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Use the Complex Number System

Evaluate the square root of a negative number

Evaluate the Square Root of a Negative Number

Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. For example, to simplify \(\sqrt{-1},\) we are looking for a real number x so that x2 = –1. Since all real numbers squared are positive numbers, there is no real number that equals –1 when squared.

Mathematicians have often expanded their numbers systems as needed. They added 0 to the counting numbers to get the whole numbers. When they needed negative balances, they added negative numbers to get the integers. When they needed the idea of parts of a whole they added fractions and got the rational numbers. Adding the irrational numbers allowed numbers like \(\sqrt{5}.\) All of these together gave us the real numbers and so far in your study of mathematics, that has been sufficient.

But now we will expand the real numbers to include the square roots of negative numbers. We start by defining the imaginary unit \(i\) as the number whose square is –1.

We will use the imaginary unit to simplify the square roots of negative numbers.

We will use this definition in the next example. Be careful that it is clear that the i is not under the radical. Sometimes you will see this written as \(\sqrt{\text{-}b}=i\sqrt{b}\) to emphasize the i is not under the radical. But the \(\sqrt{\text{-}b}=\sqrt{b}\ i\) is considered standard form.

Example

Try it.

Write each expression in terms of i and simplify if possible:

ⓐ \(\sqrt{-25}\) ⓑ \(\sqrt{-7}\) ⓒ \(\sqrt{-12}.\)

Solution

ⓐ

\(\sqrt{-25}\)
Use the definition of the square root of negative numbers.\(\sqrt{25}\ i\)
Simplify.\(5i\)
ⓑ
\(\sqrt{-7}\)
Use the definition of the square root of negative numbers.\(\sqrt{7}i\)
Simplify.Be careful that it is clear that \(i\) is not under the radical sign.
ⓒ
\(\sqrt{-12}\)
Use the definition of the square root of negative numbers.\(\sqrt{12}\ i\)
Simplify \(\sqrt{12}.\)\(2\sqrt{3}\ i\)

A complex number is in standard form when written as \(a+bi,\) where a and b are real numbers.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Add or Subtract Complex Numbers

We are now ready to perform the operations of addition, subtraction, multiplication and division on the complex numbers—just as we did with the real numbers.

Adding and subtracting complex numbers is much like adding or subtracting like terms. We add or subtract the real parts and then add or subtract the imaginary parts. Our final result should be in standard form.

Example

Try it.

Add: \(\sqrt{-12}+\sqrt{-27}.\)

Solution

\(\sqrt{-12}+\sqrt{-27}\)
Use the definition of the square root of negative numbers.\(\sqrt{12}\ i+\sqrt{27}\ i\)
Simplify the square roots.\(2\sqrt{3}\ i+3\sqrt{3}\ i\)
Add.\(5\sqrt{3}\ i\)

Remember to add both the real parts and the imaginary parts in this next example.

Example

Try it.

Simplify: ⓐ \((4-3i)+(5+6i)\) ⓑ \((2-5i)-(5-2i).\)

Solution

ⓐ

\((4-3i)+(5+6i)\)
Use the Associative Property to put the real
parts and the imaginary parts together.		\((4+5)+(-3i+6i)\)
Simplify.\(9+3i\)
ⓑ
\((2-5i)-(5-2i)\)
Distribute.\(2-5i-5+2i\)
Use the Associative Property to put the real
parts and the imaginary parts together.		\(2-5-5i+2i\)
Simplify.\(-3-3i\)

Multiply Complex Numbers

Multiplying complex numbers is also much like multiplying expressions with coefficients and variables. There is only one special case we need to consider. We will look at that after we practice in the next two examples.

Example

Try it.

Multiply: \(2i(7-5i).\)

Solution

\(2i(7-5i)\)
Distribute.\(14i-10{i}^{2}\)
Simplify \({i}^{2}.\)\(14i-10(-1)\)
Multiply.\(14i+10\)
Write in standard form.\(10+14i\)

In the next example, we multiply the binomials using the Distributive Property or FOIL.

Example

Try it.

Multiply: \((3+2i)(4-3i).\)

Solution

\((3+2i)(4-3i)\)
Use FOIL.\(12-9i+8i-6{i}^{2}\)
Simplify \({i}^{2}\) and combine like terms.\(12-i-6(-1)\)
Multiply.\(12-i+6\)
Combine the real parts.\(18-i\)

In the next example, we could use FOIL or the Product of Binomial Squares Pattern.

Example

Try it.

