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Complex Numbers

Add and subtract complex numbers.

Complex Numbers

Use the product property to simplify radical expressions (IA 8.2.1)
Evaluate the square root of a negative number (IA 8.8.1)

Example

Try it.

Simplify: ⓐ \(\sqrt{500}\) ⓑ \(\sqrt[3]{16}\) ⓒ \(\sqrt[4]{243}.\)

Solution

ⓐ

	\(\sqrt{500}\)
Rewrite the radicand as a product using the largest perfect square factor.	\(\sqrt{100\cdot 5}\)
Rewrite the radical as the product of two radicals.	\(\sqrt{100}\cdot \sqrt{5}\)
Simplify.	\(10\sqrt{5}\)

ⓑ

	\(\sqrt[3]{16}\)
Rewrite the radicand as a product using the largest perfect cube factor.	\(\sqrt[3]{8\cdot 2}\)
Rewrite the radical as the product of two radicals.	\(\sqrt[3]{8}\cdot \sqrt[3]{{2}^{}}\)
Simplify.	\(2\ \sqrt[3]{{2}^{}}\)

ⓒ

	\(\sqrt[4]{243}\)
Rewrite the radicand as a product using the largest perfect fourth power factor.	\(\sqrt[4]{81\cdot {3}^{}}\)
Rewrite the radical as the product of two radicals.	\(\sqrt[4]{81}\cdot \sqrt[4]{3}\)
Simplify.	\(3\ \sqrt[4]{{3}^{}}\)

ⓓ

	\(\sqrt[5]{{x}^{8}}\)
Rewrite the radicand as a product using the largest perfect fifth power factor.	\(\sqrt[5]{{x}^{5}\cdot {x}^{3}}\)
Rewrite the radical as the product of two radicals.	\(\sqrt[5]{{x}^{5}}\sqrt[5]{{x}^{3}}\)
Simplify.	\(x\sqrt[5]{{x}^{3}}\)

Simplify a radical expression using the Product Property.

Try it.

\(\sqrt{54}\)

Try it.

\(\sqrt{125}\)

Try it.

\(\sqrt{25{x}^{3}}\)

Try it.

\(\sqrt[3]{625}\)

Try it.

\(\sqrt[3]{128}\)

Try it.

\(\sqrt[4]{{x}^{10}}\)

Try it.

\(\sqrt[4]{16{r}^{9}}\)

Try it.

\(\sqrt[4]{81{s}^{10}{q}^{4}}\)

Condensed: the full section is in OpenStax College Algebra 2e.

Expressing Square Roots of Negative Numbers as Multiples of

We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number \(i\) is defined as the square root of \(-1.\)

\[\sqrt{-1}=i\]

So, using properties of radicals,

\[{i}^{2}={(\sqrt{-1})}^{2}=-1\]

We can write the square root of any negative number as a multiple of \(i.\) Consider the square root of \(-49.\)

\[\begin{array}{lll}\sqrt{-49} & = & \sqrt{49⋅(-1)} \\ & = & \sqrt{49}\sqrt{-1} \\ & = & 7i\end{array}\]

We use \(7i\) and not \(-7i\) because the principal root of \(49\) is the positive root.

A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written \(a+bi\) where \(a\) is the real part and \(b\) is the imaginary part. For example, \(5+2i\) is a complex number. So, too, is \(3+4i\sqrt{3}.\)

Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers.

Example

Try it.

Express \(\sqrt{-9}\) in standard form.

Solution

\[\begin{array}{lll}\sqrt{-9} & = & \sqrt{9}\sqrt{-1} \\ & = & 3i\end{array}\]

In standard form, this is \(0+3i.\)

Plotting a Complex Number on the Complex Plane

We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs \((a,b),\) where \(a\) represents the coordinate for the horizontal axis and \(b\) represents the coordinate for the vertical axis.

Let’s consider the number \(-2+3i.\) The real part of the complex number is \(-2\) and the imaginary part is 3. We plot the ordered pair \((-2,3)\) to represent the complex number \(-2+3i,\) as shown in .

Example

Try it.

Plot the complex number \(3-4i\) on the complex plane.

