Complex number | NEB Class 11 Mathematics Notes

Chapter Overview

Namaste, dear students! Today we are diving into a fascinating world where numbers are not just 'real.' In your earlier classes, you might have been stuck when solving equations like $x^2 + 4 = 0$. You would say, 'solving $x^2 = -4$ is impossible because no real number squared is negative!'.

In the 16th century, mathematicians like Cardano and Euler decided to break this barrier. They introduced an 'imaginary unit' $i$ such that $i^2 = -1$. This allowed us to solve every quadratic equation. A Complex Number is simply a combination of a real number and an imaginary number.

Importance: In the NEB examinations, this chapter is a highly accessible scoring section. It forms the foundation for higher engineering mathematics and physics.

Real-Life Applications: Understanding complex numbers is crucial for calculating AC circuit impedance in Electrical Engineering, processing voice/data in Mobile Signal Processing, and modeling wave mechanics in Physics.

What this chapter covers

Definition of Complex Numbers as Ordered PairsAlgebraic Operations (Sum, Product, Quotient)The Cyclic Powers of Iota (i)Geometric Representation on the Argand PlaneModulus and Conjugate PropertiesMultiplicative Inverse and its DerivationAlgebraic and Inspection Methods for Square RootsTriangle Inequality and Locus Problems

Learning Objectives

1

Define a complex number as an ordered pair and convert it to standard algebraic form.

2

Perform basic arithmetic operations including sum, product, and quotients of complex numbers.

3

Understand the geometric representation of complex numbers in the Argand Plane.

4

Calculate and apply the conjugate, modulus, and their corresponding properties to simplify expressions.

5

Determine the algebraic square root of any complex number using rigorous step-by-step methods.

6

Solve equations involving complex variables, identify locus equations, and establish basic inequalities.

Key Definitions

Complex Number ($z$)

An ordered pair $(x, y)$ of two real numbers $x$ and $y$, usually denoted in standard form by $z = x + iy$.

Real Part & Imaginary Part

For a complex number $z = x + iy$, $x$ is the Real Part denoted as $Re(z)$, and $y$ is the Imaginary Part denoted as $Im(z)$.

Purely Imaginary Number

A complex number where the real part is strictly zero, represented in the form $(0, y)$ or $iy$ (where $y \neq 0$).

Argand Plane

A coordinate plane where the horizontal X-axis represents real numbers and the vertical Y-axis represents imaginary numbers.

Conjugate ($\bar{z}$)

If $z = x + iy$, its conjugate is $\bar{z} = x - iy$. Geometrically, it represents a reflection of $z$ across the real axis.

Modulus ($|z|$)

The distance of the complex number $z = x + iy$ from the origin in the Argand plane, defined as $|z| = \sqrt{x^2 + y^2}$.

Understanding the Powers of $i$ (Iota)

Key Insight

Since the imaginary unit is defined by $i = \sqrt{-1}$ (or $i^2 = -1$), higher powers of $i$ cycle continuously through four distinct values:

$i^1 = i$
$i^2 = -1$
$i^3 = i^2 \cdot i = -i$
$i^4 = (i^2)^2 = 1$

To simplify any power $i^n$, divide $n$ by $4$ and evaluate using the remainder $r$: $i^n = i^r$, where $r \in \{0, 1, 2, 3\}$.

Example: For $i^{102}$, we divide $102$ by $4$ to get a remainder of $2$. Therefore, $i^{102} = i^2 = -1$.

The Algebra of Complex Numbers

Key Insight

Arithmetic with complex numbers is highly intuitive and mirrors the algebra of real binomials, with the crucial substitution $i^2 = -1$.

Equality: $(a, b) = (c, d)$ if and only if $a = c$ AND $b = d$.
Addition: Combine the real and imaginary parts separately: $(a, b) + (c, d) = (a+c, b+d)$.
Multiplication: $(a, b) \cdot (c, d) = (ac - bd, ad + bc)$.

