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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.
- If b is a positive real number, then \(\sqrt{\text{-}b}=\sqrt{b}i\)
- Product of Complex Conjugates
- If a, b are real numbers, then
\((a-bi)(a+bi)={a}^{2}+{b}^{2}\)
- If a, b are real numbers, then
- How to Divide Complex Numbers
- Write both the numerator and denominator in standard form.
- Multiply the numerator and denominator by the complex conjugate of the denominator.
- Simplify and write the result in standard form.
Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.
Tababar (40)
Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.
-
Given the numbers \(-4,-\sqrt{7},0.\overset{-}{5},\frac{7}{3},3,\sqrt{81},\) list the ⓐ rational numbers, ⓑ irrational numbers, ⓒ real numbers.
Jawaabta muuji
ⓐ \(-4,0.\overset{-}{5},\frac{7}{3},3,\sqrt{81};\) ⓑ \(\sqrt{7};\) ⓒ \(-4,-\sqrt{7},0.\overset{-}{5},\frac{7}{3},3,\sqrt{81}\)
-
Multiply: \((x-3)(2x+5).\)
Jawaabta muuji
\(2{x}^{2}-x-15\)
-
Rationalize the denominator:\(\frac{\sqrt{5}}{\sqrt{5}-\sqrt{3}}.\)
Jawaabta muuji
\(\frac{5+\sqrt{15}}{2}\)
-
Write each expression in terms of i and simplify if possible:
ⓐ \(\sqrt{-25}\) ⓑ \(\sqrt{-7}\) ⓒ \(\sqrt{-12}.\)
Jawaabta muuji
ⓐ
ⓑ\(\sqrt{-25}\) 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 \(\sqrt{12}.\) \(2\sqrt{3}\ i\) -
Write each expression in terms of i and simplify if possible:
ⓐ \(\sqrt{-81}\) ⓑ \(\sqrt{-5}\) ⓒ \(\sqrt{-18}.\)
Jawaabta muuji
ⓐ \(9i\) ⓑ \(\sqrt{5}i\) ⓒ \(3\sqrt{2}i\)
-
Write each expression in terms of i and simplify if possible:
ⓐ \(\sqrt{-36}\) ⓑ \(\sqrt{-3}\) ⓒ \(\sqrt{-27}.\)
Jawaabta muuji
ⓐ \(6i\) ⓑ \(\sqrt{3}i\) ⓒ \(3\sqrt{3}i\)
-
Add: \(\sqrt{-12}+\sqrt{-27}.\)
Jawaabta muuji
\(\sqrt{-12}+\sqrt{-27}\) 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. \(5\sqrt{3}\ i\) -
Add: \(\sqrt{-8}+\sqrt{-32}.\)
Jawaabta muuji
\(6\sqrt{2}i\)
-
Add: \(\sqrt{-27}+\sqrt{-48}.\)
Jawaabta muuji
\(7\sqrt{3}i\)
-
Simplify: ⓐ \((4-3i)+(5+6i)\) ⓑ \((2-5i)-(5-2i).\)
Jawaabta muuji
ⓐ
ⓑ\((4-3i)+(5+6i)\) 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. \(-3-3i\) -
Simplify: ⓐ \((2+7i)+(4-2i)\) ⓑ \((8-4i)-(2-i).\)
Jawaabta muuji
ⓐ \(6+5i\) ⓑ \(6-3i\)
-
Simplify: ⓐ \((3-2i)+(-5-4i)\) ⓑ \((4+3i)-(2-6i).\)
Jawaabta muuji
ⓐ \(-2-6i\) ⓑ \(2+9i\)
-
Multiply: \(2i(7-5i).\)
Jawaabta muuji
\(2i(7-5i)\) Distribute. \(14i-10{i}^{2}\) Simplify \({i}^{2}.\) \(14i-10(-1)\) Multiply. \(14i+10\) Write in standard form. \(10+14i\) -
Multiply: \(4i(5-3i).\)
Jawaabta muuji
\(12+20i\)
-
Multiply: \(-3i(2+4i).\)
Jawaabta muuji
\(12-6i\)
-
Multiply: \((3+2i)(4-3i).\)
Jawaabta muuji
\((3+2i)(4-3i)\) 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. \(18-i\) -
Multiply: \((5-3i)(-1-2i).\)
Jawaabta muuji
\(-11-7i\)
-
Multiply: \((-4-3i)(2+i).\)
Jawaabta muuji
\(-5-10i\)
-
Multiply: \({(3+2i)}^{2}\)
Jawaabta muuji
Use the Product of Binomial Squares Pattern, \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}.\) Simplify. Simplify \({i}^{2}.\) Simplify. -
