Complex Numbers Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Complex Numbers.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Numbers of the form a+bia + bi where a,ba, b are real and i=โˆ’1i = \sqrt{-1}; they extend the real numbers to solve x2=โˆ’1x^2 = -1.

Extending numbers into a second dimension to solve equations like x2=โˆ’1x^2 = -1.

Read the full concept explanation โ†’

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: A complex number a+bia+bi adds a sideways imaginary axis so that x2=โˆ’1x^2=-1 finally has a solution.

Common stuck point: The procedure for complex numbers is the easy part; the trap is forgetting i2=โˆ’1i^2 = -1 and leaving it unsimplified. Asking "Does the problem require the square root of a negative number, i=โˆ’1i=\sqrt{-1}?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Does the problem require the square root of a negative number, i=โˆ’1i=\sqrt{-1}?

Worked Examples

Example 1

easy
Simplify i2i^2, i3i^3, and i4i^4.

Answer

i2=โˆ’1,i3=โˆ’i,i4=1i^2 = -1, \quad i^3 = -i, \quad i^4 = 1

First step

1
i2=โˆ’1i^2 = -1 by definition of the imaginary unit.

Full solution

  1. 2
    i3=i2โ‹…i=(โˆ’1)โ‹…i=โˆ’ii^3 = i^2 \cdot i = (-1) \cdot i = -i.
  2. 3
    i4=i3โ‹…i=(โˆ’i)โ‹…i=โˆ’i2=โˆ’(โˆ’1)=1i^4 = i^3 \cdot i = (-i) \cdot i = -i^2 = -(-1) = 1.
The powers of ii cycle with period 4: i,โˆ’1,โˆ’i,1,i,โˆ’1,โˆ’i,1,โ€ฆi, -1, -i, 1, i, -1, -i, 1, \ldots Knowing this cycle allows rapid simplification of any power of ii by finding the remainder when the exponent is divided by 4.

Example 2

medium
Multiply (3+2i)(1โˆ’i)(3 + 2i)(1 - i) and write the result in standard form a+bia + bi.

Example 3

medium
Multiply (4+3i)(2โˆ’i)(4 + 3i)(2 - i).

Example 4

medium
Find the product of z=5+12iz = 5 + 12i with its conjugate zห‰\bar{z}.

Example 5

hard
Find the reciprocal of 3โˆ’4i3 - 4i in a+bia + bi form.

Example 6

challenge
Show that โˆฃz1z2โˆฃ=โˆฃz1โˆฃโ€‰โˆฃz2โˆฃ|z_1 z_2| = |z_1|\,|z_2| for z1=1+2iz_1 = 1 + 2i and z2=3โˆ’iz_2 = 3 - i.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Add (4+3i)+(2โˆ’5i)(4 + 3i) + (2 - 5i).

Example 2

medium
Simplify i23i^{23}.

Example 3

easy
What is i2i^2?

Example 4

easy
Write โˆ’9\sqrt{-9} in the form bibi.

Example 5

easy
Identify the real and imaginary parts of 4+7i4 + 7i.

Example 6

easy
Compute i4i^4.

Example 7

easy
Add (2+3i)+(1+5i)(2 + 3i) + (1 + 5i).

Example 8

easy
Compute the magnitude โˆฃ3+4iโˆฃ|3 + 4i|.

Example 9

easy
Compute i3i^3.

Example 10

easy
Subtract (5+2i)โˆ’(3+6i)(5 + 2i) - (3 + 6i).

Example 11

medium
Multiply (2+i)(3+2i)(2 + i)(3 + 2i).

Example 12

medium
Compute the modulus โˆฃ5โˆ’12iโˆฃ|5 - 12i|.

Example 13

medium
Compute i10i^{10}.

Example 14

medium
Find the complex conjugate of 7โˆ’3i7 - 3i and its product with the original.

Example 15

medium
Solve x2=โˆ’25x^2 = -25 over the complex numbers.

Example 16

medium
Compute (1+i)2(1 + i)^2.

Example 17

medium
Add (0+4i)+(6+0i)(0 + 4i) + (6 + 0i) and give the real and imaginary parts.

Example 18

medium
Is โˆฃโˆ’3โˆ’4iโˆฃ|{-3} - 4i| equal to โˆฃ3+4iโˆฃ|3 + 4i|? Justify.

Example 19

challenge
Divide 4+2i1+i\frac{4 + 2i}{1 + i} and write in a+bia+bi form.

Example 20

challenge
Compute i1+i2+i3+i4i^{1} + i^2 + i^3 + i^4.

Example 21

challenge
If โˆฃa+biโˆฃ=10|a + bi| = 10 and a=6a = 6 with b>0b>0, find bb.

Example 22

medium
Multiply iโ‹…(2+3i)i\cdot(2 + 3i).

Example 23

easy
Write โˆ’16\sqrt{-16} in the form bibi.

Example 24

easy
Add (7+2i)+(โˆ’3+4i)(7 + 2i) + (-3 + 4i).

Example 25

easy
Subtract (8โˆ’i)โˆ’(3+4i)(8 - i) - (3 + 4i).

Example 26

easy
Compute the magnitude โˆฃ6+8iโˆฃ|6 + 8i|.

Example 27

easy
Find the complex conjugate of โˆ’2+5i-2 + 5i.

Example 28

medium
Compute (3โˆ’2i)2(3 - 2i)^2.

Example 29

medium
Compute โˆฃ7โˆ’24iโˆฃ|7 - 24i|.

Example 30

medium
Solve x2+16=0x^2 + 16 = 0 over the complex numbers.

Example 31

medium
Solve x2โˆ’2x+5=0x^2 - 2x + 5 = 0 over the complex numbers.

Example 32

medium
Compute (2+i)+(3โˆ’4i)โˆ’(1+2i)(2 + i) + (3 - 4i) - (1 + 2i).

Example 33

medium
Multiply iโ‹…(4โˆ’5i)i\cdot(4 - 5i).

Example 34

hard
Divide 3+2i1โˆ’i\frac{3 + 2i}{1 - i} and write in a+bia + bi form.

Example 35

hard
Compute (1+i)4(1 + i)^4.

Example 36

hard
If z=2+3iz = 2 + 3i, compute z2โˆ’4z+13z^2 - 4z + 13.

Example 37

hard
If โˆฃa+biโˆฃ=13|a + bi| = 13 and a=5a = 5 with b>0b > 0, find bb.

Example 38

challenge
Find all complex zz with z2=โˆ’1+i3โ‹…0z^2 = -1 + i\sqrt{3} \cdot 0... actually, find all complex zz with z2=2iz^2 = 2i.

Background Knowledge

These ideas may be useful before you work through the harder examples.

real numbersquadratic formula