Algebraic expansion trick: $(a + ib)(c + id) = ac + iad + ibc + i^2bd$. Substituting $i^2 = -1$ yields $(ac - bd) + i(ad + bc)$.

Formula Sheet & Proof Derivations

Multiplicative Inverse

z^{-1} = \left( \frac{x}{x^2+y^2}, \frac{-y}{x^2+y^2} \right)

Provides the unique complex coordinates $z^{-1}$ such that $z \cdot z^{-1} = (1, 0)$.

1Let $z = (a, b)$ be a non-zero complex number. We want to find $z^{-1} = (x, y)$ such that $z \cdot z^{-1} = (1, 0)$.

2By the definition of complex multiplication: $(ax - by, ay + bx) = (1, 0)$.

3Equating components gives two linear equations in terms of $x$ and $y$:
1) $ax - by = 1$
2) $ay + bx = 0$

4From equation (2), express $y$ in terms of $x$: $y = \frac{-bx}{a}$.

5Substitute this expression for $y$ into equation (1):
$$ ax - b\left(\frac{-bx}{a}\right) = 1 \implies ax + \frac{b^2x}{a} = 1 $$

6Find a common denominator: $a^2x + b^2x = a \implies x(a^2 + b^2) = a$.

7Solving for $x$ and $y$ yields:
$$ x = \frac{a}{a^2+b^2} \quad \text{and} \quad y = \frac{-b}{a^2+b^2} $$

8Thus, the multiplicative inverse is $z^{-1} = \left( \frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2} \right)$.

Square Root of a Complex Number

\sqrt{a + ib} = \pm \left[ \sqrt{\frac{|z|+a}{2}} + i \frac{b}{|b|} \sqrt{\frac{|z|-a}{2}} \right]

A systematic formula where the sign factor $\frac{b}{|b|}$ aligns the sign of the imaginary term with the sign of $b$.

1Let $\sqrt{a + ib} = x + iy$. Squaring both sides yields: $a + ib = (x^2 - y^2) + i(2xy)$.

2By equating real and imaginary parts, we obtain:
1) $x^2 - y^2 = a$
2) $2xy = b$

3Recall the algebraic identity: $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$.

4Substitute equations (1) and (2) into the identity:
$$ (x^2 + y^2)^2 = a^2 + b^2 = |z|^2 \implies x^2 + y^2 = |z| $$

5Solve the system of equations by addition and subtraction:
Addition: $2x^2 = |z| + a \implies x = \pm \sqrt{\frac{|z|+a}{2}}$
Subtraction: $2y^2 = |z| - a \implies y = \pm \sqrt{\frac{|z|-a}{2}}$

6Since the product $2xy = b$, $x$ and $y$ must have the same sign if $b > 0$, and opposite signs if $b < 0$. We reflect this constraint using the factor $\frac{b}{|b|}$.

Question Recognition Guide

Use this strategic flowchart to quickly recognize and attack complex number problems in your exams.

Does the question ask you to 'Express in $a + ib$ form'?

Yes

Rationalize the denominator. Multiply both the numerator and the denominator by the conjugate of the denominator.

No

Continue to next step ↓

Does the question require you to 'Find $x$ and $y$'?

Yes

Simplify both sides into standard form. Equate the real parts together, and equate the imaginary parts together to solve for the variables.

No

Continue to next step ↓

Does the question ask to 'Find the Square Root'?

Yes

If the numbers are simple, use the inspection method to make a perfect square. Otherwise, apply the algebraic formula.

No

Identify if it is a locus question like $|z - z_0| = r$, which traces out a Circle with center $z_0$ and radius $r$.

Memory Tricks

Trick: Incorporate these conceptual shortcuts to streamline calculations under exam pressure.

