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Complex numbers

i² = −1, arithmetic in the complex plane, modulus and conjugates.

Allowing one new number, i with i² = −1, makes every polynomial equation solvable. Complex numbers add like vectors and multiply by multiplying lengths and adding angles; the modulus |a + bi| = √(a² + b²) is the distance from 0. The solver evaluates the arithmetic exactly and the quadratic pages show where complex roots come from.

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 a 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 negative 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 –25.

\[\begin{array}{l}\begin{array}{l} \\ \sqrt{-25}=\sqrt{25⋅(-1)}\end{array} \\ =\sqrt{25}\sqrt{-1} \\ =5i\end{array}\]

We use \(5i\) and not \(-\text{5}i\) because the principal root of \(25\) 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 \(bi\) is the imaginary part. For example, \(5+2i\) is a complex number. So, too, is \(3+4\sqrt{3}i.\)

Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers.

Example

Try it.

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

Solution

\[\sqrt{-9}=\sqrt{9}\sqrt{-1}=3i\]

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 \(3i.\) 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 \(-4i.\) We plot the ordered pair \((3,-4)\)

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 combine the imaginary parts.

Example

Try it.

Add \(3-4i\) and \(2+5i.\)

Solution

We add the real parts and add the imaginary parts.

\[\begin{array}{l}(a+bi)+(c+di)=(a+c)+(b+d)i \\ (3-4i)+(2+5i)=(3+2)+(-4+5)i \\ =5+i\end{array}\]

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.

Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,

Example

Try it.

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

Solution

Distribute the 4.

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

Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get

\[(a+bi)(c+di)=ac+adi+bci+bd{i}^{2}\]

Because \({i}^{2}=-1,\) we have

\[(a+bi)(c+di)=ac+adi+bci-bd\]

To simplify, we combine the real parts, and we combine the imaginary parts.

\[(a+bi)(c+di)=(ac-bd)+(ad+bc)i\]

Example

Try it.

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

Solution

Use \((a+bi)(c+di)=(ac-bd)+(ad+bc)i\)

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

Dividing Complex Numbers

Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. 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 \(a-bi.\)

Note that complex conjugates have a reciprocal 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 Precalculus 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 itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of \(i.\)

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

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: \(35=4⋅8+3.\)

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

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.
To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator.
The powers of \(i\) are cyclic, repeating every fourth one.

Complex Numbers

Try it.

Explain how to add complex numbers.

Solution

Add the real parts together and the imaginary parts together.

Try it.

What is the basic principle in multiplication of complex numbers?

Try it.

Give an example to show the product of two imaginary numbers is not always imaginary.

Solution

\(i\) times \(i\) equals –1, which is not imaginary. (answers vary)

Try it.

What is a characteristic of the plot of a real number in the complex plane?

For the following exercises, evaluate the algebraic expressions.

Try it.

\(\text{If}\ f(x)={x}^{2}+x-4,\) evaluate \(f(2i).\)

Solution

\(-8+2i\)

Try it.

\(\text{If}\ f(x)={x}^{3}-2,\) evaluate \(f(i).\)

Try it.

\(\text{If}\ f(x)={x}^{2}+3x+5,\) evaluate \(f(2+i).\)

Solution

\(14+7i\)

Try it.

\(\text{If}\ f(x)=2{x}^{2}+x-3,\) evaluate \(f(2-3i).\)

Try it.

\(\text{If}\ f(x)=\frac{x+1}{2-x},\) evaluate \(f(5i).\)

Solution

\(-\frac{23}{29}+\frac{15}{29}i\)

Try it.

\(\text{If}\ f(x)=\frac{1+2x}{x+3},\) evaluate \(f(4i).\)

For the following exercises, determine the number of real and nonreal solutions for each quadratic function shown.

For the following exercises, plot the complex numbers on the complex plane.

For the following exercises, perform the indicated operation and express the result as a simplified complex number.

Try it.

\((3+2i)+(5-3i)\)

Solution

\(8-i\)

Try it.

\((-2-4i)+(1+6i)\)

Try it.

\((-5+3i)-(6-i)\)

Solution

\(-11+4i\)

Try it.

