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Complex Numbers & Argand Geometry

Axiomatic Definition

A complex number $z \in \mathbb{C}$ is an ordered pair of real numbers $(x, y) \in \mathbb{R}^2$, written in algebraic form as $z = x + iy$, where $i = \sqrt{-1}$ denotes the imaginary unit satisfying $i^2 = -1$.


🚦 Step 0 Decision Tree: Complex Algebra vs Geometry

🚦 Step 0 Decision TreeDeciding Approach: Algebraic Substitution (x+iy) vs Argand Vector Rotation

Click on your mathematical problem pattern below to instantly reveal the optimal, lowest-algebra solution path:

📐Optimal Solution Blueprint for: Definite Integral with Trigonometric Fraction
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Step 1: Check King’s Symmetry:Apply King’s property: I=abf(a+bx)dxI = \int_a^b f(a+b-x) dx. Write 2I=ab[f(x)+f(a+bx)]dx2I = \int_a^b [f(x) + f(a+b-x)] dx.
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Step 2: Add Integrals:Summing the integrands usually cancels the denominator completely, leaving ab1dx=ba\int_a^b 1 dx = b - a.
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Step 3: Solve for I:2I=ba    I=ba22I = b - a \implies I = \frac{b - a}{2}. Done in 3 lines without finding an antiderivative!
⚡ Pro-Tip / Exam Shortcut: For 0πxf(sinx)dx\int_0^\pi x f(\sin x) dx, King’s rule instantly eliminates the linear xx term, converting it to π20πf(sinx)dx\frac{\pi}{2} \int_0^\pi f(\sin x) dx.

1. Algebraic & Polar Representations

Every complex number z=x+iy can be uniquely expressed in:

z=r(cosθ+isinθ)=reiθ

where:

  • Modulus: |z|=r=x2+y20
  • Principal Argument: Arg(z)=θ(π,π]

Fundamental Properties of Modulus and Conjugate

For z,z1,z2C:

  1. zz¯=|z|2
  2. z1±z2=z¯1±z¯2 and z1z2=z¯1z¯2
  3. |z1z2|=|z1||z2| and |z1z2|=|z1||z2| (z20)
  4. Triangle Inequalities:||z1||z2|||z1±z2||z1|+|z2|Equality holds on the right iff Arg(z1)=Arg(z2) (origin, z1,z2 collinear in same ray).

2. Rotation Theorem (Coni's Theorem)

Vector Rotation in the Argand Plane

Let $z_1, z_2, z_3$ be complex numbers representing points $A, B, C$. The directed vector $\vec{BA} = z_1 - z_2$ rotated through an angle $\alpha$ counter-clockwise to align along $\vec{BC} = z_3 - z_2$ satisfies: $$\frac{z_3 - z_2}{z_1 - z_2} = \frac{|z_3 - z_2|}{|z_1 - z_2|} e^{i\alpha}$$

Condition for Equilateral Triangle

Points z1,z2,z3 form the vertices of an equilateral triangle iff:

z12+z22+z32=z1z2+z2z3+z3z11z1z2+1z2z3+1z3z1=0

3. 4-Tier Scaffolded Hint Drill

💡 4-Tier Scaffolded Hint SystemOlympiad Drill: Unimodular Sum Inverse
Try solving with Hint 1 before unlocking deeper tiers!