Steps to Remember:

Multiplying by $i$: Geometrically, multiplying any complex number by $i$ rotates its position counter-clockwise by exactly $90^\circ$ in the Argand plane.
Conjugate Sign Swap: Think of conjugate as 'Change'. Only change the sign of the imaginary ($i$) term, leaving the real term completely untouched.
Product of Conjugates: Multiplying a complex number by its conjugate always eliminates the imaginary unit: $(a+ib)(a-ib) = a^2 + b^2$.

Mnemonic: Conjugate = Change i's sign. Product of Conjugates = Sum of Squares (No i!).

Common Mistakes in NEB Exams

Avoid these typical student errors to preserve marks in the complex numbers evaluation.

The Imaginary Square Trap: Neglecting that $i^2 = -1$. Students frequently write $(i)^2 = 1$ during algebra, which reverses addition and subtraction signs.
Improper Conjugation: Changing the sign of the real part instead of the imaginary part. For example, the conjugate of $z = -3 + 4i$ is $\bar{z} = -3 - 4i$, NOT $+3 + 4i$.
Ignoring Signs in Square Roots: When $b$ (the imaginary part) is negative in $\sqrt{a + ib}$, $x$ and $y$ must have opposite signs. Ensure you track the sign of $b$.

Solved Examples (Easy → Hard)

1

Simplify the power of iota: $i^{102}$.

Easy

Solution

Let us divide the exponent by 4 to find the position in the cyclic pattern of iota.

We write: $102 = 4 \times 25 + 2$.
Using indices rules:
$$ i^{102} = (i^4)^{25} \cdot i^2 $$
Since $i^4 = 1$ and $i^2 = -1$, substitute these values:
$(1)^{25} \cdot (-1) = 1 \cdot (-1) = -1$.

∴ -1

2

Express the quotient $\frac{2 + i}{1 - i}$ in standard $a+ib$ form.

Medium

Solution

To eliminate the imaginary unit from the denominator, multiply both numerator and denominator by the conjugate of $(1 - i)$, which is $(1 + i)$.

$$ \frac{2 + i}{1 - i} = \frac{(2 + i)(1 + i)}{(1 - i)(1 + i)} $$
Multiply the numerator terms:
$(2)(1) + (2)(i) + (i)(1) + i^2 = 2 + 3i - 1 = 1 + 3i$.

Multiply the denominator terms (using conjugate sum of squares):
$1^2 + 1^2 = 2$.

Now, split the real and imaginary components:
$$ \frac{1 + 3i}{2} = \frac{1}{2} + i\frac{3}{2} $$

∴ 0.5 + 1.5i

3

Find the square root of $7 + 24i$ by the inspection method.

Hard

Solution

We wish to write $7 + 24i$ in the perfect square form $(x + iy)^2 = (x^2 - y^2) + 2ixy$.

First, analyze the imaginary term: $2xy = 24 \implies xy = 12$.
Next, analyze the real term: $x^2 - y^2 = 7$.

We seek integer factors of $12$ whose squares subtract to $7$. The factors $4$ and $3$ work perfectly since:
$4^2 - 3^2 = 16 - 9 = 7$.

Rewrite the original expression using these values:
$$ 7 + 24i = 16 - 9 + 24i = 4^2 + (3i)^2 + 2(4)(3i) = (4 + 3i)^2 $$
Taking the square root on both sides yields the final value.

∴ ±(4 + 3i)

MCQ Practice

Mixed Level — MCQ

1

What is the value of $i^{403}$?

Correct!

Incorrect. Correct: $-i$

Solution

Divide the exponent 403 by 4. The quotient is 100 and the remainder is 3. Since $i^4 = 1$, we can rewrite the expression as $i^{403} = (i^4)^{100} \cdot i^3 = 1 \cdot (-i) = -i$.

∴ $-i$

2

If $z = (0, -1)$, then $z^2$ is:

Correct!

Incorrect. Correct: $(-1, 0)$

Solution

Represent $z$ in standard form: $z = 0 - i = -i$. Squaring it yields $z^2 = (-i)^2 = i^2 = -1$. Writing this as an ordered pair gives $(-1, 0)$.