Multiply: \({(3+2i)}^{2}\)

Solution
Use the Product of Binomial Squares Pattern, \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}.\)
Simplify.
Simplify \({i}^{2}.\)
Simplify.

Since the square root of a negative number is not a real number, when we have the square roots of two negative numbers, we cannot use the Product Property for Radicals. In order to multiply square roots of negative numbers we should first write them as complex numbers, using \(\sqrt{\text{-}b}=\sqrt{b}i.\) This is one place students tend to make errors, so be careful when you see multiplying with a negative square root.

Example

Try it.

Multiply: \(\sqrt{-36}\cdot \sqrt{-4}.\)

Solution

To multiply square roots of negative numbers, we first write them as complex numbers.

\(\sqrt{-36}\cdot \sqrt{-4}\)
Write as complex numbers using \(\sqrt{\text{-}b}=\sqrt{b}i.\)\(\sqrt{36}\ i\cdot \sqrt{4}\ i\)
Simplify.\(6i\cdot 2i\)
Multiply.\(12{i}^{2}\)
Simplify \({i}^{2}\) and multiply.\(-12\)

In the next example, each binomial has a square root of a negative number. Before multiplying, each square root of a negative number must be written as a complex number.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Divide Complex Numbers

Dividing complex numbers is much like rationalizing a denominator. We want our result to be in standard form with no imaginary numbers in the denominator.

How to Divide Complex Numbers

Try it.

Divide: \(\frac{4+3i}{3-4i}.\)

Solution

We summarize the steps here.

Example

Try it.

Divide, writing the answer in standard form: \(\frac{-3}{5+2i}.\)

Solution

\(\frac{-3}{5+2i}\)
Multiply the numerator and denominator by the
complex conjugate of the denominator.		\(\frac{-3(5-2i)}{(5+2i)(5-2i)}\)
Multiply in the numerator and use the Product of
Complex Conjugates Pattern in the denominator.		\(\frac{-15+6i}{{5}^{2}+{2}^{2}}\)
Simplify.\(\frac{-15+6i}{29}\)
Write in standard form.\(-\frac{15}{29}+\frac{6}{29}i\)

Be careful as you find the conjugate of the denominator.

Example

Try it.

Divide: \(\frac{5+3i}{4i}.\)

Solution

\(\frac{5+3i}{4i}\)
Write the denominator in standard form.\(\frac{5+3i}{0+4i}\)
Multiply the numerator and denominator by
the complex conjugate of the denominator.		\(\frac{(5+3i)(0-4i)}{(0+4i)(0-4i)}\)
Simplify.\(\frac{(5+3i)(-4i)}{(4i)(-4i)}\)
Multiply.\(\frac{-20i-12{i}^{2}}{-16{i}^{2}}\)
Simplify the \({i}^{2}.\)\(\frac{-20i+12}{16}\)
Rewrite in standard form.\(\ \frac{12}{16}-\frac{20}{16}i\)
Simplify the fractions.\(\ \frac{3}{4}-\frac{5}{4}i\)

Simplify Powers of

The powers of \(i\) make an interesting pattern that will help us simplify higher powers of i. Let’s evaluate the powers of \(i\) to see the pattern.

\[\begin{array}{llllllllll}{i}^{1} & & & \ {i}^{2} & & & \ {i}^{3} & & & \ {i}^{4} \\ i & & & \ -1 & & & \ {i}^{2}\cdot i & & & \ {i}^{2}\cdot {i}^{2} \\ & & & & & & \ -1\cdot i & & & \ (-1)(-1) \\ & & & & & & \ -i & & & \ 1 \\ \\ \\ {i}^{5} & & & \ {i}^{6} & & & \ {i}^{7} & & & \ {i}^{8} \\ {i}^{4}\cdot i & & & \ {i}^{4}\cdot {i}^{2} & & & \ {i}^{4}\cdot {i}^{3} & & & \ {i}^{4}\cdot {i}^{4} \\ 1\cdot i & & & \ 1\cdot {i}^{2} & & & \ 1\cdot {i}^{3} & & & \ 1\cdot 1 \\ i & & & \ {i}^{2} & & & \ {i}^{3} & & & \ 1 \\ & & & \ -1 & & & \ -i\end{array}\]

We summarize this now.

\[\begin{array}{llllllll}{i}^{1} & = & i & & & \ {i}^{5} & = & i \\ {i}^{2} & = & -1 & & & \ {i}^{6} & = & -1 \\ {i}^{3} & = & \text{-}i & & & \ {i}^{7} & = & \text{-}i \\ {i}^{4} & = & 1 & & & \ {i}^{8} & = & 1\end{array}\]

If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of i. Since i4 = 1, we rewrite each power, in, as a product using i4 to a power and another power of i.

