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Complex Number Formulas — Rules, Table & Examples

Key Formula

Addition: (a+bi)+(c+di)=(a+c)+(b+d)iSubtraction: (a+bi)(c+di)=(ac)+(bd)iMultiplication: (a+bi)(c+di)=(acbd)+(ad+bc)iDivision: a+bic+di=(ac+bd)+(bcad)ic2+d2Modulus: a+bi=a2+b2Conjugate: a+bi=abi\begin{gathered}\text{Addition: } (a+bi)+(c+di) = (a+c)+(b+d)i\\\text{Subtraction: } (a+bi)-(c+di) = (a-c)+(b-d)i\\\text{Multiplication: } (a+bi)(c+di) = (ac-bd)+(ad+bc)i\\\text{Division: } \frac{a+bi}{c+di} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\\\text{Modulus: } |a+bi| = \sqrt{a^2+b^2}\\\text{Conjugate: } \overline{a+bi} = a-bi\end{gathered}
Where:
  • aa = Real part of the first complex number
  • bb = Imaginary part of the first complex number
  • cc = Real part of the second complex number
  • dd = Imaginary part of the second complex number
  • ii = The imaginary unit, where i² = −1

Worked Example

Problem: Multiply the complex numbers (3 + 2i) and (1 − 4i).
Step 1: Apply the multiplication formula (a+bi)(c+di)=(acbd)+(ad+bc)i(a+bi)(c+di) = (ac - bd) + (ad + bc)i. Here a=3a=3, b=2b=2, c=1c=1, d=4d=-4.
Step 2: Compute the real part: acbdac - bd.
(3)(1)(2)(4)=3+8=11(3)(1) - (2)(-4) = 3 + 8 = 11
Step 3: Compute the imaginary part: ad+bcad + bc.
(3)(4)+(2)(1)=12+2=10(3)(-4) + (2)(1) = -12 + 2 = -10
Step 4: Combine the real and imaginary parts.
(3+2i)(14i)=1110i(3+2i)(1-4i) = 11 - 10i
Answer: (3+2i)(14i)=1110i(3+2i)(1-4i) = 11 - 10i

Another Example

This example demonstrates division, which requires multiplying by the complex conjugate — a technique that is conceptually different from multiplication alone.

Problem: Divide (5 + 10i) by (3 + 4i).
Step 1: Write the division and identify that you must multiply numerator and denominator by the conjugate of the denominator, 3+4i=34i\overline{3+4i} = 3-4i.
5+10i3+4i34i34i\frac{5+10i}{3+4i} \cdot \frac{3-4i}{3-4i}
Step 2: Expand the denominator using the difference-of-squares pattern. Since zzˉ=c2+d2z \cdot \bar{z} = c^2 + d^2:
(3+4i)(34i)=32+42=9+16=25(3+4i)(3-4i) = 3^2 + 4^2 = 9 + 16 = 25
Step 3: Expand the numerator using the multiplication formula:
(5+10i)(34i)=(15+40)+(3020)i... let’s compute carefully:(5+10i)(3-4i) = (15+40) + (30-20)i \cdot \text{... let's compute carefully:}
Step 4: Numerator real part: ac+bd=(5)(3)+(10)(4)=15+40=55ac + bd = (5)(3) + (10)(4) = 15 + 40 = 55. Numerator imaginary part: bcad=(10)(3)(5)(4)=3020=10bc - ad = (10)(3) - (5)(4) = 30 - 20 = 10.
(5+10i)(34i)=55+10i(5+10i)(3-4i) = 55 + 10i
Step 5: Divide each part by 25.
55+10i25=5525+1025i=115+25i\frac{55 + 10i}{25} = \frac{55}{25} + \frac{10}{25}i = \frac{11}{5} + \frac{2}{5}i
Answer: 5+10i3+4i=115+25i\dfrac{5+10i}{3+4i} = \dfrac{11}{5} + \dfrac{2}{5}i

Frequently Asked Questions

What is the formula for multiplying two complex numbers?
To multiply (a+bi)(c+di)(a+bi)(c+di), use the formula (acbd)+(ad+bc)i(ac - bd) + (ad + bc)i. This comes from distributing (FOIL) and replacing i2i^2 with 1-1. The real part is acbdac - bd and the imaginary part is ad+bcad + bc.
How do you divide complex numbers?
To divide a+bic+di\frac{a+bi}{c+di}, multiply both the numerator and denominator by the conjugate of the denominator, cdic - di. This makes the denominator a real number c2+d2c^2 + d^2, and you can then separate the result into real and imaginary parts.
What is the modulus of a complex number and when do you use it?
The modulus of a+bia + bi is a+bi=a2+b2|a+bi| = \sqrt{a^2 + b^2}. It represents the distance from the origin to the point (a,b)(a, b) on the complex plane. You use it when converting to polar form, comparing magnitudes, or solving equations involving absolute values of complex numbers.

Rectangular Form (a + bi) vs. Polar Form (r(cos θ + i sin θ))

Rectangular Form (a + bi)Polar Form (r(cos θ + i sin θ))
Representationa + bi with real and imaginary partsr(cos θ + i sin θ) or r·cis θ
Best for addition/subtractionYes — simply add or subtract real and imaginary partsNo — must convert to rectangular first
Best for multiplication/divisionWorks but requires FOIL and simplificationYes — multiply moduli and add angles
Best for powers and rootsTedious for large exponentsYes — use De Moivre's Theorem: rⁿ cis(nθ)
Key formula(a+bi)(c+di) = (ac−bd)+(ad+bc)ir₁·r₂ · cis(θ₁ + θ₂)

Why It Matters

Complex number formulas appear throughout Algebra 2, precalculus, and physics courses whenever you work with quadratic equations that have no real solutions, AC circuit analysis, or signal processing. Mastering these formulas — especially conjugate multiplication for division — is essential for standardized tests and for later topics like De Moivre's Theorem and Euler's formula. They also form the algebraic backbone for understanding transformations in the complex plane.

Common Mistakes

Mistake: Forgetting that i² = −1 when multiplying, which leads to a sign error in the real part.
Correction: When you FOIL (a+bi)(c+di)(a+bi)(c+di), the term bdi2bdi^2 equals bd-bd, not +bd+bd. Always replace i2i^2 with 1-1 before combining terms.
Mistake: Dividing complex numbers by separately dividing real and imaginary parts (e.g., writing (6+4i)/(3+2i) = 2+2i).
Correction: Division does not work component-wise. You must multiply numerator and denominator by the conjugate of the denominator to clear the imaginary part from the denominator first.

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