∴ $(-1, 0)$

3

The conjugate of $\frac{1}{i}$ is:

Correct!

Incorrect. Correct: $i$

Solution

First, simplify the term: $\frac{1}{i} = \frac{1 \cdot i}{i \cdot i} = \frac{i}{i^2} = \frac{i}{-1} = -i$. The conjugate of $-i$ is obtained by reversing the sign of the imaginary component, yielding $+i$.

∴ $i$

4

If $|z-1| = 2$, the locus of $z$ is a:

Correct!

Incorrect. Correct: Circle

Solution

Let $z = x+iy$. The equation becomes $|(x-1) + iy| = 2$. Squaring both sides yields $(x-1)^2 + y^2 = 2^2 = 4$, which is the standard algebraic equation of a circle.

∴ Circle

5

The value of $(1+i)^2$ is:

Correct!

Incorrect. Correct: $2i$

Solution

Use the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ to expand $(1+i)^2$. This gives $1 + 2i + i^2 = 1 + 2i - 1 = 2i$.

∴ $2i$

6

The multiplicative inverse of $(1, 1)$ is:

Correct!

Incorrect. Correct: $(0.5, -0.5)$

Solution

For $z = (1, 1)$, we use the formula $z^{-1} = \left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right)$. This yields $\left(\frac{1}{1+1}, \frac{-1}{1+1}\right) = (0.5, -0.5)$.

∴ $(0.5, -0.5)$

7

The real part of $i(2-3i)$ is:

Correct!

Incorrect. Correct: $3$

Solution

Distribute $i$ into the expression: $i(2 - 3i) = 2i - 3i^2$. Since $i^2 = -1$, this simplifies to $2i - 3(-1) = 3 + 2i$. The real component is $3$.

∴ $3$

8

What is the value of $|3-4i| \times |1+i|$?

Correct!

Incorrect. Correct: $5\sqrt{2}$

Solution

Calculate the individual moduli: $|3-4i| = \sqrt{3^2 + (-4)^2} = 5$ and $|1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$. Multiplying them yields $5\sqrt{2}$.

∴ $5\sqrt{2}$

9

If $z \cdot \bar{z} = 0$, then $z$ is:

Correct!

Incorrect. Correct: $0$

Solution

Recall that $z \cdot \bar{z} = |z|^2 = x^2 + y^2$. For the sum of two real squares to equal 0, both real terms must be 0. Thus $x=0$ and $y=0$, which means $z=0$.

∴ $0$

10

The square root of $-i$ is:

Correct!

Incorrect. Correct: $\pm \frac{1-i}{\sqrt{2}}$

Solution

Let us check by squaring the candidate: $\left(\pm \frac{1-i}{\sqrt{2}}\right)^2 = \frac{1 - 2i + i^2}{2} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i$. Since the squared value matches, this is the correct root.

∴ $\pm \frac{1-i}{\sqrt{2}}$

Short Questions (2 Marks Each)

1

Evaluate: $i^{10} + i^{20} + i^{30}$.

[2 marks]

Solution

Let us evaluate each power of iota by dividing the exponents by 4:
1) $i^{10} = (i^4)^2 \cdot i^2 = (1)^2 \cdot (-1) = -1$
2) $i^{20} = (i^4)^5 = (1)^5 = 1$
3) $i^{30} = (i^4)^7 \cdot i^2 = (1)^7 \cdot (-1) = -1$

Now, sum the individual terms: $-1 + 1 - 1 = -1$.

∴ -1

2

Find $x$ and $y$ if $(x+iy)(1+i) = 2 + 4i$.

