MathsCBSE Class 11 · Complex Numbers and Quadratic Equations

Powers of i and Algebra of Complex Numbers (Class 11)

Q: Why do powers of i repeat every four?

Because i⁴ = 1. After four multiplications you are back at 1, so the cycle i, −1, −i, 1 repeats.

Updated 17 October 2026 · Fact-checked

The powers of i repeat every four steps: i, −1, −i, 1. To find i^n, divide n by 4 and use the remainder (negative n: add a multiple of 4 first). Two complex numbers are equal only when their real parts and imaginary parts are equal, so an equation in x + iy splits into two real equations. Divide by multiplying top and bottom by the conjugate of the denominator.

Powers of i repeat every four steps, so divide n by 4 and use the remainder; any four consecutive powers add to 0.

Illustration drawn for QuestionBuddy.

Understand Powers of i and Algebra of Complex Numbers

A complex number $z = a + ib$ with $i^{2} = - 1$ obeys ordinary algebra, with $i^{2}$ replaced by $- 1$ whenever it appears.

Cycle of powers. $i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1$ , then it repeats. For any integer $n$ , $i^{n} = i^{r}$ where $r$ is the remainder of $n$ on division by $4$ taken from ${0, 1, 2, 3}$ . For a negative power add a multiple of 4: $i^{- 57} = i^{- 57 + 60} = i^{3} = - i$ . Any four consecutive powers add to $0$ , which is why long sums like $i + i^{2} + \dots + i^{101}$ collapse to the few terms left over.

Equality. $a + ib = c + i d$ exactly when $a = c$ and $b = d$ . This is the main tool: every complex equation gives two real equations, one from the real parts and one from the imaginary parts.

Operations. Add and subtract part by part; multiply like binomials: $(a + ib) (c + i d) = (a c - b d) + i (a d + b c)$ . Useful squares: $(1 + i)^{2} = 2 i$ , $(1 - i)^{2} = - 2 i$ , so $\frac{1 + i}{1 - i} = i$ .

Division and inverse. Multiply the numerator and denominator by the conjugate of the denominator:
$ $\frac{a + ib}{c + i d} = \frac{( a + ib ) ( c - i d )}{c ^{2} + d ^{2}} .$ $
The multiplicative inverse of $z \neq = 0$ is $z^{- 1} = \frac{z ˉ}{∣ z ∣ ^{2}} = \frac{a - ib}{a ^{2} + b ^{2}}$ .

Equations with an unknown complex number. If a question gives $(x + i y)^{2} = p + i q$ , expand and equate: $x^{2} - y^{2} = p$ and $2 x y = q$ . A useful extra relation comes from $(x^{2} + y^{2})^{2} = (x^{2} - y^{2})^{2} + (2 x y)^{2}$ . For an equation such as $z^{2} = \overset{z}{ˉ}$ , write $z = x + i y$ , equate real and imaginary parts, and handle the cases $y = 0$ and $y \neq = 0$ separately.

Order is not defined. There is no meaning to $z_{1} < z_{2}$ for complex numbers; inequalities are only used for real quantities like $∣ z ∣$ or $Re z$ . Also, $a b = ab$ fails when both $a, b$ are negative: $- 2 - 3 = - 6$ , not $6$ .

Powers of i and Algebra of Complex Numbers: formulas to remember

Cycle of powers of i: $i^{4} k = 1, i^{(} 4 k + 1) = i, i^{(} 4 k + 2) = - 1, i^{(} 4 k + 3) = - i$; Reduce the exponent modulo 4.
Equality: $a + ib = c + i d i f an d o n l y i f a = c an d b = d$; Gives two real equations.
Product: $(a + ib) (c + i d) = (a c - b d) + i (a d + b c)$; Replace i squared by −1.
Quotient: $(a + ib) / (c + i d) = (a + ib) (c - i d) / (c^{2} + d^{2})$; Multiply by the conjugate of the denominator.
Inverse: $1/ z = (a - ib) / (a^{2} + b^{2})$; Valid for z not equal to 0.
Square of 1 + i: $(1 + i)^{2} = 2 ian d (1 - i)^{2} = - 2 i$; So (1 + i)/(1 − i) = i.

How to solve Powers of i and Algebra of Complex Numbers questions

Use these steps for powers of i, equations in x and y, and divisions.

