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10 Class Math Chapter wise Class test 2027 | Chapter 1 Test 3 Complex Numbers | free pdf download
Class 10 Math Chapter 1 Complex Numbers Test 3 Excercise 3.3 and Review Exercise is prepared for students studying Class 10 Mathematics. This Complex Numbers Chapter 1 Test contains objective-type MCQs, short-answer questions, and subjective questions based on important concepts of Unit 1. Students can use this Class 10 Math Chapter Wise Test for classroom assessment, revision, practice, and examination preparation. The complete Complex Numbers Test PDF is available for download and printing.
Unit Wise Class Test
T-3
Unit 1: Complex Numbers
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer.(1 × 5 = 5)
(i) z = x + iy is called a complex number where i =
By definition, the imaginary unit is i = √−1 (so i² = −1). Ans: (c)
(ii) Additive inverse of 1 + i is
Additive inverse of z is −z, because z + (−z) = 0.
−(1 + i) = −1 − i Ans: (a)
−(1 + i) = −1 − i Ans: (a)
(iii) If (2 + i)i = x + 2i, then x =
Step 1: Expand: (2 + i)i = 2i + i²
Step 2: Since i² = −1: = −1 + 2i
Step 3: Compare with x + 2i: real parts give x = −1
Ans: (a)
Step 2: Since i² = −1: = −1 + 2i
Step 3: Compare with x + 2i: real parts give x = −1
Ans: (a)
(iv) If z = x + yi then z + z̄ =
z̄ = x − yi
z + z̄ = (x + yi) + (x − yi) = 2x Ans: (a)
z + z̄ = (x + yi) + (x − yi) = 2x Ans: (a)
(v) z z̄ =
z z̄ = (x + yi)(x − yi) = x² − y²i² = x² + y²
And |z|² = (√(x² + y²))² = x² + y². So z z̄ = |z|² Ans: (b)
And |z|² = (√(x² + y²))² = x² + y². So z z̄ = |z|² Ans: (b)
Unit Wise Class Test
T-3
Unit 1: Complex Numbers
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions:(6 × 2 = 12)
(i) Find the modulus of the following complex numbers: 4 + 3i
Step 1: For z = x + iy, |z| = √(x² + y²). Here x = 4, y = 3
Step 2: |z| = √(16 + 9) = √25
Ans: |z| = 5
Step 2: |z| = √(16 + 9) = √25
Ans: |z| = 5
(ii) If z1 = 2 + 7i and z2 = 4 − 3i, then verify that: z̄1 + z̄2 = z̄1 + z̄2
(The paper prints the same expression on both sides; the usual statement is conj(z₁ + z₂) = z̄₁ + z̄₂, verified below.)
Step 1: z₁ + z₂ = (2 + 7i) + (4 − 3i) = 6 + 4i
Step 2: LHS = conj(z₁ + z₂) = 6 − 4i
Step 3: z̄₁ = 2 − 7i and z̄₂ = 4 + 3i
Step 4: RHS = z̄₁ + z̄₂ = (2 − 7i) + (4 + 3i) = 6 − 4i
LHS = RHS = 6 − 4i. Hence verified ✓
Step 1: z₁ + z₂ = (2 + 7i) + (4 − 3i) = 6 + 4i
Step 2: LHS = conj(z₁ + z₂) = 6 − 4i
Step 3: z̄₁ = 2 − 7i and z̄₂ = 4 + 3i
Step 4: RHS = z̄₁ + z̄₂ = (2 − 7i) + (4 + 3i) = 6 − 4i
LHS = RHS = 6 − 4i. Hence verified ✓
(iii) If z = 5 − 2i, then verify that: z̄̄ = z
Step 1: z = 5 − 2i → z̄ = 5 + 2i (change sign of imaginary part)
Step 2: Conjugate again: conj(z̄) = 5 − 2i
5 − 2i = z. Hence verified ✓
Step 2: Conjugate again: conj(z̄) = 5 − 2i
5 − 2i = z. Hence verified ✓
(iv) If z = 4 − 3i, then verify that: |z| = |−z| = |z̄| = |−z̄|
Step 1: |z| = |4 − 3i| = √(16 + 9) = 5
Step 2: −z = −4 + 3i → |−z| = √(16 + 9) = 5
Step 3: z̄ = 4 + 3i → |z̄| = √(16 + 9) = 5
Step 4: −z̄ = −4 − 3i → |−z̄| = √(16 + 9) = 5
All four equal 5. Hence verified ✓
Step 2: −z = −4 + 3i → |−z| = √(16 + 9) = 5
Step 3: z̄ = 4 + 3i → |z̄| = √(16 + 9) = 5
Step 4: −z̄ = −4 − 3i → |−z̄| = √(16 + 9) = 5
All four equal 5. Hence verified ✓
(v) Define a complex number.