[2 marks]

Solution

Isolate $x + iy$ by dividing both sides by $1 + i$:
$$ x + iy = \frac{2 + 4i}{1 + i} $$
Multiply the numerator and denominator by the conjugate $(1 - i)$:
$$ x + iy = \frac{(2 + 4i)(1 - i)}{(1 + i)(1 - i)} = \frac{2 - 2i + 4i - 4i^2}{1^2 + 1^2} = \frac{2 + 2i + 4}{2} = \frac{6 + 2i}{2} = 3 + i $$
Equating components yields $x = 3$ and $y = 1$.

∴ x = 3, y = 1

3

Find the modulus of $\frac{3+i}{3-i}$.

[2 marks]

Solution

Using properties of modulus, the modulus of a quotient is equal to the quotient of the moduli:
$$ \left| \frac{3+i}{3-i} \right| = \frac{|3+i|}{|3-i|} $$
Evaluate the numerator and denominator lengths:
$$ |3+i| = \sqrt{3^2 + 1^2} = \sqrt{10} \quad \text{and} \quad |3-i| = \sqrt{3^2 + (-1)^2} = \sqrt{10} $$
Therefore, the fraction simplifies to $\frac{\sqrt{10}}{\sqrt{10}} = 1$.

∴ 1

4

Express $\sqrt{-16} + \sqrt{-25}$ in standard $a+ib$ form.

[2 marks]

Solution

Convert the radical expressions using the imaginary unit $i = \sqrt{-1}$:
$$ \sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i $$
$$ \sqrt{-25} = \sqrt{25} \cdot \sqrt{-1} = 5i $$
Summing these terms: $4i + 5i = 9i$. Represented in standard form: $0 + 9i$.

∴ 0 + 9i

5

Show that $z + \bar{z}$ is always real.

[2 marks]

Solution

Let $z = x + iy$ be a complex number. Its conjugate is $\bar{z} = x - iy$.
Add the two expressions together:
$$ z + \bar{z} = (x + iy) + (x - iy) = 2x $$
Since $x$ is defined as a real number, the sum $2x$ contains no imaginary component and is strictly real.

∴ z + z̄ = 2x, which is purely real.

6

Find the conjugate of $(2+3i)^2$.

[2 marks]

Solution

First, expand the expression using the binomial formula:
$$ (2+3i)^2 = 2^2 + 2(2)(3i) + (3i)^2 = 4 + 12i - 9 = -5 + 12i $$
Next, obtain the conjugate by reversing the sign of the imaginary coefficient:
$$ \overline{-5 + 12i} = -5 - 12i $$

∴ -5 - 12i

7

Simplify $i^{-35}$.

[2 marks]

Solution

Express the term with a positive exponent: $\frac{1}{i^{35}}$.
Divide the exponent $35$ by $4$: $35 = 4 \times 8 + 3$.
$$ \frac{1}{i^{35}} = \frac{1}{i^3} $$
Substitute $i^3 = -i$:
$$ \frac{1}{-i} = \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-(-1)} = i $$

∴ i

8

If $z = (3, 4)$, find its multiplicative inverse $z^{-1}$.

[2 marks]

Solution

We represent the coordinate $(3, 4)$ in formula form where $a = 3$ and $b = 4$. The sum of squares is:
$$ a^2 + b^2 = 3^2 + 4^2 = 25 $$
Apply the multiplicative inverse formula:
$$ z^{-1} = \left( \frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2} \right) = \left( \frac{3}{25}, \frac{-4}{25} \right) $$

∴ (3/25, -4/25)

9

Prove that $|z_1 z_2| = |z_1| |z_2|$ for $z_1 = i$ and $z_2 = 1+i$.

[2 marks]

Solution

First, calculate the product of the two complex numbers:
$$ z_1 z_2 = i(1+i) = i + i^2 = -1 + i $$
Find the modulus of the product:
$$ |z_1 z_2| = |-1 + i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} $$
Calculate the individual moduli:
$$ |z_1| = |i| = 1 \quad \text{and} \quad |z_2| = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} $$
Product of the moduli: $1 \cdot \sqrt{2} = \sqrt{2}$. Since $|z_1 z_2| = |z_1||z_2| = \sqrt{2}$, the identity is verified.