1For $i^{n}$ , reduce $n$ to a remainder $0$ to $3$ (add multiples of $4$ if $n$ is negative) and read the value from the cycle.
2For a sum of consecutive powers, drop complete blocks of four and add what is left.
3For an equation, move everything to the form (real part) $+ i$ (imaginary part) on each side.
4Equate real parts and imaginary parts to get two real equations; solve them.
5For a quotient, multiply numerator and denominator by the conjugate of the denominator and simplify using $i^{2} = - 1$ .
6Check by substituting your answer back into the original equation.

Powers of i and Algebra 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.

Rotation by 90 degrees is multiplication by i

Suppose a security camera is mounted on a motor that turns it 90 degrees anticlockwise every second. Treat its direction as a complex number: east is $1$ , and each turn multiplies by $i$ .

The camera starts facing east. In which direction does it face after 203 seconds?

Try it first, then show the solutionSolution

After $n$ seconds the direction is $i^{n}$ , so we need $i^{203}$ .
$203 = 4 \times 50 + 3$ , so $i^{203} = i^{3} = - i$ .
The number $- i$ lies on the negative imaginary axis, which is the south direction.

Answer: South.

Memory hook: Four quarter-turns bring you back: the powers of i repeat every four.

Division gives current in an AC circuit

Suppose a circuit has voltage $V = (8 + 6 i)$ volts and impedance $Z = (2 + i)$ ohms, treated as complex numbers. Current is $I = V / Z$ .

Find $I$ in the form $a + ib$ and its size $∣ I ∣$ .

Try it first, then show the solutionSolution

$I = \frac{( 8 + 6 i ) ( 2 - i )}{( 2 + i ) ( 2 - i )} = \frac{16 - 8 i + 12 i - 6 i ^{2}}{5} = \frac{22 + 4 i}{5}$ .
So $I = 4.4 + 0.8 i$ amperes.
$∣ I ∣ = 4. 4^{2} + 0. 8^{2} = 19.36 + 0.64 = 20 = 25 \approx 4.47$ .

Answer: $I = 4.4 + 0.8 i$ ; $∣ I ∣ = 25 \approx 4.47$ A.

Memory hook: Dividing complex numbers means rationalising with the conjugate.

Tricks to remember Powers of i and Algebra of Complex Numbers

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

Remainder rule
Divide the power by 4: remainder 0, 1, 2, 3 gives 1, i, −1, −i.
Use it when: Any power of i.
Block of four
i^k + i^(k+1) + i^(k+2) + i^(k+3) = 0 for every k.
Use it when: Long sums of powers of i.
Two equations from one
One complex equation means two real equations: real parts match and imaginary parts match.
Use it when: Finding real x and y.
The i-fraction
(1 + i)/(1 − i) = i and (1 − i)/(1 + i) = −i.
Use it when: Powers of such fractions are powers of i.

Powers of i and Algebra of Complex Numbers: solved examples

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

Example 1: Negative power · 2 marks · Very short answerBoard level

Find the value of $i^{- 57}$ .

Try it first, then show the solutionSolution

$i^{- 57} = i^{- 57 + 60} = i^{3}$ , since adding a multiple of 4 does not change the value. (1 mark)
$i^{3} = - i$ . (1 mark)

Answer: $- i$

Example 2: Real x and y · 3 marks · Short answerBoard level

Find real numbers $x$ and $y$ such that $(x + i y) (2 - 3 i) = 4 + i$ .

Try it first, then show the solutionSolution

$x + i y = \frac{4 + i}{2 - 3 i} = \frac{( 4 + i ) ( 2 + 3 i )}{13}$ . (1 mark)
Numerator: $8 + 12 i + 2 i + 3 i^{2} = 5 + 14 i$ . (1 mark)
So $x = \frac{5}{13}, y = \frac{14}{13}$ . Check: $(\frac{5}{13} + \frac{14}{13} i) (2 - 3 i) = \frac{10 + 42 + 13 i}{13} = 4 + i$ . (1 mark)

Answer: $x = \frac{5}{13}, y = \frac{14}{13}$

Example 3: Size after simplifying · 3 marks · Short answerHarder board level

If $a + ib = \frac{( 1 + i ) ^{3}}{2 - i}$ , find $a$ , $b$ and $a^{2} + b^{2}$ .