Multiply using the Binomial Squares pattern: \({(-2-5i)}^{2}.\)
Jawaabta muuji
\(-21+20i\)
-
Multiply using the Binomial Squares pattern: \({(-5+4i)}^{2}.\)
Jawaabta muuji
\(9-40i\)
-
Multiply: \(\sqrt{-36}\cdot \sqrt{-4}.\)
Jawaabta muuji
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. \(-12\) -
Multiply: \(\sqrt{-49}\cdot \sqrt{-4}.\)
Jawaabta muuji
\(-14\)
-
Multiply: \(\sqrt{-36}\cdot \sqrt{-81}.\)
Jawaabta muuji
\(-54\)
-
Multiply: \((3-\sqrt{-12})(5+\sqrt{-27}).\)
Jawaabta muuji
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. \(33-\sqrt{3}\ i\) -
Multiply: \((4-\sqrt{-12})(3-\sqrt{-48}).\)
Jawaabta muuji
\(-12-22\sqrt{3}i\)
-
Multiply: \((-2+\sqrt{-8})(3-\sqrt{-18}).\)
Jawaabta muuji
\(6+12\sqrt{2}i\)
-
Multiply: \((3-2i)(3+2i).\)
Jawaabta muuji
\((3-2i)(3+2i)\) Use FOIL. \(9+6i-6i-4{i}^{2}\) Combine like terms and simplify \({i}^{2}.\) \(9-4(-1)\) Multiply and combine like terms. 13 -
Multiply: \((4-3i)\cdot (4+3i).\)
Jawaabta muuji
25
-
Multiply: \((-2+5i)\cdot (-2-5i).\)
Jawaabta muuji
29
-
Multiply using the Product of Complex Conjugates Pattern: \((8-2i)(8+2i).\)
Jawaabta muuji
Use the Product of Complex Conjugates Pattern,
\((a-bi)(a+bi)={a}^{2}+{b}^{2}.\)Simplify the squares. Add. -
Multiply using the Product of Complex Conjugates Pattern: \((3-10i)(3+10i).\)
Jawaabta muuji
109
-
Multiply using the Product of Complex Conjugates Pattern: \((-5+4i)(-5-4i).\)
Jawaabta muuji
41
-
Divide: \(\frac{4+3i}{3-4i}.\)
-
Divide: \(\frac{2+5i}{5-2i}.\)
Jawaabta muuji
i
-
Divide: \(\frac{1+6i}{6-i}.\)
Jawaabta muuji
i
-
Divide, writing the answer in standard form: \(\frac{-3}{5+2i}.\)
Jawaabta muuji
\(\frac{-3}{5+2i}\) 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. \(-\frac{15}{29}+\frac{6}{29}i\) -
Divide, writing the answer in standard form: \(\frac{4}{1-4i}.\)
Jawaabta muuji
\(\frac{4}{17}+\frac{16}{17}i\)
-
Divide, writing the answer in standard form: \(\frac{-2}{-1+2i}.\)
Jawaabta muuji
\(\frac{2}{5}+\frac{4}{5}i\)
-
Divide: \(\frac{5+3i}{4i}.\)
Jawaabta muuji
\(\frac{5+3i}{4i}\) 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. \(\ \frac{3}{4}-\frac{5}{4}i\)
A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.
Ku soo biir Ku soo galCalaamadaha la isticmaalay halkan
Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.
Sida loo: Use the Complex Number System
- Evaluate the square root of a negative number
- Add and subtract complex numbers
- Multiply complex numbers
- Divide complex numbers
- Simplify powers of
- Write both the numerator and denominator in standard form.
- Multiply the numerator and denominator by the complex conjugate of the denominator.
- Simplify and write the result in standard form.
Su'aalaha dadku waydiiyaan
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.
Qaybaha ee boggaan waxaa la beddelay ka OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.
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