∴ Proved; both sides equal √2.

10

Find the additive inverse of $-5+2i$.

[2 marks]

Solution

The additive inverse of any complex number $z$ is $-z$.
Multiply the given expression by $-1$:
$$ -(-5 + 2i) = 5 - 2i $$
Verify the result by adding them: $(-5+2i) + (5-2i) = 0 + 0i$.

∴ 5 - 2i

Long Questions (3-5 Marks Each)

1

If $x+iy = \sqrt{\frac{a+ib}{c+id}}$, prove that $(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$.

[5 marks]

Solution

Given the equation:
$$ x + iy = \sqrt{\frac{a+ib}{c+id}} $$
Take the absolute value (modulus) on both sides:
$$ |x + iy| = \left| \sqrt{\frac{a+ib}{c+id}} \right| $$
By properties of exponents and limits, we can shift the modulus inside the square root:
$$ |x + iy| = \sqrt{\left| \frac{a+ib}{c+id} \right|} $$
Expand the left and right expressions using standard modulus definitions:
$$ \sqrt{x^2 + y^2} = \sqrt{\frac{|a+ib|}{|c+id|}} = \sqrt{\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}} $$
Square both sides to eliminate the outer radicals:
$$ x^2 + y^2 = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} $$
Square both sides once more to obtain the final relation:
$$ (x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2} $$

∴ Proved using modulus properties and successive squaring.

2

Find the square root of $5 + 12i$ using the algebraic method.

[4 marks]

Solution

Let us set up the algebraic system. Let $\sqrt{5+12i} = x+iy$.
Squaring both sides yields $5 + 12i = (x^2 - y^2) + i(2xy)$.
Equating real and imaginary parts:
1) $x^2 - y^2 = 5$
2) $2xy = 12$

Now, utilize the key algebraic identity to find $x^2 + y^2$:
$$ (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 $$
$$ (x^2 + y^2)^2 = 5^2 + 12^2 = 25 + 144 = 169 \implies x^2 + y^2 = 13 $$
Now we have a system of simple linear equations for the squares:
Add: $2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$
Subtract: $2y^2 = 8 \implies y^2 = 4 \implies y = \pm 2$

Since the product $2xy = 12$ is positive, $x$ and $y$ must share the same sign. Thus, the square roots are $\pm(3 + 2i)$.

∴ ±(3 + 2i)

3

Prove the Triangle Inequality: $|z_1 + z_2| \leq |z_1| + |z_2|$.

[5 marks]

Solution

Let us utilize the property $z \cdot \bar{z} = |z|^2$ to expand the expression:
$$ |z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = (z_1 + z_2)(\bar{z}_1 + \bar{z}_2) $$
Distribute the terms carefully:
$$ = z_1\bar{z}_1 + z_1\bar{z}_2 + z_2\bar{z}_1 + z_2\bar{z}_2 = |z_1|^2 + |z_2|^2 + (z_1\bar{z}_2 + \overline{z_1\bar{z}_2}) $$
The sum of a complex number and its conjugate equals twice its real component:
$$ = |z_1|^2 + |z_2|^2 + 2Re(z_1\bar{z}_2) $$
Since the real part of any complex number is always less than or equal to its absolute value ($Re(w) \leq |w|$):
$$ |z_1 + z_2|^2 \leq |z_1|^2 + |z_2|^2 + 2|z_1\bar{z}_2| $$
Using properties of conjugates where $|\bar{z}| = |z|$:
$$ |z_1 + z_2|^2 \leq |z_1|^2 + |z_2|^2 + 2|z_1||z_2| = (|z_1| + |z_2|)^2 $$
Taking the square root on both sides yields the Triangle Inequality:
$$ |z_1 + z_2| \leq |z_1| + |z_2| $$

∴ Proved using the algebraic identities of conjugate multiplication.