We rewrite it in the form \({i}^{n}={({i}^{4})}^{q}\cdot {i}^{r},\) where the exponent, q, is the quotient of n divided by 4 and the exponent, r, is the remainder from this division. For example, to simplify i57, we divide 57 by 4 and we get 14 with a remainder of 1. In other words, \(57=4\cdot 14+1.\) So we write \({i}^{57}={({1}^{4})}^{14}\cdot {i}^{1}\) and then simplify from there.

Example

Try it.

Simplify: \({i}^{86}.\)

Solution

\({i}^{86}\)
Divide 86 by 4 and rewrite \({i}^{86}\) in the
\({i}^{n}={({i}^{4})}^{q}\cdot {i}^{r}\) form.		\({({1}^{4})}^{21}\cdot {i}^{2}\)
Simplify.\({(1)}^{21}\cdot (-1)\)
Simplify.\(-1\)

Key Concepts

  • Square Root of a Negative Number
    • If b is a positive real number, then \(\sqrt{\text{-}b}=\sqrt{b}i\)
      \(a+bi\)
      \(b=0\)\(\begin{array}{l} \\ a+0\cdot i \\ \\ a\end{array}\)Real number
      \(b\ne 0\)\(a+bi\)Imaginary number
      \(a=0\)\(\begin{array}{l}0+bi \\ \\ \\ bi\end{array}\)Pure imaginary number
    • A complex number is in standard form when written as a + bi, where a, b are real numbers.
  • Product of Complex Conjugates
    • If a, b are real numbers, then
      \((a-bi)(a+bi)={a}^{2}+{b}^{2}\)
  • How to Divide Complex Numbers
    1. Write both the numerator and denominator in standard form.
    2. Multiply the numerator and denominator by the complex conjugate of the denominator.
    3. Simplify and write the result in standard form.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Given the numbers \(-4,-\sqrt{7},0.\overset{-}{5},\frac{7}{3},3,\sqrt{81},\) list the ⓐ rational numbers, ⓑ irrational numbers, ⓒ real numbers.
    If you missed this problem, review .

    Lafunua jibu

    ⓐ \(-4,0.\overset{-}{5},\frac{7}{3},3,\sqrt{81};\) ⓑ \(\sqrt{7};\) ⓒ \(-4,-\sqrt{7},0.\overset{-}{5},\frac{7}{3},3,\sqrt{81}\)

  2. Multiply: \((x-3)(2x+5).\)
    If you missed this problem, review .

    Lafunua jibu

    \(2{x}^{2}-x-15\)

  3. Rationalize the denominator:\(\frac{\sqrt{5}}{\sqrt{5}-\sqrt{3}}.\)
    If you missed this problem, review .

    Lafunua jibu

    \(\frac{5+\sqrt{15}}{2}\)

  4. Write each expression in terms of i and simplify if possible:

    ⓐ \(\sqrt{-25}\) ⓑ \(\sqrt{-7}\) ⓒ \(\sqrt{-12}.\)

    Lafunua jibu

    ⓐ

    \(\sqrt{-25}\)
    Use the definition of the square root of negative numbers.\(\sqrt{25}\ i\)
    Simplify.\(5i\)
    ⓑ
    \(\sqrt{-7}\)
    Use the definition of the square root of negative numbers.\(\sqrt{7}i\)
    Simplify.Be careful that it is clear that \(i\) is not under the radical sign.
    ⓒ
    \(\sqrt{-12}\)
    Use the definition of the square root of negative numbers.\(\sqrt{12}\ i\)
    Simplify \(\sqrt{12}.\)\(2\sqrt{3}\ i\)

  5. Write each expression in terms of i and simplify if possible:

    ⓐ \(\sqrt{-81}\) ⓑ \(\sqrt{-5}\) ⓒ \(\sqrt{-18}.\)

    Lafunua jibu

    ⓐ \(9i\) ⓑ \(\sqrt{5}i\) ⓒ \(3\sqrt{2}i\)

  6. Write each expression in terms of i and simplify if possible:

    ⓐ \(\sqrt{-36}\) ⓑ \(\sqrt{-3}\) ⓒ \(\sqrt{-27}.\)

    Lafunua jibu

    ⓐ \(6i\) ⓑ \(\sqrt{3}i\) ⓒ \(3\sqrt{3}i\)