Solution

The real part of the complex number is \(3,\) and the imaginary part is –4. We plot the ordered pair \((3,-4)\) as shown in .

Adding and Subtracting Complex Numbers

Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and then combine the imaginary parts.

Example

Try it.

Add or subtract as indicated.

ⓐ \((3-4i)+(2+5i)\)
ⓑ \((-5+7i)-(-11+2i)\)

Solution

We add the real parts and add the imaginary parts.

ⓐ
\(\begin{array}{lll}(3-4i)+(2+5i) & = & 3-4i+2+5i \\ & = & 3+2+(-4i)+5i \\ & = & (3+2)+(-4+5)i \\ & = & 5+i\end{array}\)
ⓑ
\(\begin{array}{lll}(-5+7i)-(-11+2i) & = & -5+7i+11-2i \\ & = & -5+11+7i-2i \\ & = & (-5+11)+(7-2)i \\ & = & 6+5i\end{array}\)

Multiplying Complex Numbers

Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.

Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, \(3(6+2i)\) :

Example

Try it.

Find the product \(4(2+5i).\)

Solution

Distribute the 4.

\[\begin{array}{lll}4(2+5i) & = & (4⋅2)+(4⋅5i) \\ & = & 8+20i\end{array}\]

Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. The difference with complex numbers is that when we get a squared term, \({i}^{2},\) it equals \(-1.\)

\[\begin{array}{llll}(a+bi)(c+di) & = & ac+adi+bci+bd{i}^{2} & \\ & = & ac+adi+bci-bd & \ {i}^{2}=-1 \\ & = & (ac-bd)+(ad+bc)i & \ \text{Group real terms and imaginary terms}.\end{array}\]

Example

Try it.

Multiply: \((4+3i)(2-5i).\)

Solution

\[\begin{array}{lll}(4+3i)(2-5i) & = & 4(2)-4(5i)+3i(2)-(3i)(5i) \\ & = & 8-20i+6i-15({i}^{2}) \\ & = & (8+15)+(-20+6)i \\ & = & 23-14i\end{array}\]

Dividing Complex Numbers

Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form \(a+bi.\) We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of \(a+bi\) is \(a-bi.\) For example, the product of \(a+bi\) and \(a-bi\) is

\[\begin{array}{lll}(a+bi)(a-bi) & = & {a}^{2}-abi+abi-{b}^{2}{i}^{2} \\ & = & {a}^{2}+{b}^{2}\end{array}\]

The result is a real number.

Note that complex conjugates have an opposite relationship: The complex conjugate of \(a+bi\) is \(a-bi,\) and the complex conjugate of \(a-bi\) is \(a+bi.\) Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose we want to divide \(c+di\) by \(a+bi,\) where neither \(a\) nor \(b\) equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.

\[\frac{c+di}{a+bi}\ \text{where }a\ne 0\ \text{and }b\ne 0\]

Multiply the numerator and denominator by the complex conjugate of the denominator.

\[\frac{(c+di)}{(a+bi)}⋅\frac{(a-bi)}{(a-bi)}=\frac{(c+di)(a-bi)}{(a+bi)(a-bi)}\]

Apply the distributive property.

\[=\frac{ca-cbi+adi-bd{i}^{2}}{{a}^{2}-abi+abi-{b}^{2}{i}^{2}}\]

Simplify, remembering that \({i}^{2}=-1.\)

\[\begin{array}{l}=\frac{ca-cbi+adi-bd(-1)}{{a}^{2}-abi+abi-{b}^{2}(-1)} \\ =\frac{(ca+bd)+(ad-cb)i}{{a}^{2}+{b}^{2}}\end{array}\]

Example

Try it.

Find the complex conjugate of each number.

ⓐ \(2+i\sqrt{5}\)
ⓑ \(-\frac{1}{2}i\)

Solution

ⓐThe number is already in the form \(a+bi.\) The complex conjugate is \(a-bi,\) or \(2-i\sqrt{5}.\)
ⓑWe can rewrite this number in the form \(a+bi\) as \(0-\frac{1}{2}i.\) The complex conjugate is \(a-bi,\) or \(0+\frac{1}{2}i.\) This can be written simply as \(\frac{1}{2}i.\)

Condensed: the full section is in OpenStax College Algebra 2e.

