Class 11 Maths · Original EZT practice
Complex Numbers (Class 11 Chapter 4): summary and practice
Complex numbers begin with one bold invention, i^2 = -1, and grow into a full arithmetic: add, multiply, divide, measure size with the modulus, flip signs with the conjugate, and plot everything on the Argand plane. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Complex numbers and i
A complex number is a + ib with a and b real and i^2 = -1; a is the real part and b the imaginary part. Powers of i cycle: i^3 = -i, i^4 = 1, and then the pattern repeats.
2. Algebra of complex numbers
Add and subtract component-wise; multiply like binomials using i^2 = -1; divide by multiplying numerator and denominator by the conjugate. The chapter also gives square roots of a negative real number, like sqrt(-3) = i sqrt3, and warns that sqrt(a) sqrt(b) = sqrt(ab) fails when both are negative.
3. Modulus and conjugate
The modulus |a + bi| = sqrt(a^2 + b^2) measures distance from the origin; the conjugate flips the sign of the imaginary part. Together they give the reciprocal: 1/z = conjugate of z divided by |z|^2.
4. Argand plane and polar form
Plot a + bi as the point (a, b). Polar form writes z = r(cos theta + i sin theta) with r the modulus and theta the argument - and the quadrant of the point decides theta, not the formula alone.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Add and multiply complex numbers
Choose the method: Combine components; replace i^2 with -1.
EZT practice question
Let z = 3 + 4i and w = 1 - 2i. Find z + w and z w.
Show the worked solution
- z + w = (3 + 1) + (4 - 2)i = 4 + 2i.
- z w = 3 - 6i + 4i - 8i^2 = 3 - 2i + 8 = 11 - 2i.
Answer: z + w = 4 + 2i and z w = 11 - 2i.
Check: Multiplication works like binomial expansion with one extra step: i^2 becomes -1.
Watch for this: Watch the last term: -8i^2 is +8, not -8.
Now try it yourself
Compute (2 + i)(2 - i).
Check your answer
5.
4 - i^2 = 4 + 1 = 5 - a conjugate pair always multiplies to a real number.
Divide complex numbers
Choose the method: Multiply top and bottom by the conjugate of the denominator.
EZT practice question
Let z = 3 + 4i and w = 1 - 2i. Find z / w.
Show the worked solution
- Multiply by (1 + 2i)/(1 + 2i): numerator (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i^2 = -5 + 10i.
- Denominator (1 - 2i)(1 + 2i) = 1 + 4 = 5.
- So z/w = (-5 + 10i)/5.
Answer: z / w = -1 + 2i.
Check: Verify: (-1 + 2i)(1 - 2i) = -1 + 2i + 2i + 4 = 3 + 4i - back to z.
Watch for this: The denominator becomes |w|^2, a real number - do not leave i downstairs.
Now try it yourself
Find 1/(3 + 4i).
Check your answer
(3 - 4i)/25.
1/z = conjugate over |z|^2 = (3 - 4i)/25.
Reduce powers of i
Choose the method: The cycle i, -1, -i, 1 repeats every 4.
EZT practice question
Evaluate i^5, i^6 and i^23.
Show the worked solution
- i^5 = i^4 x i = i.
- i^6 = i^4 x i^2 = -1.
- i^23 = i^20 x i^3 = 1 x (-i).
Answer: i^5 = i, i^6 = -1 and i^23 = -i.
Check: Divide the exponent by 4 and read the remainder: 0 gives 1, 1 gives i, 2 gives -1, 3 gives -i.
Watch for this: Do not multiply the base i repeatedly - use the cycle.
Now try it yourself
Evaluate i^10.
Check your answer
-1.
10 = 4 x 2 + 2, and i^2 = -1.
Compute modulus and conjugate
Choose the method: |a + bi| = sqrt(a^2 + b^2); conjugate flips the i-sign.
EZT practice question
For z = 3 + 4i, find |z| and the conjugate of z.