Try it first, then show the solutionSolution

$(1 + i)^{3} = (1 + i) (2 i) = - 2 + 2 i$ . (1 mark)
$\frac{- 2 + 2 i}{2 - i} = \frac{( - 2 + 2 i ) ( 2 + i )}{5} = \frac{- 4 - 2 i + 4 i + 2 i ^{2}}{5} = \frac{- 6 + 2 i}{5}$ . (1 mark)
$a = - \frac{6}{5}, b = \frac{2}{5}$ , so $a^{2} + b^{2} = \frac{36 + 4}{25} = \frac{8}{5}$ . (Check: $∣1 + i ∣^{3} /∣2 - i ∣$ squared $= 8/5$ .) (1 mark)

Answer: $a = - \frac{6}{5}, b = \frac{2}{5}, a^{2} + b^{2} = \frac{8}{5}$

Example 4: A complex number whose square is given · 3 marks · Short answerHarder board level

Find real $x, y$ if $(x + i y)^{2} = 5 + 12 i$ .

Try it first, then show the solutionSolution

Expand: $x^{2} - y^{2} + 2 i x y = 5 + 12 i$ , so $x^{2} - y^{2} = 5$ and $2 x y = 12$ . (1 mark)
$(x^{2} + y^{2})^{2} = (x^{2} - y^{2})^{2} + (2 x y)^{2} = 25 + 144 = 169$ , so $x^{2} + y^{2} = 13$ (it is positive). (1 mark)
Adding and subtracting: $x^{2} = 9$ , $y^{2} = 4$ . Since $x y = 6 > 0$ , $x$ and $y$ have the same sign. (1 mark)

Answer: $x + i y = 3 + 2 i$ or $- 3 - 2 i$

Example 5: Case study: a circuit · 4 marks · Case-basedHOTS / competency

Suppose a lab circuit has a supply of $V = 10$ volts, and two parts with impedances $Z_{1} = (1 + 2 i)$ ohm and $Z_{2} = (3 - 2 i)$ ohm are in series, so $Z = Z_{1} + Z_{2}$ , and current $I = V / Z$ .
(i) Find $Z$ . (1 mark)
(ii) Find $I$ as $a + ib$ . (1 mark)
(iii) Find $∣ I ∣$ . (1 mark)
(iv) Find $1/ Z$ and verify that $Z \cdot \frac{1}{Z} = 1$ . (1 mark)

Try it first, then show the solutionSolution

(i) $Z = (1 + 2 i) + (3 - 2 i) = 4$ . (1 mark)
(ii) $I = 10/4 = 2.5$ , that is $2.5 + 0 i$ . (1 mark)
(iii) $∣ I ∣ = 2.5$ A. (1 mark)
(iv) $1/ Z = \frac{1}{4}$ and $4 \times \frac{1}{4} = 1$ . The imaginary parts cancelled, so the circuit behaves like a pure resistor. (1 mark)

Answer: (i) 4 ohm (ii) 2.5 A (iii) 2.5 A (iv) 1/4

Where students lose marks in Powers of i and Algebra of Complex Numbers

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

Mistake: Using the remainder wrongly for a negative power
Students only flip the sign of the power instead of reducing it.
Wrong working
$i^{- 58} = - i^{58} = - i^{2} = 1$ .
Right working
Add a multiple of 4: $i^{- 58} = i^{- 58 + 60} = i^{2} = - 1$ .
Do this: Always add the nearest multiple of 4 that makes the power non-negative.
Usually costs: The answer mark.
Mistake: Not equating real and imaginary parts separately
The two sides are compared as whole numbers.
Wrong working
$x + i y = 5 + 14 i$ hence " $x = 5 + 14 i$ ".
Right working
$x = 5, y = 14$ (real and imaginary parts match).
Do this: Write "Re: ..., Im: ..." for the two equations.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the conjugate while dividing
Students multiply only the numerator.
Wrong working
$\frac{4 + i}{2 - 3 i} = \frac{( 4 + i ) ( 2 + 3 i )}{2 - 3 i}$ .
Right working
Multiply both numerator and denominator: denominator becomes $(2 - 3 i) (2 + 3 i) = 13$ .
Do this: Write the fraction $\frac{d ˉ}{d ˉ}$ explicitly.
Usually costs: The whole division.
Mistake: Taking √a·√b = √(ab) for negative a and b
The rule only works when at least one is non-negative.
Wrong working
$- 2 - 3 = 6$ .
Right working
$- 2 - 3 = i 2 \cdot i 3 = - 6$ .
Do this: Pull out $i$ first, then multiply.
Usually costs: 1 mark.