Simplifying Powers of

The powers of \(i\) are cyclic. Let’s look at what happens when we raise \(i\) to increasing powers.

\[\begin{array}{l}{i}^{1}=i \\ {i}^{2}=-1 \\ {i}^{3}={i}^{2}⋅i=-1⋅i=-i \\ {i}^{4}={i}^{3}⋅i=-i⋅i=-{i}^{2}=-(-1)=1 \\ {i}^{5}={i}^{4}⋅i=1⋅i=i\end{array}\]

We can see that when we get to the fifth power of \(i,\) it is equal to the first power. As we continue to multiply \(i\) by increasing powers, we will see a cycle of four. Let’s examine the next four powers of \(i.\)

\[\begin{array}{l}{i}^{6}={i}^{5}⋅i=i⋅i={i}^{2}=-1 \\ {i}^{7}={i}^{6}⋅i={i}^{2}⋅i={i}^{3}=-i \\ {i}^{8}={i}^{7}⋅i={i}^{3}⋅i={i}^{4}=1 \\ {i}^{9}={i}^{8}⋅i={i}^{4}⋅i={i}^{5}=i\end{array}\]

The cycle is repeated continuously: \(i,-1,-i,1,\) every four powers.

Example

Try it.

Evaluate: \({i}^{35}.\)

Solution

Since \({i}^{4}=1,\) we can simplify the problem by factoring out as many factors of \({i}^{4}\) as possible. To do so, first determine how many times 4 goes into 35: \(35=4⋅8+3.\)

\[{i}^{35}={i}^{4⋅8+3}={i}^{4⋅8}⋅{i}^{3}={({i}^{4})}^{8}⋅{i}^{3}={1}^{8}⋅{i}^{3}={i}^{3}=-i\]

Key Concepts

The square root of any negative number can be written as a multiple of \(i.\)
To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.
Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts.
Complex numbers can be multiplied and divided.
- To multiply complex numbers, distribute just as with polynomials. See and .
- To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See and .
The powers of \(i\) are cyclic, repeating every fourth one.

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Cevabı göster

ⓐ

	\(\sqrt{500}\)
Rewrite the radicand as a product using the largest perfect square factor.	\(\sqrt{100\cdot 5}\)
Rewrite the radical as the product of two radicals.	\(\sqrt{100}\cdot \sqrt{5}\)
Simplify.	\(10\sqrt{5}\)

ⓑ

	\(\sqrt[3]{16}\)
Rewrite the radicand as a product using the largest perfect cube factor.	\(\sqrt[3]{8\cdot 2}\)
Rewrite the radical as the product of two radicals.	\(\sqrt[3]{8}\cdot \sqrt[3]{{2}^{}}\)
Simplify.	\(2\ \sqrt[3]{{2}^{}}\)

ⓒ

	\(\sqrt[4]{243}\)
Rewrite the radicand as a product using the largest perfect fourth power factor.	\(\sqrt[4]{81\cdot {3}^{}}\)
Rewrite the radical as the product of two radicals.	\(\sqrt[4]{81}\cdot \sqrt[4]{3}\)
Simplify.	\(3\ \sqrt[4]{{3}^{}}\)

ⓓ

	\(\sqrt[5]{{x}^{8}}\)
Rewrite the radicand as a product using the largest perfect fifth power factor.	\(\sqrt[5]{{x}^{5}\cdot {x}^{3}}\)
Rewrite the radical as the product of two radicals.	\(\sqrt[5]{{x}^{5}}\sqrt[5]{{x}^{3}}\)
Simplify.	\(x\sqrt[5]{{x}^{3}}\)

