4.2 Complex Number Operations and Equations

4.2 Complex Number Operations and Equations

The system of complex numbers $\mathbb{C}$ forms an algebraically closed field. This means that not only can we perform standard arithmetic operations (addition, subtraction, multiplication, division) following the standard algebraic laws, but every non-constant polynomial equation with complex coefficients also has at least one root in $\mathbb{C}$ (the Fundamental Theorem of Algebra). In this section, we explore arithmetic operations, geometric transformations, powers, roots, and polynomial equations over $\mathbb{C}$.

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1. Basic Arithmetic Operations on Complex Numbers

Let $z_1 = a + bi$ and $z_2 = c + di$ ($a, b, c, d \in \mathbb{R}$).

Addition and Subtraction

Addition and subtraction are performed component-wise by grouping real and imaginary parts: $z_1 + z_2 = (a + c) + (b + d)i$ $z_1 - z_2 = (a - c) + (b - d)i$

Geometric Interpretation: In the complex plane, complex addition corresponds to vector addition $\vec{OP_1} + \vec{OP_2} = \vec{OP}$, obeying the parallelogram rule. Complex subtraction $z_1 - z_2$ corresponds to the vector $\vec{P_2 P_1}$, whose length $|z_1 - z_2|$ represents the Euclidean distance between $P_1$ and $P_2$.

Multiplication

Multiplication follows the distributive property, replacing $i^2$ with $-1$: $z_1 z_2 = (a + bi)(c + di) = ac + ad i + bc i + bd i^2 = (ac - bd) + (ad + bc)i$

Division

To divide $z_1$ by $z_2 \neq 0$, we rationalize the denominator by multiplying both numerator and denominator by the complex conjugate $\bar{z}_2 = c - di$: $\frac{z_1}{z_2} = \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}i$

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2. Algebraic Properties of Powers of $i$

The powers of the imaginary unit $i$ display a strict 4-periodicity: $i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$ In general, for any integer $k \in \mathbb{Z}$ and remainder $r \in {0, 1, 2, 3}$: $i^{4k + r} = i^r$

Crucial Summation Property:

The sum of any four consecutive integer powers of $i$ is identically zero: $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0 \quad (\forall n \in \mathbb{Z})$ Gaokao Technique: High-degree power expressions like $(1 + i)^n$ or $\sum_{k=1}^{100} i^k$ are simplified by using $i^4 = 1$, $(1 + i)^2 = 2i$, and $(1 - i)^2 = -2i$.

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3. Geometric Meaning of Multiplication and Rotation

Using exponential form $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$: $z_1 z_2 = (r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = (r_1 r_2) e^{i(\theta_1 + \theta_2)}$

Rotation-Dilation Theorem:

Multiplying a complex number $z$ by $w = r e^{i\theta}$ performs a rotation-dilation transformation in the complex plane:

1. Modulus Scaling: The distance from the origin is multiplied by $r = |w|$.
2. Angular Rotation: The vector is rotated counterclockwise by an angle $\theta = \arg(w)$.

Special Cases:

• Multiplying by $i = 1 \cdot e^{i\pi/2}$: Rotates the vector counterclockwise by $90^\circ$ ($\pi/2$ radians) without altering its length.
• Multiplying by $-1 = 1 \cdot e^{i\pi}$: Rotates the vector by $180^\circ$ ($\pi$ radians), equivalent to reflection through the origin.

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4. De Moivre's Theorem and $n$-th Roots

De Moivre's Theorem

For any real number $\theta$ and any integer $n \in \mathbb{Z}$: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ In exponential form, this is simply $(e^{i\theta})^n = e^{in\theta}$.

Roots of Complex Numbers

To solve $w^n = z = r e^{i\theta}$ for $w$: $w_k = \sqrt[n]{r} e^{i \frac{\theta + 2k\pi}{n}} = \sqrt[n]{r} \left( \cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta + 2k\pi}{n} \right), \quad k = 0, 1, \dots, n-1$

Roots of Unity ($z^n = 1$):

The $n$-th roots of unity are $\omega_k = e^{i \frac{2k\pi}{n}}$ for $k = 0, 1, \dots, n-1$. In the complex plane, these $n$ roots form the vertices of a regular $n$-gon inscribed in the unit circle.

