Complex Numbers — MHT-CET Maths

Formulas (6)

• Powers of i and the Standard Form x + iy · Cycle of powers of i
  $$i^2 = -1,\quad i^3 = -i,\quad i^4 = 1,\qquad i^n = i^{\,n \bmod 4}$$
• The Conjugate: Rationalising a Denominator and Equating Conjugates · Conjugate identities
  $$z\bar z = |z|^2 \qquad \frac{z_1}{z_2} = \frac{z_1\bar z_2}{|z_2|^2} \qquad z_1 = \bar z_2 \iff \operatorname{Re}z_1 = \operatorname{Re}z_2,\ \operatorname{Im}z_1 = -\operatorname{Im}z_2$$
• Purely Real or Purely Imaginary: Set the Other Part to Zero · Real and imaginary conditions
  $$z \text{ purely imaginary} \iff \operatorname{Re}z = 0 \qquad z \text{ purely real} \iff \operatorname{Im}z = 0 \qquad \sin^2\theta = \sin^2\alpha \Rightarrow \theta = n\pi \pm \alpha$$
• Solving for z: Put z = x + iy and Equate Parts · Equating parts
  $$a + ib = c + id \iff a = c \text{ and } b = d \qquad \frac{1}{1 - i} = \frac{1 + i}{2}$$
• A Polynomial at a Complex x: Use Its Minimal Quadratic, and Cube Expansions · Minimal quadratic and the cube
  $$x = a + ib \Rightarrow x^2 - 2ax + (a^2 + b^2) = 0 \qquad (p + iq)^3 = (p^3 - 3pq^2) + i(3p^2q - q^3)$$
• Cube Roots of Unity: ω³ = 1 and 1 + ω + ω² = 0 · The two facts
  $$\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0, \qquad \omega^n = \omega^{\,n \bmod 3}, \qquad \bar\omega = \omega^2 = \frac{1}{\omega}$$

Watch out for (6)

• i³ = i, and other lapses in the cycle
  → Powers of i and the Standard Form x + iy
• Conjugating only the numerator
  → The Conjugate: Rationalising a Denominator and Equating Conjugates
• Setting the imaginary part to zero for 'purely imaginary'
  → Purely Real or Purely Imaginary: Set the Other Part to Zero
• Dividing by 1 − i without rationalising
  → Solving for z: Put z = x + iy and Equate Parts
• Substituting the complex number directly
  → A Polynomial at a Complex x: Use Its Minimal Quadratic, and Cube Expansions
• Stopping at ω⁴
  → Cube Roots of Unity: ω³ = 1 and 1 + ω + ω² = 0

Formulas (5)

• Modulus and Its Properties: |z₁z₂| = |z₁||z₂| · Modulus rules
  $$|z_1z_2| = |z_1||z_2| \qquad \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \qquad |z^n| = |z|^n \qquad |\sqrt z| = \sqrt{|z|}$$
• The Argument: Reference Angle Plus the Quadrant · Argument rules
  $$\arg(z_1z_2) = \arg z_1 + \arg z_2 \qquad \arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2 \qquad \arg\bar z = -\arg z$$
• Polar Form and De Moivre: Powers, Rotations and sin θ + i cos θ · De Moivre and the rotation by i
  $$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta \qquad iz \text{ is } z \text{ rotated by } \tfrac{\pi}{2} \qquad \sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta)$$
• |z| + z = a + ib: Equate the Imaginary Part, Then Solve for |z| · Closed form
  $$|z| + z = a + ib \ \Rightarrow\ \operatorname{Im}z = b,\quad |z| = \frac{a^2 + b^2}{2a}$$
• Find z From a Given Modulus: Simplify, Then Fix the Parameter · Modulus first, algebra second
  $$\left|\frac{(1 + i)^2}{a - i}\right| = \frac{2}{\sqrt{a^2 + 1}} \qquad |2 + \cos\theta + i\sin\theta|^2 = 5 + 4\cos\theta$$

Watch out for (5)

• Expanding the product to find its modulus
  → Modulus and Its Properties: |z₁z₂| = |z₁||z₂|
• The argument from the ratio alone
  → The Argument: Reference Angle Plus the Quadrant
• Applying De Moivre to sin θ + i cos θ
  → Polar Form and De Moivre: Powers, Rotations and sin θ + i cos θ
• Answering the twin sitting's value
  → |z| + z = a + ib: Equate the Imaginary Part, Then Solve for |z|
• z or z̄?
  → Find z From a Given Modulus: Simplify, Then Fix the Parameter

Formulas (4)

• |z − a| = r Is a Circle: Centre a, Radius r · Circle in the Argand plane
  $$|z - a| = r \iff \text{circle, centre } a,\ \text{radius } r \qquad \left|\frac{z - a}{z - b}\right| = k \ne 1 \iff \text{circle (Apollonius)}$$
• |z − a| = |z − b| Is the Perpendicular Bisector of ab · Line from equal distances
  $$|z - a| = |z - b| \iff \text{perpendicular bisector of } a, b \qquad |z - a| - |z - b| = |a - b| \ \text{ is a ray (degenerate hyperbola)}$$
• Re of a Quotient Equals Zero: Rationalise, Then Read the Circle · Purely imaginary quotient
  $$\operatorname{Re}\frac{z - 1}{2z + 1} = 0 \iff 2x^2 + 2y^2 - x - 1 = 0 \iff \left(x - \tfrac14\right)^2 + y^2 = \tfrac{9}{16}$$
• Greatest and Least |z| on a Disc: |a| + r and |a| − r · Extreme modulus on a disc
  $$|z - a| \le r:\quad |a| - r \le |z| \le |a| + r \qquad \max|z| - \min|z| = 2r \ (|a| \ge r)$$

Watch out for (4)

• Reading |z + 1| as centred at +1
  → |z − a| = r Is a Circle: Centre a, Radius r
• Calling every modulus locus a circle
  → |z − a| = |z − b| Is the Perpendicular Bisector of ab
• Reporting the radius squared
  → Re of a Quotient Equals Zero: Rationalise, Then Read the Circle
• Answering 2√5 for the difference
  → Greatest and Least |z| on a Disc: |a| + r and |a| − r