Addition, subtraction, and multiplication of complex numbers

LVL: FREE

MODULE: Pre-Calculus (Functions and Series)

[EXEC: MICRO_CORE]

✖️ 1. Adding and subtracting complex numbers algebraically by combining like parts

➕ Adding and Subtracting Complex Numbers

  • Treat real parts and imaginary parts as separate groups.
  • Add or subtract the real parts together.
  • Add or subtract the imaginary parts together.
  • Write the result in standard form a+bia + bi.

Example: (3+4i)+(2−5i)=(3+2)+(4−5)i=5−i(3 + 4i) + (2 - 5i) = (3+2) + (4-5)i = 5 - i

💡 Think of it like combining x-terms and y-terms in algebra.

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1. Adding and subtracting complex numbers algebraically by combining like parts

Adding and Subtracting Complex Numbers

Complex numbers in the form a+bia + bi are added or subtracted by combining their real parts separately from their imaginary parts. This operation treats real and imaginary components as independent dimensions that do not mix.

Intuition: Just as vectors are added component-wise, complex numbers combine horizontally (real axis) and vertically (imaginary axis) independently.

Core Rules:

  • For (a+bi)+(c+di)(a + bi) + (c + di), compute (a+c)+(b+d)i(a + c) + (b + d)i
  • For (a+bi)−(c+di)(a + bi) - (c + di), compute (a−c)+(b−d)i(a - c) + (b - d)i
  • Real parts combine with real parts only; imaginary parts combine with imaginary parts only
  • The result is always another complex number in standard form

Consequence: Addition and subtraction preserve the algebraic structure, making complex numbers a closed system under these operations.

Example: (3+5i)+(2−7i)=(3+2)+(5−7)i=5−2i(3 + 5i) + (2 - 7i) = (3 + 2) + (5 - 7)i = 5 - 2i

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✖️ 2. Evaluating higher powers of ii using the repetitive 4-step cycle

🔄 Powers of i

  • The powers of ii repeat every 4 steps: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1.
  • To find ini^n, divide the exponent by 4 and use the remainder.
  • Remainder 0 gives 1, remainder 1 gives ii, remainder 2 gives −1-1, remainder 3 gives −i-i.

Example: i23=i20+3=(i4)5⋅i3=1⋅(−i)=−ii^{23} = i^{20+3} = (i^4)^5 \cdot i^3 = 1 \cdot (-i) = -i

💡 The cycle is: i → -1 → -i → 1 → repeat forever.

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2. Evaluating higher powers of ii using the repetitive 4-step cycle

Powers of the Imaginary Unit

The imaginary unit ii satisfies i2=−1i^2 = -1, which generates a repeating cycle of four values for successive powers: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1, then the pattern repeats.

Intuition: Each multiplication by ii represents a 90-degree counterclockwise rotation in the complex plane, so four rotations return to the starting position.

Core Rules:

  • Compute ini^n by finding the remainder when nn is divided by 4
  • If n≡0(mod4)n \equiv 0 \pmod{4}: in=1i^n = 1
  • If n≡1(mod4)n \equiv 1 \pmod{4}: in=ii^n = i
  • If n≡2(mod4)n \equiv 2 \pmod{4}: in=−1i^n = -1
  • If n≡3(mod4)n \equiv 3 \pmod{4}: in=−ii^n = -i

Consequence: Any power of ii reduces to one of four values, enabling simplification of expressions involving high exponents.

Example: i23=i23 mod 4=i3=−ii^{23} = i^{23 \bmod 4} = i^3 = -i

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✖️ 3. Multiplying complex numbers and introducing its interpretation as scaling and rotation

✖️ Multiplying Complex Numbers

  • Use FOIL just like binomials: First, Outer, Inner, Last.
  • Remember that i2=−1i^2 = -1 when simplifying.
  • Combine real parts and imaginary parts at the end.
  • Geometrically, multiplication scales and rotates the complex plane.

Example: (2+3i)(1−4i)=2−8i+3i−12i2=2−5i+12=14−5i(2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i^2 = 2 - 5i + 12 = 14 - 5i

💡 FOIL works, but remember i2i^2 becomes −1-1.

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3. Multiplying complex numbers and introducing its interpretation as scaling and rotation

Multiplication of Complex Numbers

Multiplying (a+bi)(c+di)(a + bi)(c + di) uses the distributive property (FOIL) and the identity i2=−1i^2 = -1 to obtain (ac−bd)+(ad+bc)i(ac - bd) + (ad + bc)i. Geometrically, this operation combines scaling by magnitude and rotation by argument.

Intuition: Multiplication stretches (or shrinks) one complex number by the magnitude of the other while rotating it by the other's angle from the positive real axis.