Special Case: Cube Roots of Unity ($z^3 = 1$):

The three roots are $1, \omega, \omega^2$, where: $\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \frac{2\pi}{3}}, \quad \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i = e^{i \frac{4\pi}{3}}$ They satisfy the fundamental algebraic identities: $1 + \omega + \omega^2 = 0, \quad \omega^3 = 1, \quad \bar{\omega} = \omega^2$

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5. Complex Quadratic Equations

Consider the quadratic equation with real coefficients: $ax^2 + bx + c = 0 \quad (a, b, c \in \mathbb{R}, a \neq 0)$ The discriminant is $\Delta = b^2 - 4ac$.

• If $\Delta > 0$: Two distinct real roots $x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$.
• If $\Delta = 0$: One repeated real root $x_1 = x_2 = -\frac{b}{2a}$.
• If $\Delta < 0$: Two complex conjugate roots: $x_{1,2} = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}$

Vieta's Formulas continue to hold over $\mathbb{C}$: $z_1 + z_2 = -\frac{b}{a}, \quad z_1 z_2 = \frac{c}{a}$

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6. Worked Gaokao Exam Examples

Example 1: Quotient Simplification and Complex Equation

Problem: Simplify $z = \frac{(1 + 2i)(1 - i)}{1 + i}$, express $z$ in standard form $a + bi$, and solve for $w \in \mathbb{C}$ such that $w \cdot z + \bar{w} = 4 + i$.

Solution:

1. Simplify numerator: $(1 + 2i)(1 - i) = 1 - i + 2i - 2i^2 = 1 + i + 2 = 3 + i$
2. Perform division: $z = \frac{3 + i}{1 + i} = \frac{(3 + i)(1 - i)}{(1 + i)(1 - i)} = \frac{3 - 3i + i - i^2}{1^2 - i^2} = \frac{4 - 2i}{2} = 2 - i$
3. Solve equation for $w$: Let $w = x + yi$ ($x, y \in \mathbb{R}$). Then $\bar{w} = x - yi$. Substitute $z = 2 - i$ into $w \cdot z + \bar{w} = 4 + i$: $(x + yi)(2 - i) + (x - yi) = 4 + i$ Expand the product: $(2x + x i + 2y i - y i^2) + x - y i = (2x + y) + (2y - x)i + x - y i$ Group real and imaginary parts: $(3x + y) + (y - x)i = 4 + i$
4. Equate real and imaginary parts: $\begin{cases} 3x + y = 4 \\ -x + y = 1 \end{cases}$
5. Subtract second equation from first: $4x = 3 \implies x = \frac{3}{4}$ Substitute back to find $y$: $y = 1 + x = 1 + \frac{3}{4} = \frac{7}{4}$
6. Thus, $w = \frac{3}{4} + \frac{7}{4}i$.

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Example 2: High-Power Evaluation and Cyclicity

Problem: Evaluate the sum $S = \sum_{k=1}^{50} \left(\frac{1 + i}{\sqrt{2}}\right)^{2k}$.

Solution:

1. Let $u = \frac{1 + i}{\sqrt{2}}$. Notice that: $u^2 = \left(\frac{1 + i}{\sqrt{2}}\right)^2 = \frac{1 + 2i + i^2}{2} = \frac{2i}{2} = i$
2. Therefore, the term in the sum is $u^{2k} = (u^2)^k = i^k$.
3. The sum $S$ simplifies to: $S = \sum_{k=1}^{50} i^k = i + i^2 + i^3 + i^4 + \dots + i^{50}$
4. Since every 4 consecutive powers of $i$ sum to 0 ($i + i^2 + i^3 + i^4 = 0$): $50 = 4 \times 12 + 2$ The first 48 terms group into 12 sets of 4, each summing to 0.
5. The remaining terms are the 49th and 50th terms: $S = i^{49} + i^{50} = i^1 + i^2 = i - 1 = -1 + i$