  7. Add: \(\sqrt{-12}+\sqrt{-27}.\)

    Lafunua jibu

    \(\sqrt{-12}+\sqrt{-27}\)
    Use the definition of the square root of negative numbers.\(\sqrt{12}\ i+\sqrt{27}\ i\)
    Simplify the square roots.\(2\sqrt{3}\ i+3\sqrt{3}\ i\)
    Add.\(5\sqrt{3}\ i\)

  8. Add: \(\sqrt{-8}+\sqrt{-32}.\)

    Lafunua jibu

    \(6\sqrt{2}i\)

  9. Add: \(\sqrt{-27}+\sqrt{-48}.\)

    Lafunua jibu

    \(7\sqrt{3}i\)

  10. Simplify: ⓐ \((4-3i)+(5+6i)\) ⓑ \((2-5i)-(5-2i).\)

    Lafunua jibu

    ⓐ

    \((4-3i)+(5+6i)\)
    Use the Associative Property to put the real
    parts and the imaginary parts together.    		\((4+5)+(-3i+6i)\)
    Simplify.\(9+3i\)
    ⓑ
    \((2-5i)-(5-2i)\)
    Distribute.\(2-5i-5+2i\)
    Use the Associative Property to put the real
    parts and the imaginary parts together.    		\(2-5-5i+2i\)
    Simplify.\(-3-3i\)

  11. Simplify: ⓐ \((2+7i)+(4-2i)\) ⓑ \((8-4i)-(2-i).\)

    Lafunua jibu

    ⓐ \(6+5i\) ⓑ \(6-3i\)

  12. Simplify: ⓐ \((3-2i)+(-5-4i)\) ⓑ \((4+3i)-(2-6i).\)

    Lafunua jibu

    ⓐ \(-2-6i\) ⓑ \(2+9i\)

  13. Multiply: \(2i(7-5i).\)

    Lafunua jibu

    \(2i(7-5i)\)
    Distribute.\(14i-10{i}^{2}\)
    Simplify \({i}^{2}.\)\(14i-10(-1)\)
    Multiply.\(14i+10\)
    Write in standard form.\(10+14i\)

  14. Multiply: \(4i(5-3i).\)

    Lafunua jibu

    \(12+20i\)

  15. Multiply: \(-3i(2+4i).\)

    Lafunua jibu

    \(12-6i\)

  16. Multiply: \((3+2i)(4-3i).\)

    Lafunua jibu

    \((3+2i)(4-3i)\)
    Use FOIL.\(12-9i+8i-6{i}^{2}\)
    Simplify \({i}^{2}\) and combine like terms.\(12-i-6(-1)\)
    Multiply.\(12-i+6\)
    Combine the real parts.\(18-i\)

  17. Multiply: \((5-3i)(-1-2i).\)

    Lafunua jibu

    \(-11-7i\)

  18. Multiply: \((-4-3i)(2+i).\)

    Lafunua jibu

    \(-5-10i\)

  19. Multiply: \({(3+2i)}^{2}\)

    Lafunua jibu
    Use the Product of Binomial Squares Pattern, \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}.\)
    Simplify.
    Simplify \({i}^{2}.\)
    Simplify.

  20. Multiply using the Binomial Squares pattern: \({(-2-5i)}^{2}.\)

    Lafunua jibu

    \(-21+20i\)

  21. Multiply using the Binomial Squares pattern: \({(-5+4i)}^{2}.\)

    Lafunua jibu

    \(9-40i\)

  22. Multiply: \(\sqrt{-36}\cdot \sqrt{-4}.\)

    Lafunua jibu

    To multiply square roots of negative numbers, we first write them as complex numbers.

    \(\sqrt{-36}\cdot \sqrt{-4}\)
    Write as complex numbers using \(\sqrt{\text{-}b}=\sqrt{b}i.\)\(\sqrt{36}\ i\cdot \sqrt{4}\ i\)
    Simplify.\(6i\cdot 2i\)
    Multiply.\(12{i}^{2}\)
    Simplify \({i}^{2}\) and multiply.\(-12\)

  23. Multiply: \(\sqrt{-49}\cdot \sqrt{-4}.\)

    Lafunua jibu

    \(-14\)

  24. Multiply: \(\sqrt{-36}\cdot \sqrt{-81}.\)

    Lafunua jibu

    \(-54\)

  25. Multiply: \((3-\sqrt{-12})(5+\sqrt{-27}).\)

    Lafunua jibu

    To multiply square roots of negative numbers, we first write them as complex numbers.