4

If $z_1 = 1-i$ and $z_2 = -2+4i$, find $Im\left(\frac{z_1 z_2}{\bar{z}_1}\right)$.

[4 marks]

Solution

First, calculate the product in the numerator $z_1 z_2$:
$$ z_1 z_2 = (1-i)(-2+4i) = -2 + 4i + 2i - 4i^2 = -2 + 6i + 4 = 2 + 6i $$
Next, write down the conjugate of the denominator: $\bar{z}_1 = \overline{1-i} = 1+i$.
Now, perform the division by rationalizing the denominator:
$$ \frac{z_1 z_2}{\bar{z}_1} = \frac{2+6i}{1+i} = \frac{(2+6i)(1-i)}{(1+i)(1-i)} $$
$$ = \frac{2 - 2i + 6i - 6i^2}{1^2 + 1^2} = \frac{2 + 4i + 6}{2} = \frac{8 + 4i}{2} = 4 + 2i $$
Extract the imaginary part of this simplified result: $Im(4+2i) = 2$.

∴ 2

5

Solve for $x$ and $y$: $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$.

[4 marks]

Solution

To solve, find a common denominator for the terms on the left side of the equation:
$$ \frac{(x-1)(3-i) + (y-1)(3+i)}{(3+i)(3-i)} = i $$
Expand both numerator products and the denominator product:
$$ \frac{(3x - x i - 3 + i) + (3y + y i - 3 - i)}{3^2 + 1^2} = i $$
$$ \frac{(3x + 3y - 6) + i(y - x)}{10} = i $$
Multiply both sides by 10 to clear the denominator:
$$ (3x + 3y - 6) + i(y - x) = 10i $$
Equating real parts: $3x + 3y - 6 = 0 \implies x + y = 2$
Equating imaginary parts: $y - x = 10 \implies -x + y = 10$

Solve this system of two linear equations by addition:
$$ 2y = 12 \implies y = 6 $$
Substitute $y=6$ into the first equation: $x + 6 = 2 \implies x = -4$.

∴ x = -4, y = 6

6

Verify the associative property of complex multiplication: $(z_1 z_2) z_3 = z_1 (z_2 z_3)$ for general coordinates.

[5 marks]

Solution

Let $z_1 = a+ib$, $z_2 = c+id$, and $z_3 = e+if$ be any three complex numbers.

Evaluate the left-hand side (LHS):
$$ z_1 z_2 = (ac-bd) + i(ad+bc) $$
$$ (z_1 z_2) z_3 = [ (ac-bd)e - (ad+bc)f ] + i[ (ac-bd)f + (ad+bc)e ] $$
$$ = (ace - bde - adf - bcf) + i(acf - bdf + ade + bce) $$

Evaluate the right-hand side (RHS):
$$ z_2 z_3 = (ce-df) + i(cf+de) $$
$$ z_1 (z_2 z_3) = [ a(ce-df) - b(cf+de) ] + i[ b(ce-df) + a(cf+de) ] $$
$$ = (ace - adf - bcf - bde) + i(bce - bdf + acf + ade) $$

Comparing the expanded terms, we see that LHS and RHS are identical, verifying associativity.

∴ Verified algebraically term-for-term.

7

Find the square root of the quotient expression: $\frac{1+i}{1-i}$.

[4 marks]

Solution

First, simplify the complex expression inside the square root by multiplying by the conjugate of the denominator:
$$ \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1^2 + 1^2} = \frac{2i}{2} = i $$
Now we need to find the square root of $i$. Let $\sqrt{i} = x+iy$.
Squaring both sides yields $i = (x^2 - y^2) + i(2xy)$.
Equating parts: $x^2 - y^2 = 0$ and $2xy = 1$.
Since $x^2 = y^2$ and $xy > 0$, we have $x = y$.
Substitute $x=y$ into $2xy = 1$:
$$ 2x^2 = 1 \implies x = \pm \frac{1}{\sqrt{2}} $$
Since $x=y$, we get both roots: $\pm \left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) = \pm \frac{1+i}{\sqrt{2}}$.

