |z| = √(a² + b²), with z z̄ = |z|² and |z₁z₂| = |z₁||z₂|.

Understand Algebra, Conjugate and Modulus of Complex Numbers

Powers of i. $i^{1} = i$ , $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ , and the pattern repeats every four powers: $i^n = i^{n mod 4}$ . So $i^{27} = i^{3} = - i$ and $i^{- 1} = - i$ .

Operations. For $z_{1} = a + ib$ and $z_{2} = c + i d$ : $z_{1} \pm z_{2} = (a \pm c) + i (b \pm d)$ ; $z_{1} z_{2} = (a c - b d) + i (a d + b c)$ . Two complex numbers are equal if their real parts are equal and their imaginary parts are equal.

Conjugate. $\overset{z}{ˉ} = a - ib$ . Properties: $z + \overset{z}{ˉ} = 2 a$ (real), $z - \overset{z}{ˉ} = 2 ib$ (purely imaginary), $z \overset{z}{ˉ} = a^{2} + b^{2}$ , $\overline{z_{1} + z_{2}} = \overset{z_{1}}{ˉ} + \overset{z_{2}}{ˉ}$ , $\overline{z_{1} z_{2}} = \overset{z_{1}}{ˉ} \overset{z_{2}}{ˉ}$ . A number is real if and only if $z = \overset{z}{ˉ}$ .

Modulus. $∣ z ∣ = a^{2} + b^{2} \geq 0$ . Properties: $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ , $∣ z_{1} / z_{2} ∣ = ∣ z_{1} ∣/∣ z_{2} ∣$ , $∣ z ∣ = ∣ \overset{z}{ˉ} ∣$ , and $∣ z_{1} + z_{2} ∣ \leq ∣ z_{1} ∣ + ∣ z_{2} ∣$ .

Division and inverse. $\frac{z _{1}}{z _{2}} = \frac{z _{1} z _{2} ˉ}{∣ z _{2} ∣ ^{2}}$ . The multiplicative inverse of z is $\frac{1}{z} = \frac{z ˉ}{∣ z ∣ ^{2}}$ .

Solving equations. Write the unknown as $x + i y$ , simplify both sides to the form $A + i B$ , and equate real and imaginary parts to get two equations in x and y.

Algebra, Conjugate and Modulus of Complex Numbers: formulas to remember

Product: $$(a + ib)(c + id) = (ac - bd) + i(ad + bc)$$; Use $i^{2} = - 1$ .
Conjugate and modulus: $z \bar{z} = |z|^2 = a^2 + b^2$ | $|z_1 z_2| = |z_1||z_2|$; Conjugate of a product is the product of conjugates.
Quotient: $\frac{z_1}{z_2} = \frac{z_1 \bar{z_2}}{|z_2|^2}$; Rationalise using the conjugate.
Inverse: $\frac{1}{z} = \frac{\bar{z}}{|z|^2}$; z ≠ 0.

Algebra, Conjugate and Modulus of Complex Numbers in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Impedance of a circuit

Suppose in an AC circuit the voltage is $V = 4 + i$ volts (as a complex quantity) and the impedance is $Z = 2 - 3 i$ ohms.

Find the current $I = V / Z$ .

Try it first, then show the solutionSolution

$I = \frac{4 + i}{2 - 3 i} = \frac{( 4 + i ) ( 2 + 3 i )}{( 2 - 3 i ) ( 2 + 3 i )}$ .
Numerator: $8 + 12 i + 2 i + 3 i^{2} = 5 + 14 i$ ; denominator: $4 + 9 = 13$ .
$I = \frac{5}{13} + \frac{14}{13} i \approx 0.385 + 1.077 i$ amperes.

Answer: 5/13 + (14/13)i amperes.

Memory hook: Multiply the top and bottom by the conjugate of the denominator.

Tricks to remember Algebra, Conjugate and Modulus of Complex Numbers

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide by the conjugate
To divide, multiply numerator and denominator by the conjugate of the denominator.
Use it when: Quotients.
i cycles in fours
Reduce the exponent modulo 4.
Use it when: Powers of i.

