Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

Complex numbers extend the real number system by adding the imaginary unit ii, where i2=1i^2 = -1. This cheat sheet helps students recognize, graph, simplify, and operate with numbers in the form a+bia + bi. It is useful for algebra, precalculus, and any topic involving roots, quadratics, vectors, or rotations.

Students need these rules because complex numbers appear whenever equations have no real solutions.

Key Facts

  • A complex number in standard form is z=a+biz = a + bi, where aa is the real part and bb is the imaginary part.
  • The imaginary unit is defined by i=1i = \sqrt{-1} and i2=1i^2 = -1.
  • Powers of ii repeat in a cycle: i1=ii^1 = i, i2=1i^2 = -1, i3=ii^3 = -i, and i4=1i^4 = 1.
  • To add or subtract complex numbers, combine real parts and imaginary parts: (a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i.
  • To multiply complex numbers, distribute and use i2=1i^2 = -1: (a+bi)(c+di)=(acbd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i.
  • The complex conjugate of z=a+biz = a + bi is z=abi\overline{z} = a - bi.
  • The modulus of z=a+biz = a + bi is z=a2+b2|z| = \sqrt{a^2 + b^2}.
  • To divide complex numbers, multiply by the conjugate of the denominator: a+bic+dicdicdi\frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}.

Vocabulary

Complex Number
A number that can be written as a+bia + bi, where aa and bb are real numbers and i2=1i^2 = -1.
Imaginary Unit
The number ii defined by i=1i = \sqrt{-1}, so i2=1i^2 = -1.
Real Part
The real part of z=a+biz = a + bi is aa, written as Re(z)=a\operatorname{Re}(z) = a.
Imaginary Part
The imaginary part of z=a+biz = a + bi is bb, written as Im(z)=b\operatorname{Im}(z) = b.
Complex Conjugate
The complex conjugate of a+bia + bi is abia - bi, which changes the sign of the imaginary part.
Modulus
The modulus z|z| is the distance from the origin to the point z=a+biz = a + bi on the Argand plane.

Common Mistakes to Avoid

  • Forgetting that i2=1i^2 = -1, which makes products like (3i)(4i)(3i)(4i) equal to 12-12, not 12i12i or 12i212i^2.
  • Combining real and imaginary terms as if they are like terms, which is wrong because 5+2i5 + 2i cannot simplify to 7i7i or 77.
  • Using the wrong sign for the conjugate, which is wrong because the conjugate of 4+7i-4 + 7i is 47i-4 - 7i, not 47i4 - 7i.
  • Dividing without rationalizing the complex denominator, which leaves a denominator containing ii instead of converting the quotient to standard form a+bia + bi.
  • Graphing a+bia + bi as (b,a)(b, a), which is wrong because the real part aa is the horizontal coordinate and the imaginary part bb is the vertical coordinate.

Practice Questions

  1. 1 Simplify (43i)+(7+9i)(4 - 3i) + (7 + 9i) and write the answer in standard form a+bia + bi.
  2. 2 Multiply (2+5i)(34i)(2 + 5i)(3 - 4i) and simplify using i2=1i^2 = -1.
  3. 3 Find the modulus and conjugate of z=6+8iz = -6 + 8i.
  4. 4 Explain why multiplying a complex number by its conjugate always gives a real number.