∴ ±(1+i)/√2

8

If $(a+ib)(c+id)(e+if)(g+ih) = A+iB$, prove $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2$.

[4 marks]

Solution

Given the algebraic product equation:
$$ (a+ib)(c+id)(e+if)(g+ih) = A+iB $$
Apply the modulus operator on both sides of the equation:
$$ |(a+ib)(c+id)(e+if)(g+ih)| = |A+iB| $$
Recall that the modulus of a product equals the product of the individual moduli:
$$ |a+ib| \cdot |c+id| \cdot |e+if| \cdot |g+ih| = |A+iB| $$
Substitute the modulus formulas:
$$ \sqrt{a^2+b^2} \cdot \sqrt{c^2+d^2} \cdot \sqrt{e^2+f^2} \cdot \sqrt{g^2+h^2} = \sqrt{A^2+B^2} $$
Square both sides of the equation to eliminate the radicals:
$$ (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2 $$

∴ Proved using the product property of moduli.

9

Express $\frac{(1+i)^3}{1-i}$ in the form $a+ib$.

[4 marks]

Solution

First, expand the numerator using the binomial expansion of $(a+b)^3$:
$$ (1+i)^3 = 1^3 + 3(1^2)(i) + 3(1)(i^2) + i^3 $$
$$ = 1 + 3i - 3 - i = -2 + 2i $$
Now substitute this expression back and rationalize the denominator by multiplying by the conjugate $(1+i)$:
$$ \frac{-2+2i}{1-i} = \frac{(-2+2i)(1+i)}{(1-i)(1+i)} $$
$$ = \frac{-2 - 2i + 2i + 2i^2}{1^2 + 1^2} = \frac{-2 - 2}{2} = \frac{-4}{2} = -2 $$
Represented in standard form: $-2 + 0i$.

∴ -2 + 0i

10

Show that the sum of the positive square roots of positive and negative imaginary units yields $\sqrt{-i} + \sqrt{i} = \sqrt{2}$.

[4 marks]

Solution

Using the algebraic derivations for square roots, the principal square roots are:
$$ \sqrt{i} = \frac{1+i}{\sqrt{2}} \quad \text{and} \quad \sqrt{-i} = \frac{1-i}{\sqrt{2}} $$
Add the two principal roots together:
$$ \sqrt{-i} + \sqrt{i} = \frac{1-i}{\sqrt{2}} + \frac{1+i}{\sqrt{2}} $$
Combine terms under a common denominator:
$$ = \frac{1 - i + 1 + i}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} $$

∴ Proved; sum yields √2.

One-Page Revision Cheat Sheet

1

Complex Number Definition: Represented as $z = x+iy$ or $(x,y)$, uniting real and imaginary components.

2

Algebraic Operations: Add/subtract components individually. Multiply using expansion with the identity $i^2 = -1$. Divide using rationalization.

3

The Conjugate Concept: Flip only the sign of the imaginary coefficient. Geometrically, it reflects the coordinate over the X-axis.

4

The Modulus Formula: Calculates origin distance: $|z| = \sqrt{x^2 + y^2}$. Note that $z \cdot \bar{z} = |z|^2$.

5

Square Root Equation: $\pm \left[ \sqrt{\frac{|z|+a}{2}} + i \frac{b}{|b|} \sqrt{\frac{|z|-a}{2}} \right]$ matches signs dynamically with the sign of the imaginary coefficient $b$.

6

The Inverse iota Trick: Whenever you encounter $\frac{1}{i}$ in an expression, write it instantly as $-i$.

7

Common Sign Error: Never write $i^2 = 1$; always evaluate it strictly as $-1$ to prevent systematic errors in algebraic steps.