Algebra, Conjugate and Modulus of Complex Numbers: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Power of i · 1 mark · MCQBoard level

The value of $i^{27}$ is:
(a) $i$
(b) $- i$
(c) $1$
(d) $- 1$

Try it first, then show the solutionSolution

$27 = 4 \times 6 + 3$ , so $i^{27} = i^{3} = - i$ .

Answer: (b) $- i$

Example 2: Product and modulus · 3 marks · Short answerBoard level

Find $(2 + 3 i) (4 - i)$ in the form $a + ib$ and its modulus.

Try it first, then show the solutionSolution

$(2 + 3 i) (4 - i) = 8 - 2 i + 12 i - 3 i^{2}$ . (1 mark)
$= 8 + 3 + 10 i = 11 + 10 i$ . (1 mark)
$∣ z ∣ = 121 + 100 = 221 \approx 14.87$ . (1 mark)

Answer: 11 + 10i; modulus √221.

Example 3: Quotient · 3 marks · Short answerHarder board level

Express $\frac{3 + 2 i}{1 - 2 i}$ in the form $a + ib$ and find its modulus.

Try it first, then show the solutionSolution

Multiply by the conjugate: $\frac{( 3 + 2 i ) ( 1 + 2 i )}{1 + 4}$ . (1 mark)
$= \frac{3 + 6 i + 2 i + 4 i ^{2}}{5} = \frac{- 1 + 8 i}{5}$ . (1 mark)
Modulus $= \frac{1 + 64}{5} = \frac{65}{5} \approx 1.612$ . (1 mark)

Answer: −1/5 + (8/5)i; modulus √65/5.

Example 4: Solving for x and y · 4 marks · Short answerHarder board level

Find real x and y if $(x + i y) (2 - 3 i) = 4 + i$ . Also find the multiplicative inverse of $3 + 4 i$ .

Try it first, then show the solutionSolution

$x + i y = \frac{4 + i}{2 - 3 i} = \frac{( 4 + i ) ( 2 + 3 i )}{13} = \frac{5 + 14 i}{13}$ . (1.5 marks)
Equating parts: $x = \frac{5}{13}$ and $y = \frac{14}{13}$ . (0.5 mark)
$\frac{1}{3 + 4 i} = \frac{3 - 4 i}{25}$ . (1 mark)
Check: $(3 + 4 i) (3 - 4 i) = 25$ . (1 mark)

Answer: x = 5/13, y = 14/13; inverse (3 − 4i)/25.

Example 5: Properties of modulus · 5 marks · Long answerHOTS / competency

Take $z_{1} = 3 + 4 i$ and $z_{2} = 1 - i$ . Verify $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ and $\overline{z_{1} z_{2}} = \overset{z_{1}}{ˉ} \overset{z_{2}}{ˉ}$ . Show that if $∣ z ∣ = 1$ then $z + \frac{1}{z}$ is real, by taking $z = \frac{3 + 4 i}{5}$ .

Try it first, then show the solutionSolution

$z_{1} z_{2} = (3 + 4 i) (1 - i) = 3 - 3 i + 4 i + 4 = 7 + i$ ; $∣ z_{1} z_{2} ∣ = 50 = 52$ . (1.5 marks)
$∣ z_{1} ∣∣ z_{2} ∣ = 5 \times 2$ , equal. (0.5 mark)
$\overline{z_{1} z_{2}} = 7 - i$ and $\overset{z_{1}}{ˉ} \overset{z_{2}}{ˉ} = (3 - 4 i) (1 + i) = 3 + 3 i - 4 i + 4 = 7 - i$ , equal. (1.5 marks)
For $∣ z ∣ = 1$ , $1/ z = \overset{z}{ˉ} = \frac{3 - 4 i}{5}$ ; $z + 1/ z = \frac{6}{5}$ , a real number. (1.5 marks)

Answer: Verified; z + 1/z = 6/5.

Where students lose marks in Algebra, Conjugate and Modulus of Complex Numbers

The slips examiners see most often, and what to write instead.