Exam tips

Show the reduction of the power (for example " $203 = 4 \times 50 + 3$ ") so the method mark is visible.
For "find real $x$ , $y$ " questions, always write the two real equations before solving.
Rationalise with the conjugate and write the denominator as a real number before splitting into real and imaginary parts.
Check your final complex number by substituting it back, especially for $(x + i y)^{2} = \dots$ questions where the signs of $x$ and $y$ are linked.
Remember that complex numbers have no order; do not write $z_{1} < z_{2}$ .

Practice questions on Powers of i and Algebra 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 value of $i^{49} + i^{68} + i^{89} + i^{110}$ is:
(a) $2 i$
(b) 0
(c) $2 + 2 i$
(d) $1 + i$
Hint
Reduce each exponent modulo 4.
Answer
(a). $i^{49} = i$ , $i^{68} = 1$ , $i^{89} = i$ , $i^{110} = i^{2} = - 1$ ; sum $= 2 i$ .
Q2 · 1 mark · MCQHarder board level
The least positive integer $n$ for which $(\frac{1 + i}{2})^{n} = 1$ is:
(a) 2
(b) 4
(c) 8
(d) 16
Hint
Square the base first.
Answer
(c). $(\frac{1 + i}{2})^{2} = \frac{2 i}{2} = i$ , so the base to the power $n$ equals $1$ only when $n = 8$ (powers $1, \dots, 7$ give $\frac{1 + i}{2}, i, \dots$ , none equal to 1).
Q3 · 3 marks · Short answerHarder board level
Find the multiplicative inverse of $(2 - 3 i)^{2}$ .
Hint
Square first, then use $\overset{z}{ˉ} /∣ z ∣^{2}$ .
Answer
$(2 - 3 i)^{2} = 4 - 12 i + 9 i^{2} = - 5 - 12 i$ . $∣ z ∣^{2} = 25 + 144 = 169$ . Inverse $= \frac{- 5 + 12 i}{169}$ .
Q4 · 3 marks · Short answerHarder board level
Find the least positive integer $n$ for which $(1 + i)^{2 n} = (1 - i)^{2 n}$ .
Hint
$(1 + i)^{2} = 2 i$ and $(1 - i)^{2} = - 2 i$ .
Answer
$(2 i)^{n} = (- 2 i)^{n}$ , so $i^{n} = (- 1)^{n} i^{n}$ , giving $(- 1)^{n} = 1$ . Hence $n$ is even; the least positive value is $n = 2$ .
Q5 · 4 marks · Short answerHOTS / competency
Find all complex numbers $z$ such that $z^{2} = - \overset{z}{ˉ}$ .
Hint
Write $z = x + i y$ and treat $y = 0$ and $y \neq = 0$ separately.
Answer
$x^{2} - y^{2} = - x$ and $2 x y = y$ . If $y = 0$ : $x^{2} = - x$ , so $z = 0$ or $- 1$ . If $y \neq = 0$ : $x = \frac{1}{2}$ , then $\frac{1}{4} - y^{2} = - \frac{1}{2}$ , $y^{2} = \frac{3}{4}$ . So $z = 0, - 1, \frac{1}{2} + \frac{3}{2} i, \frac{1}{2} - \frac{3}{2} i$ .

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.What is i^4?Show answer
1.
2.What is i^(−1)?Show answer
−i.
3.When are a + ib and c + id equal?Show answer
When a = c and b = d.
4.What is the inverse of z = a + ib?Show answer
(a − ib)/(a² + b²).
5.What is (1 + i)/(1 − i)?Show answer
i.

Powers of i and Algebra of Complex Numbers: frequently asked questions

Why do powers of i repeat every four?

Because i⁴ = 1. After four multiplications you are back at 1, so the cycle i, −1, −i, 1 repeats.

Can I compare two complex numbers with < or >?

No. Complex numbers are not ordered. You can compare their moduli or real parts, which are real numbers.

How do I find x and y from an equation in complex numbers?

Simplify both sides to the form real + i × imaginary, then equate real parts and imaginary parts to get two real equations.

Is the square root of a complex number in the 2026-27 syllabus?

The chapter page notes that square roots of complex numbers and quadratic equations with complex roots are not examined in the annual exam. Equating parts to solve (x + iy)² = p + iq remains a good skill.