1st Year Math Chapter 1 Exercise 1.2 Notes: Square Root of a Complex Number (Punjab Board 2026-27)
Complete concept notes, formulas, and solved examples for ICS/FSc Part 1 Mathematics, Punjab Board (PECTAA), Single National Curriculum 2026-27 session.
Exercise 1.2 follows the basic operations covered in Exercise 1.1 and teaches how to find the square root of a complex number algebraically — writing √(a + bi) in the standard form x + yi.
What Does This Exercise Cover?
This exercise applies the definitions of real part, imaginary part, and modulus from Exercise 1.1 to solve a new type of problem: expressing the square root of any complex number as another complex number in standard form.
Key Concepts
1. Setting Up the Problem
To find the square root of z = a + bi, assume √(a + bi) = x + yi, where x and y are real numbers to be found.
2. Squaring and Comparing
Squaring both sides gives (x + yi)² = x² − y² + 2xyi. Setting this equal to a + bi and comparing real and imaginary parts gives two equations: x² − y² = a and 2xy = b.
3. Using the Modulus
A third equation, x² + y² = √(a² + b²), comes from taking the modulus of both sides of the original equation. Combining this with x² − y² = a allows x² and y² to be solved directly, without needing to solve the system by substitution.
4. Determining the Sign
Once x² and y² are found, taking square roots gives the possible values of x and y — but only one sign combination is valid. The equation 2xy = b determines whether x and y should have the same sign (b positive) or opposite signs (b negative).
Formulas for √(a + bi) = x + yi
These formulas summarize the algebraic method described above:
| Quantity | Formula |
| Modulus of a + bi | |z| = √(a² + b²) |
| x² | x² = (|z| + a) / 2 |
| y² | y² = (|z| − a) / 2 |
| Sign rule | x and y share the same sign if b > 0; opposite signs if b < 0 |
Solved Examples
Example 1: Find the square root of 3 + 4i.
Here a = 3, b = 4, so |z| = √(3² + 4²) = √25 = 5.
x² = (5 + 3)/2 = 4, so x = ±2.
y² = (5 − 3)/2 = 1, so y = ±1.
Since b = 4 is positive, x and y take the same sign.
Answer: √(3 + 4i) = ±(2 + i)
Example 2: Find the square root of −5 + 12i.
Here a = −5, b = 12, so |z| = √((−5)² + 12²) = √169 = 13.
x² = (13 + (−5))/2 = 4, so x = ±2.
y² = (13 − (−5))/2 = 9, so y = ±3.
Since b = 12 is positive, x and y take the same sign.
Answer: √(−5 + 12i) = ±(2 + 3i)
Sample MCQs
1. To find the square root of a + bi, we assume it equals:
a) x + y b) x + yi c) xy d) x − yi
Answer: b) x + yi
2. After squaring x + yi, the imaginary part 2xy is set equal to:
a) a b) b c) x² − y² d) x² + y²
Answer: b) b
3. The equation x² + y² = √(a² + b²) is obtained from:
a) The real part only b) The imaginary part only c) Taking the modulus of both sides d) Adding a and b
Answer: c) Taking the modulus of both sides
4. The square root of 3 + 4i is:
a) ±(2 + i) b) ±(1 + 2i) c) ±(4 + 3i) d) ±(3 + 2i)
Answer: a) ±(2 + i)
5. If 2xy is negative, x and y must have:
a) The same sign b) Opposite signs c) Both be zero d) Both be equal
Answer: b) Opposite signs
Important Short Questions
- Explain the algebraic method used to find the square root of a complex number.
- What two equations result from equating real and imaginary parts of (x + yi)² = a + bi?
- How is the value of |a + bi| used when finding a square root?
- How is the sign relationship between x and y determined?
- Find the square root of −7 − 24i using the method described in this exercise.
Important Long Questions
- Find the square root of 8 − 6i, showing every step of the algebraic method.
- Derive the formulas x² = (|z| + a)/2 and y² = (|z| − a)/2 starting from (x + yi)² = a + bi.
- Find the square root of −5 + 12i and verify your answer by squaring it.
- Explain why every non-zero complex number has exactly two square roots that differ only in sign.
How to Approach This Exercise Effectively
- Always compute |z| = √(a² + b²) first — every other value in the method depends on it.
- Keep a should the two equations x²−y²=a and x²+y²=|z| side by side; adding and subtracting them directly gives x² and y².
- Double-check the sign of x and y using 2xy = b before writing the final answer.
- Verify your answer by squaring it back — (x + yi)² should equal the original complex number exactly.
- Practice with both positive and negative values of b, since the sign rule is a common source of mistakes.
FAQs
Q: Why does a complex number have two square roots?
A: Just like real numbers, squaring either +k or −k gives the same positive result, so every non-zero complex number has two square roots that are negatives of each other.
Q: Why do we need the modulus in this method?
A: The modulus provides a third equation (x² + y² = |z|) that, combined with x² − y² = a, lets us solve for x² and y² directly by addition and subtraction.
Q: Can this method fail for some complex numbers?
A: No — every complex number a + bi has a well-defined square root using this method, since |z| ≥ |a| always, keeping both x² and y² non-negative.
Q: Is there a quicker way than this algebraic method?
A: The polar form (covered in Exercise 1.5) offers an alternative approach using angles, but the algebraic method in this exercise is the standard technique expected in exams.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 1: Complex Numbers, Exercise 1.2.