\(\sqrt{54}\)
\(\sqrt{125}\)
\(\sqrt{25{x}^{3}}\)
\(\sqrt[3]{625}\)
\(\sqrt[3]{128}\)
\(\sqrt[4]{{x}^{10}}\)
\(\sqrt[4]{16{r}^{9}}\)
\(\sqrt[4]{81{s}^{10}{q}^{4}}\)
Simplify: ⓐ \(\sqrt{-64}\) ⓑ \(\sqrt{-98}\) ⓒ \(\sqrt{-243}\)
Cevabı göster
1. ⓐ \(\sqrt{-64}\)
  Rewrite the radicand as a product. \(\sqrt{-1⋅64}\)
  Rewrite the radical as the product of two radicals. \(\sqrt{-1}⋅\sqrt{64}\)
  Use the definition of \(i\) and simplify radical terms. \(i⋅8=8i\)
2. ⓑ \(\sqrt{-98}\)
  Rewrite the radicand as a product using the largest perfect square factor. \(\sqrt{-1⋅49⋅2}\)
  Rewrite the radical as the product of three radicals. \(\sqrt{-1}⋅\sqrt{49}⋅\sqrt{2}\)
  Use the definition of \(i\) and simplify radical terms. \(7i\sqrt{2}\)
3. ⓒ \(\sqrt{-243}\)
  Rewrite the radicand as a product using the largest perfect square factor. \(\sqrt{-1⋅81⋅3}\)
  Rewrite the radical as the product of three radicals. \(\sqrt{-1}⋅\sqrt{81}⋅\sqrt{3}\)
  Use the definition of \(i\) and simplify radical terms. \(9i\sqrt{3}\)
\(\sqrt{-100}\)
\(\sqrt{-12}\)
\(\sqrt{-75}\)
Multiply: \(\sqrt{-49}⋅\sqrt{-9}\)
Multiply: \(\sqrt{36}⋅\sqrt{-81}\)
Multiply: \((2-\sqrt{-16})(3+\sqrt{-4})\)
Multiply: \((2+\sqrt{-25})(2-\sqrt{-25})\)
Add: \((1+3i)+(7+4i)\)
Add: \((5-\sqrt{-36})+(2-\sqrt{-49})\)
Subtract: \((5-\sqrt{-36})-(2-\sqrt{-49})\)
Subtract: \((-7-\sqrt{-50})-(-32-\sqrt{-18})\)
Express \(\sqrt{-9}\) in standard form.

Cevabı göster

\[\begin{array}{lll}\sqrt{-9} & = & \sqrt{9}\sqrt{-1} \\ & = & 3i\end{array}\]
In standard form, this is \(0+3i.\)
Express \(\sqrt{-24}\) in standard form.

Cevabı göster

\(\sqrt{-24}=0+2i\sqrt{6}\)
Plot the complex number \(3-4i\) on the complex plane.

Cevabı göster

The real part of the complex number is \(3,\) and the imaginary part is –4. We plot the ordered pair \((3,-4)\) as shown in .
Plot the complex number \(-4-i\) on the complex plane.
Add or subtract as indicated.
1. ⓐ \((3-4i)+(2+5i)\)
2. ⓑ \((-5+7i)-(-11+2i)\)
Cevabı göster
We add the real parts and add the imaginary parts.
1. ⓐ
  \(\begin{array}{lll}(3-4i)+(2+5i) & = & 3-4i+2+5i \\ & = & 3+2+(-4i)+5i \\ & = & (3+2)+(-4+5)i \\ & = & 5+i\end{array}\)
2. ⓑ
  \(\begin{array}{lll}(-5+7i)-(-11+2i) & = & -5+7i+11-2i \\ & = & -5+11+7i-2i \\ & = & (-5+11)+(7-2)i \\ & = & 6+5i\end{array}\)
Subtract \(2+5i\) from \(3-4i.\)

Cevabı göster

\((3-4i)-(2+5i)=1-9i\)
Find the product \(4(2+5i).\)

Cevabı göster

Distribute the 4.
\[\begin{array}{lll}4(2+5i) & = & (4⋅2)+(4⋅5i) \\ & = & 8+20i\end{array}\]
Find the product: \(\frac{1}{2}(5-2i).\)

Cevabı göster

\(\frac{5}{2}-i\)
Multiply: \((4+3i)(2-5i).\)

Cevabı göster

\[\begin{array}{lll}(4+3i)(2-5i) & = & 4(2)-4(5i)+3i(2)-(3i)(5i) \\ & = & 8-20i+6i-15({i}^{2}) \\ & = & (8+15)+(-20+6)i \\ & = & 23-14i\end{array}\]
Multiply: \((3-4i)(2+3i).\)