Core Rules:

  • Expand using FOIL: (a+bi)(c+di)=ac+adi+bci+bdi2(a + bi)(c + di) = ac + adi + bci + bdi^2
  • Substitute i2=−1i^2 = -1 to get ac+adi+bci−bdac + adi + bci - bd
  • Combine like terms: (ac−bd)+(ad+bc)i(ac - bd) + (ad + bc)i
  • Magnitudes multiply; arguments add (polar form interpretation)

Consequence: Multiplication is not commutative in geometry (order affects rotation direction) but is commutative algebraically.

Example: (2+3i)(1+4i)=2+8i+3i+12i2=2+11i−12=−10+11i(2 + 3i)(1 + 4i) = 2 + 8i + 3i + 12i^2 = 2 + 11i - 12 = -10 + 11i

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✖️ 4. Defining the complex conjugate and executing the formal division algorithm

🔁 Complex Conjugate and Division

  • The conjugate of a+bia + bi is a−bia - bi (flip the sign of the imaginary part).
  • To divide, multiply numerator and denominator by the conjugate of the denominator.
  • This eliminates ii from the denominator (rationalization).
  • Simplify and write in standard form a+bia + bi.

Example: 3+2i1−i=(3+2i)(1+i)(1−i)(1+i)=3+5i+2i21−i2=1+5i2\frac{3 + 2i}{1 - i} = \frac{(3+2i)(1+i)}{(1-i)(1+i)} = \frac{3 + 5i + 2i^2}{1 - i^2} = \frac{1 + 5i}{2}

💡 Conjugate flips the imaginary sign; multiply to clear the denominator.

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4. Defining the complex conjugate and executing the formal division algorithm

Complex Conjugate and Division

The complex conjugate of z=a+biz = a + bi is z‾=a−bi\overline{z} = a - bi, obtained by negating the imaginary part. Division a+bic+di\frac{a + bi}{c + di} is performed by multiplying numerator and denominator by the conjugate of the denominator to eliminate imaginary parts from the denominator.

Intuition: Multiplying by the conjugate exploits the identity (c+di)(c−di)=c2+d2(c + di)(c - di) = c^2 + d^2, a real number, rationalizing the denominator.

Core Rules:

  • Conjugate property: z⋅z‾=a2+b2z \cdot \overline{z} = a^2 + b^2 (always real and non-negative)
  • Multiply both parts by c+di‾=c−di\overline{c + di} = c - di
  • Simplify: (a+bi)(c−di)c2+d2\frac{(a + bi)(c - di)}{c^2 + d^2}
  • Express in standard form x+yix + yi

Consequence: Division is always defined except when the denominator is zero (when c=d=0c = d = 0).

Example: 3+2i1+i=(3+2i)(1−i)12+12=3−3i+2i−2i22=5−i2=52−12i\frac{3 + 2i}{1 + i} = \frac{(3 + 2i)(1 - i)}{1^2 + 1^2} = \frac{3 - 3i + 2i - 2i^2}{2} = \frac{5 - i}{2} = \frac{5}{2} - \frac{1}{2}i

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✖️ 5. Applications: Calculating electrical impedance in alternating current circuit analysis

⚡ Electrical Impedance in AC Circuits

  • Impedance Z=R+jXZ = R + jX combines resistance RR and reactance XX.
  • Engineers use jj instead of ii to avoid confusion with current.
  • Add impedances in series by adding complex numbers.
  • Multiply by conjugate to find current: I=VZI = \frac{V}{Z}.

Example: If Z=3+4jZ = 3 + 4j ohms and V=10V = 10 volts, then I=103+4j=10(3−4j)25=1.2−1.6jI = \frac{10}{3+4j} = \frac{10(3-4j)}{25} = 1.2 - 1.6j amperes

💡 Complex numbers model AC circuits where phase matters.

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5. Applications: Calculating electrical impedance in alternating current circuit analysis

Electrical Impedance in AC Circuits

In alternating current (AC) analysis, impedance Z=R+jXZ = R + jX represents opposition to current flow, where RR is resistance (real part) and XX is reactance (imaginary part). Engineers use jj instead of ii to avoid confusion with current notation.

Intuition: Resistors dissipate energy (real), while capacitors and inductors store/release energy with phase shifts (imaginary), making complex numbers natural for modeling AC behavior.

Core Rules:

  • Total impedance in series: Ztotal=Z1+Z2+⋯Z_{\text{total}} = Z_1 + Z_2 + \cdots (add complex numbers)
  • Total impedance in parallel: 1Ztotal=1Z1+1Z2+⋯\frac{1}{Z_{\text{total}}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \cdots (use division algorithm)
  • Magnitude ∣Z∣=R2+X2|Z| = \sqrt{R^2 + X^2} gives overall opposition
  • Phase angle θ=arctan⁡(X/R)\theta = \arctan(X/R) indicates current-voltage timing shift

Consequence: Complex arithmetic enables precise calculation of voltage, current, and power in AC systems.

Example: For Z1=3+4jZ_1 = 3 + 4j ohms and Z2=1−2jZ_2 = 1 - 2j ohms in series, Ztotal=4+2jZ_{\text{total}} = 4 + 2j ohms.

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