Mistake: Forgetting i² = −1
Leaving i² in the product.
Do this: Replace i² by −1 immediately.
Usually costs: 2 marks.
Mistake: Not rationalising a quotient
Leaving i in the denominator.
Do this: Multiply by the conjugate.
Usually costs: 1 to 2 marks.
Mistake: Equating real with imaginary parts wrongly
Mixing the two equations.
Do this: Equate real parts to real parts and imaginary to imaginary.
Usually costs: 1 mark.

Exam tips

State the conjugate you use for rationalising.
Check with z z̄ = |z|².
Write the final answer in the form a + ib.
Reduce powers of i modulo 4.

Practice questions on Algebra, Conjugate and Modulus of Complex Numbers

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The conjugate of $3 - 5 i$ is:
(a) $- 3 + 5 i$
(b) $3 + 5 i$
(c) $- 3 - 5 i$
(d) $5 - 3 i$
Hint
Change the sign of the imaginary part.
Answer
(b) $3 + 5 i$
Q2 · 2 marks · Short answerBoard level
Find the modulus of $\frac{1 + i}{1 - i}$ .
Hint
Use $∣ z_{1} / z_{2} ∣$ .
Answer
$∣1 + i ∣ = 2$ and $∣1 - i ∣ = 2$ , so the modulus is 1 (the quotient equals i).
Q3 · 3 marks · Short answerHarder board level
Simplify $i^{- 9} + i^{12} + i^{101}$ .
Hint
Reduce each power modulo 4.
Answer
$i^{- 9} = i^{- 9 + 12} = i^{3} = - i$ ; $i^{12} = 1$ ; $i^{101} = i^{1} = i$ . Sum $= 1$ .
Q4 · 3 marks · Short answerHarder board level
If $z = 2 - i$ , find $z^{2}$ , $z \overset{z}{ˉ}$ and $\frac{1}{z}$ .
Hint
Expand carefully.
Answer
$z^{2} = 4 - 4 i + i^{2} = 3 - 4 i$ ; $z \overset{z}{ˉ} = 4 + 1 = 5$ ; $\frac{1}{z} = \frac{2 + i}{5}$ .
Q5 · 3 marks · Short answerHarder board level
Find real x and y if $\frac{x - 1}{3 + i} + \frac{y - 1}{3 - i} = i$ .
Hint
Take the common denominator 10.
Answer
$(x - 1) (3 - i) + (y - 1) (3 + i) = 10 i$ : real part $3 (x + y - 2) = 0$ so $x + y = 2$ ; imaginary part $- (x - 1) + (y - 1) = 10$ , so $y - x = 10$ . Thus $y = 6$ , $x = - 4$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose in an electric network two impedances $Z_{1} = 3 + 4 i$ and $Z_{2} = 1 - i$ ohms are in series and then used with a voltage $V = 14 + 2 i$ volts.
(i) Find the total impedance. (1 mark)
(ii) Find the magnitude of the total impedance. (1 mark)
(iii) Find the current $I = V / Z$ . (1 mark)
(iv) Find $∣ I ∣$ . (1 mark)
Hint
Series impedances add.
Answer
(i) $Z = 4 + 3 i$ . (ii) $∣ Z ∣ = 5$ . (iii) $I = \frac{( 14 + 2 i ) ( 4 - 3 i )}{25} = \frac{56 - 42 i + 8 i + 6}{25} = \frac{62 - 34 i}{25} = 2.48 - 1.36 i$ . (iv) $∣ I ∣ = \frac{∣ V ∣}{∣ Z ∣} = \frac{200}{5} = 2.828$ A.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.i²?Show answer
−1.
2.z z̄?Show answer
|z|².
3.|z₁z₂|?Show answer
|z₁||z₂|.
4.Inverse of z?Show answer
z̄/|z|².

Algebra, Conjugate and Modulus of Complex Numbers: frequently asked questions

Why can complex numbers not be ordered?

There is no consistent way to say z₁ < z₂ that respects addition and multiplication, so inequalities are only used for moduli (real numbers).

Is every real number complex?

Yes: a real a is the complex number a + 0i.

What does the conjugate represent?

The reflection of the point in the real axis of the Argand plane.