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Mathematics · Ch 4 — Complex Numbers and Quadratic Equations

Complex Numbers

4.2

Complex Numbers

The Need for a New Number

You already know that the equation x2+1=0x^2 + 1 = 0 has no real solution — no real number, when squared, gives −1-1. To solve such equations, we extend the real number system by introducing a new symbol.

We denote −1\sqrt{-1} by the symbol ii. This gives us the defining property:

i2=−1i^2 = -1

So ii is a solution of x2+1=0x^2 + 1 = 0. With this single new number, we can now build an entirely new system of numbers.

Definition of a Complex Number

A complex number is any number of the form a+iba + ib, where aa and bb are real numbers, and i=−1i = \sqrt{-1}.

For example, each of the following is a complex number:

  • 2+i32 + i3
  • (−1)+i3(-1) + i\sqrt{3}
  • 14−114i\frac{1}{4} - \frac{11}{4}i
Note

The form a+iba + ib is called the standard form or Cartesian form of a complex number. The order matters — aa comes first, then ibib.

Real and Imaginary Parts

For a complex number z=a+ibz = a + ib:

  • aa is called the real part of zz, written as Re(z)\text{Re}(z)
  • bb is called the imaginary part of zz, written as Im(z)\text{Im}(z)
Watch out

The imaginary part is bb, not ibib. If z=2+i5z = 2 + i5, then Im(z)=5\text{Im}(z) = 5, not i5i5.

Example: If z=2+i5z = 2 + i5, then Re(z)=2\text{Re}(z) = 2 and Im(z)=5\text{Im}(z) = 5.

Equality of Two Complex Numbers

Two complex numbers z1=a+ibz_1 = a + ib and z2=c+idz_2 = c + id are equal if and only if their real parts are equal and their imaginary parts are equal. That is:

a+ib=c+id  ⟺  a=c and b=da + ib = c + id \iff a = c \text{ and } b = d

This is a crucial property — a single equation in complex numbers gives two separate equations in real numbers.

Important

Equality of complex numbers gives two real equations simultaneously. This is how we solve for unknown real numbers in complex equations.

Worked Example (from the textbook)

Example 1: If 4x+i(3x−y)=3+i(−6)4x + i(3x - y) = 3 + i(-6), where xx and yy are real numbers, find the values of xx and yy.

Solution:

We have:

4x+i(3x−y)=3+i(−6)4x + i(3x - y) = 3 + i(-6)

Equating the real parts:

4x=34x = 3

Equating the imaginary parts:

3x−y=−63x - y = -6

From the first equation:

x=34x = \frac{3}{4}

Substitute into the second equation:

3(34)−y=−63\left(\frac{3}{4}\right) - y = -6 …