Show the worked solution
- |z| = sqrt(9 + 16) = sqrt(25) = 5.
- Conjugate: change +4i to -4i.
Answer: |z| = 5 and the conjugate is 3 - 4i.
Check: z times its conjugate = |z|^2 = 25 - a quick self-check.
Watch for this: The conjugate flips only the imaginary part's sign, never the real part's.
Now try it yourself
Find |6 + 8i|.
Check your answer
10.
sqrt(36 + 64) = sqrt(100) = 10.
Handle square roots of negative numbers
Choose the method: sqrt(-a) = i sqrt(a) for positive a - and no combining two negatives.
EZT practice question
Write sqrt(-3) in terms of i, and explain why sqrt(-2) sqrt(-8) is not sqrt(16).
Show the worked solution
- sqrt(-3) = i sqrt3 by the chapter's definition.
- sqrt(-2) sqrt(-8) = (i sqrt2)(i sqrt8) = i^2 sqrt16 = -4, while sqrt(16) = +4 - the rule sqrt(a) sqrt(b) = sqrt(ab) needs at least one of a, b non-negative.
Answer: sqrt(-3) = i sqrt3; the product rule fails for two negatives because each factor carries an i.
Check: -4 is the correct product, as direct multiplication shows.
Watch for this: This is the chapter's own warning - applying the product rule to two negatives changes the sign of the answer.
Now try it yourself
Compute sqrt(-4) sqrt(-9).
Check your answer
-6.
(2i)(3i) = 6i^2 = -6.
Convert to polar form
Choose the method: r = |z|, theta from the quadrant.
EZT practice question
Write 1 + i in polar form.
Show the worked solution
- r = sqrt(1 + 1) = sqrt2.
- cos theta = 1/sqrt2 and sin theta = 1/sqrt2, and the point lies in the first quadrant, so theta = 45 degrees.
Answer: 1 + i = sqrt2 (cos 45 degrees + i sin 45 degrees).
Check: Expanding back: sqrt2 x (1/sqrt2) + i sqrt2 x (1/sqrt2) = 1 + i.
Watch for this: The inverse-tangent value alone can pick the wrong quadrant - check where the point lies.
Now try it yourself
Polar form of -1 + i?
Check your answer
sqrt2 (cos 135 degrees + i sin 135 degrees).
r = sqrt2 and the second-quadrant point gives theta = 135 degrees.
Try a mixed set
Choose a method or explanation before opening each answer.
1. i^2, i^3, i^4?
Show answer and reason
-1, -i, 1.
Then the cycle repeats.
2. |a + bi|?
Show answer and reason
sqrt(a^2 + b^2).
Distance from the origin.
3. 1/z in terms of conjugate?
Show answer and reason
conjugate of z divided by |z|^2.
z times its conjugate equals |z|^2.
4. sqrt(-a) for a > 0?
Show answer and reason
i sqrt(a).
The product rule sqrt(a)sqrt(b) = sqrt(ab) fails for two negatives.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Compute (2 + i)(2 - i).
- Find the modulus of 6 + 8i.
- Evaluate i^10.
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Questions about this practice page
- What is i?
- The imaginary unit, defined by i^2 = -1. Complex numbers a + ib are built from it, with a the real part and b the imaginary part.
- Why is the conjugate useful for division?
- Multiplying numerator and denominator by the conjugate turns the denominator into |z|^2, a real number - so the quotient comes out in standard a + ib form.
- What is the Argand plane?
- The picture of complex numbers as points: a + bi plots at (a, b), the modulus is its distance from the origin, and the argument is its angle from the positive real axis.
- Why does sqrt(a) sqrt(b) = sqrt(ab) fail for two negatives?
- Each negative root contributes a factor of i, and i^2 = -1 flips the sign - the chapter warns the product rule needs at least one non-negative number.
- Are these the NCERT exercise questions for Complex Numbers (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.