    \((3-\sqrt{-12})(5+\sqrt{-27})\)
    Write as complex numbers using \(\sqrt{\text{-}b}=\sqrt{b}i.\)\((3-2\sqrt{3}\ i)(5+3\sqrt{3}\ i)\)
    Use FOIL.\(15+9\sqrt{3}\ i-10\sqrt{3}\ i-6\cdot 3{i}^{2}\)
    Combine like terms and simplify \({i}^{2}.\)\(15-\sqrt{3}\ i-6\cdot (-3)\)
    Multiply and combine like terms.\(33-\sqrt{3}\ i\)

  26. Multiply: \((4-\sqrt{-12})(3-\sqrt{-48}).\)

    Lafunua jibu

    \(-12-22\sqrt{3}i\)

  27. Multiply: \((-2+\sqrt{-8})(3-\sqrt{-18}).\)

    Lafunua jibu

    \(6+12\sqrt{2}i\)

  28. Multiply: \((3-2i)(3+2i).\)

    Lafunua jibu

    \((3-2i)(3+2i)\)
    Use FOIL.\(9+6i-6i-4{i}^{2}\)
    Combine like terms and simplify \({i}^{2}.\)\(9-4(-1)\)
    Multiply and combine like terms.13

  29. Multiply: \((4-3i)\cdot (4+3i).\)

    Lafunua jibu

    25

  30. Multiply: \((-2+5i)\cdot (-2-5i).\)

    Lafunua jibu

    29

  31. Multiply using the Product of Complex Conjugates Pattern: \((8-2i)(8+2i).\)

    Lafunua jibu
    Use the Product of Complex Conjugates Pattern,
    \((a-bi)(a+bi)={a}^{2}+{b}^{2}.\)
    Simplify the squares.
    Add.

  32. Multiply using the Product of Complex Conjugates Pattern: \((3-10i)(3+10i).\)

    Lafunua jibu

    109

  33. Multiply using the Product of Complex Conjugates Pattern: \((-5+4i)(-5-4i).\)

    Lafunua jibu

    41

  34. Divide: \(\frac{4+3i}{3-4i}.\)

  35. Divide: \(\frac{2+5i}{5-2i}.\)

    Lafunua jibu

    i

  36. Divide: \(\frac{1+6i}{6-i}.\)

    Lafunua jibu

    i

  37. Divide, writing the answer in standard form: \(\frac{-3}{5+2i}.\)

    Lafunua jibu

    \(\frac{-3}{5+2i}\)
    Multiply the numerator and denominator by the
    complex conjugate of the denominator.    		\(\frac{-3(5-2i)}{(5+2i)(5-2i)}\)
    Multiply in the numerator and use the Product of
    Complex Conjugates Pattern in the denominator.    		\(\frac{-15+6i}{{5}^{2}+{2}^{2}}\)
    Simplify.\(\frac{-15+6i}{29}\)
    Write in standard form.\(-\frac{15}{29}+\frac{6}{29}i\)

  38. Divide, writing the answer in standard form: \(\frac{4}{1-4i}.\)

    Lafunua jibu

    \(\frac{4}{17}+\frac{16}{17}i\)

  39. Divide, writing the answer in standard form: \(\frac{-2}{-1+2i}.\)

    Lafunua jibu

    \(\frac{2}{5}+\frac{4}{5}i\)

  40. Divide: \(\frac{5+3i}{4i}.\)

    Lafunua jibu

    \(\frac{5+3i}{4i}\)
    Write the denominator in standard form.\(\frac{5+3i}{0+4i}\)
    Multiply the numerator and denominator by
    the complex conjugate of the denominator.    		\(\frac{(5+3i)(0-4i)}{(0+4i)(0-4i)}\)
    Simplify.\(\frac{(5+3i)(-4i)}{(4i)(-4i)}\)
    Multiply.\(\frac{-20i-12{i}^{2}}{-16{i}^{2}}\)
    Simplify the \({i}^{2}.\)\(\frac{-20i+12}{16}\)
    Rewrite in standard form.\(\ \frac{12}{16}-\frac{20}{16}i\)
    Simplify the fractions.\(\ \frac{3}{4}-\frac{5}{4}i\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
i
imaginary unit
i² = −1.
\neq
not equal
The two sides are different.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Use the Complex Number System

  1. Evaluate the square root of a negative number
  2. Add and subtract complex numbers
  3. Multiply complex numbers
  4. Divide complex numbers
  5. Simplify powers of
  6. Write both the numerator and denominator in standard form.
  7. Multiply the numerator and denominator by the complex conjugate of the denominator.
  8. Simplify and write the result in standard form.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Jaribu kufanya mambo yako mwenyewe

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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