Definition: A number of the form z = x + iy, where x and y are real numbers and i = √−1, is called a complex number.
x is the real part, Re(z) = x, and y is the imaginary part, Im(z) = y. Example: 3 + 2i
x is the real part, Re(z) = x, and y is the imaginary part, Im(z) = y. Example: 3 + 2i
(vi) Find real and imaginary parts of z = (2 + 7i)−1.
Step 1: z = 12 + 7i
Step 2: Multiply top and bottom by conjugate (2 − 7i): 2 − 7i(2 + 7i)(2 − 7i)
Step 3: Denominator: 2² − (7i)² = 4 + 49 = 53
Step 4: z = 2 − 7i53 = 253 − 753i
Ans: Re(z) = 253, Im(z) = −753
Step 2: Multiply top and bottom by conjugate (2 − 7i): 2 − 7i(2 + 7i)(2 − 7i)
Step 3: Denominator: 2² − (7i)² = 4 + 49 = 53
Step 4: z = 2 − 7i53 = 253 − 753i
Ans: Re(z) = 253, Im(z) = −753
Note: Answer the following Questions.(4 + 4 = 8)
3. Find real and imaginary parts of z = (4 + 3i)−2.
Step 1: z = 1(4 + 3i)²
Step 2: (4 + 3i)² = 16 + 24i + 9i² = 16 + 24i − 9 = 7 + 24i
Step 3: z = 17 + 24i. Multiply top and bottom by conjugate (7 − 24i): 7 − 24i(7 + 24i)(7 − 24i)
Step 4: Denominator: 7² − (24i)² = 49 + 576 = 625
Step 5: z = 7 − 24i625 = 7625 − 24625i
Ans: Re(z) = 7625, Im(z) = −24625
Step 2: (4 + 3i)² = 16 + 24i + 9i² = 16 + 24i − 9 = 7 + 24i
Step 3: z = 17 + 24i. Multiply top and bottom by conjugate (7 − 24i): 7 − 24i(7 + 24i)(7 − 24i)
Step 4: Denominator: 7² − (24i)² = 49 + 576 = 625
Step 5: z = 7 − 24i625 = 7625 − 24625i
Ans: Re(z) = 7625, Im(z) = −24625
4. Solve the given simultaneous linear equations with complex coefficients for z and w:
iz + (2 − i)w = 4 + i …(1)
iz + (3 + i)w = 3 + 3i …(2)
Step 1: Subtract (1) from (2) to eliminate iz: [(3 + i) − (2 − i)]w = (3 + 3i) − (4 + i)
Step 2: (1 + 2i)w = −1 + 2i
Step 3: w = −1 + 2i1 + 2i = (−1 + 2i)(1 − 2i)(1 + 2i)(1 − 2i)
Step 4: Numerator: −1 + 2i + 2i − 4i² = −1 + 4i + 4 = 3 + 4i. Denominator: 1 + 4 = 5
Step 5: w = 3 + 4i5 = 35 + 45i
Step 6: Put in (1): iz = (4 + i) − (2 − i)w. First (2 − i)(3 + 4i) = 6 + 8i − 3i − 4i² = 10 + 5i, so (2 − i)w = (10 + 5i)/5 = 2 + i
Step 7: iz = (4 + i) − (2 + i) = 2 → z = 2/i = 2·(−i) = −2i
Step 8: Check in (2): i(−2i) = 2 and (3 + i)(3 + 4i)/5 = (5 + 15i)/5 = 1 + 3i; sum = 3 + 3i ✓
Ans: z = −2i, w = 3 + 4i5
Step 2: (1 + 2i)w = −1 + 2i
Step 3: w = −1 + 2i1 + 2i = (−1 + 2i)(1 − 2i)(1 + 2i)(1 − 2i)
Step 4: Numerator: −1 + 2i + 2i − 4i² = −1 + 4i + 4 = 3 + 4i. Denominator: 1 + 4 = 5
Step 5: w = 3 + 4i5 = 35 + 45i
Step 6: Put in (1): iz = (4 + i) − (2 − i)w. First (2 − i)(3 + 4i) = 6 + 8i − 3i − 4i² = 10 + 5i, so (2 − i)w = (10 + 5i)/5 = 2 + i
Step 7: iz = (4 + i) − (2 + i) = 2 → z = 2/i = 2·(−i) = −2i
Step 8: Check in (2): i(−2i) = 2 and (3 + i)(3 + 4i)/5 = (5 + 15i)/5 = 1 + 3i; sum = 3 + 3i ✓
Ans: z = −2i, w = 3 + 4i5