Cevabı göster

\(18+i\)
Find the complex conjugate of each number.
1. ⓐ \(2+i\sqrt{5}\)
2. ⓑ \(-\frac{1}{2}i\)
Cevabı göster
1. ⓐThe number is already in the form \(a+bi.\) The complex conjugate is \(a-bi,\) or \(2-i\sqrt{5}.\)
2. ⓑWe can rewrite this number in the form \(a+bi\) as \(0-\frac{1}{2}i.\) The complex conjugate is \(a-bi,\) or \(0+\frac{1}{2}i.\) This can be written simply as \(\frac{1}{2}i.\)
Find the complex conjugate of \(-3+4i.\)

Cevabı göster

\(-3-4i\)
Divide: \((2+5i)\) by \((4-i).\)

Cevabı göster

We begin by writing the problem as a fraction.
\[\frac{(2+5i)}{(4-i)}\]
Then we multiply the numerator and denominator by the complex conjugate of the denominator.
\[\frac{(2+5i)}{(4-i)}\ ⋅\frac{(4+i)}{(4+i)}\]
To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL).
\[\begin{array}{llll}\frac{(2+5i)}{(4-i)}⋅\frac{(4+i)}{(4+i)} & = & \frac{8+2i+20i+5{i}^{2}}{16+4i-4i-{i}^{2}} & \\ & = & \frac{8+2i+20i+5(-1)}{16+4i-4i-(-1)} & \ \text{Because}{i}^{2}=-1. \\ & = & \frac{3+22i}{17} & \\ & = & \frac{3}{17}+\frac{22}{17}i & \ \text{Separate real and imaginary parts}.\end{array}\]
Note that this expresses the quotient in standard form.
Evaluate: \({i}^{35}.\)

Cevabı göster

Since \({i}^{4}=1,\) we can simplify the problem by factoring out as many factors of \({i}^{4}\) as possible. To do so, first determine how many times 4 goes into 35: \(35=4⋅8+3.\)
\[{i}^{35}={i}^{4⋅8+3}={i}^{4⋅8}⋅{i}^{3}={({i}^{4})}^{8}⋅{i}^{3}={1}^{8}⋅{i}^{3}={i}^{3}=-i\]
Evaluate: \({i}^{18}\)

Cevabı göster

\(-1\)
Explain how to add complex numbers.

Cevabı göster

Add the real parts together and the imaginary parts together.
What is the basic principle in multiplication of complex numbers?
Give an example to show that the product of two imaginary numbers is not always imaginary.

Cevabı göster

Possible answer: \(i\) times \(i\) equals -1, which is not imaginary.
What is a characteristic of the plot of a real number in the complex plane?

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Nasıl: Complex Numbers

Add and subtract complex numbers.
Multiply and divide complex numbers.
Simplify powers of
Use the product property to simplify radical expressions (IA 8.2.1)
Evaluate the square root of a negative number (IA 8.8.1)
Find the largest factor in the radicand that is a perfect power of the index. Rewrite the radicand as a product of two factors, using that factor.
Use the product rule to rewrite the radical as the product of two radicals.
Simplify the root of the perfect power.

Sık sorulan sorular

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.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Burada tekrar özetlenmiş ve açıklanmıştır. Hatalar bizim. Kaynağı ticari olmayan kullanım için lisanslanmış olduğu için, bu ders ücretsizdir, hiçbir şey satmaz ve aynı lisans altında paylaşılır.

More in Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Factorization of \({i}^{35}\)	\({i}^{34}⋅i\)	\({i}^{33}⋅{i}^{2}\)	\({i}^{31}⋅{i}^{4}\)	\({i}^{19}⋅{i}^{16}\)
Reduced form	\({({i}^{2})}^{17}⋅i\)	\({i}^{33}⋅(-1)\)	\({i}^{31}⋅1\)	\[{i}^{19}⋅{({i}^{4})}^{4}\]
Simplified form	\({(-1)}^{17}⋅i\)	\(-{i}^{33}\)	\({i}^{31}\)	\({i}^{19}\)