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Arithmetic Operations on Complex Numbers

CSCA Arithmetic Operations on Complex Numbers study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

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Arithmetic Operations on Complex Numbers

Operations on complex numbers combine algebra and geometry. In the CSCA exam, **Division** (rationalizing the denominator) and **Multiplication** (using $i^2=-1$) are high-frequency calculation topics. Additionally, understanding the **geometric meaning** of addition and subtraction helps solve related geometric problems.

1. Core Concepts Review

* **Definition**: $z = a + bi$ ($a, b \in \mathbb{R}$), where $i$ is the imaginary unit and $i^2 = -1$.

* **Classification**:

* $b=0$: Real number.

* $b \neq 0$: Imaginary number.

* $a=0, b \neq 0$: Pure imaginary number.

* **Equality**: $a+bi = c+di \iff a=c$ and $b=d$.

2. Rules for Arithmetic Operations

Let $z_1 = a + bi$, $z_2 = c + di$.

#### 2.1 Addition and Subtraction

Follow the principle: "Real with Real, Imaginary with Imaginary".

* **Formula**: $(a+bi) \pm (c+di) = (a \pm c) + (b \pm d)i$

* **Geometric Meaning**: Corresponds to the **Parallelogram Rule** or Triangle Rule of vectors.

#### 2.2 Multiplication

Expand like polynomial multiplication, replacing $i^2$ with $-1$.

* **Derivation**:

$$(a+bi)(c+di) = ac + adi + bci + bdi^2$$

$$= (ac - bd) + (ad + bc)i$$

#### 2.3 Division

The core of division is **Rationalizing the Denominator**.

* **Method**: Multiply both the numerator and denominator by the **Conjugate** of the denominator.

* **Complex Conjugate**: The conjugate of $z = c+di$ is $\bar{z} = c-di$. Geometrically, they are symmetric about the Real axis.

* **Key Property**: $z \cdot \bar{z} = c^2 + d^2$ (The result is a real number).

* **Process**:

$$\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2}$$

3. Practice Examples

**Example 1** (Mixed Add/Sub): Calculate $(3+2i) - (4-3i) + (-1+5i)$.

**Solution**:

Real parts: $3 - 4 + (-1) = -2$

Imaginary parts: $2 - (-3) + 5 = 2 + 3 + 5 = 10$

Result: $-2 + 10i$

**Example 2** (Multiplication): Calculate $(2-3i)(1+4i)$.

**Solution**:

Expand: $2(1) + 2(4i) - 3i(1) - 3i(4i)$

$= 2 + 8i - 3i - 12i^2$

Note that $i^2 = -1$, so $-12i^2 = +12$.

$= (2+12) + (8-3)i = 14 + 5i$

**Example 3** (Division): Calculate $\frac{1+2i}{3-4i}$.

**Solution**:

The denominator is $3-4i$, its conjugate is $3+4i$.

$$\frac{1+2i}{3-4i} = \frac{(1+2i)(3+4i)}{(3-4i)(3+4i)}$$

Denominator: $3^2 + 4^2 = 9 + 16 = 25$

Numerator: $1(3) + 1(4i) + 2i(3) + 2i(4i) = 3 + 4i + 6i - 8 = -5 + 10i$

Result: $\frac{-5+10i}{25} = -\frac{1}{5} + \frac{2}{5}i$

4. Common Errors

* **Sign Errors**: Forgetting to change the sign when calculating $i^2$ (e.g., calculating $-3i \cdot 4i$ as $-12$ instead of $+12$).

* **Conjugate Confusion**: Forgetting to multiply the numerator by the conjugate as well.

* **Order of Operations**: Always multiply/divide before adding/subtracting.