\((2-3i)-(3+2i)\)

Try it.

\((-4+4i)-(-6+9i)\)

Solution

\(2-5i\)

Try it.

\((2+3i)(4i)\)

Try it.

\((5-2i)(3i)\)

Solution

\(6+15i\)

Try it.

\((6-2i)(5)\)

Try it.

\((-2+4i)(8)\)

Solution

\(-16+32i\)

Try it.

\((2+3i)(4-i)\)

Try it.

\((-1+2i)(-2+3i)\)

Solution

\(-4-7i\)

Try it.

\((4-2i)(4+2i)\)

Try it.

\((3+4i)(3-4i)\)

Solution

Try it.

\(\frac{3+4i}{2}\)

Try it.

\(\frac{6-2i}{3}\)

Solution

\(2-\frac{2}{3}i\)

Try it.

\(\frac{-5+3i}{2i}\)

Try it.

\(\frac{6+4i}{i}\)

Solution

\(4-6i\)

Try it.

\(\frac{2-3i}{4+3i}\)

Try it.

\(\frac{3+4i}{2-i}\)

Solution

\(\frac{2}{5}+\frac{11}{5}i\)

Try it.

\(\frac{2+3i}{2-3i}\)

Try it.

\(\sqrt{-9}+3\sqrt{-16}\)

Solution

\(15i\)

Try it.

\(-\sqrt{-4}-4\sqrt{-25}\)

Try it.

\(\frac{2+\sqrt{-12}}{2}\)

Solution

\(1+i\sqrt{3}\)

Try it.

\(\frac{4+\sqrt{-20}}{2}\)

Try it.

\({i}^{8}\)

Solution

\(1\)

Try it.

\({i}^{15}\)

Try it.

\({i}^{22}\)

Solution

\(-1\)

For the following exercises, evaluate the expressions, writing the result as a simplified complex number.

Try it.

\(\frac{1}{i}+\frac{4}{{i}^{3}}\)

Solution

\(3i\)

Try it.

\(\frac{1}{{i}^{11}}-\frac{1}{{i}^{21}}\)

Try it.

\({i}^{7}(1+{i}^{2})\)

Solution

Try it.

\({i}^{-3}+5{i}^{7}\)

Try it.

\(\frac{(2+i)(4-2i)}{(1+i)}\)

Solution

5, 5i

Try it.

\(\frac{(1+3i)(2-4i)}{(1+2i)}\)

Try it.

\(\frac{{(3+i)}^{2}}{{(1+2i)}^{2}}\)

Solution

\(-2i\)

Try it.

\(\frac{3+2i}{2+i}+(4+3i)\)

Try it.

\(\frac{4+i}{i}+\frac{3-4i}{1-i}\)

Solution

\(\frac{9}{2}-\frac{9}{2}i\)

Try it.

\(\frac{3+2i}{1+2i}-\frac{2-3i}{3+i}\)

Condensed: the full section is in OpenStax Precalculus 2e.

tusaale · (1 + 2i)*(3 - i)

Precalculus · Complex-Numbers

Evaluate (1 + 2i)*(3 - i)

\left(1 + 2 i\right) \left(3 - i\right)

tallaabo tallaabo

\left(1 + 2 i\right) \left(3 - i\right) = 5 + 5 i
Evaluate.

Jawaabta muuji

5 + 5 i

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Tababar (40)

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

Jawaabta muuji

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

Jawaabta muuji

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

Jawaabta muuji

The real part of the complex number is \(3,\) and the imaginary part is \(-4i.\) We plot the ordered pair \((3,-4)\)
Plot the complex number \(-4-i\) on the complex plane.
Add \(3-4i\) and \(2+5i.\)

Jawaabta muuji

We add the real parts and add the imaginary parts.
\[\begin{array}{l}(a+bi)+(c+di)=(a+c)+(b+d)i \\ (3-4i)+(2+5i)=(3+2)+(-4+5)i \\ =5+i\end{array}\]
Subtract \(2+5i\) from \(3-4i.\)

Jawaabta muuji

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

Jawaabta muuji

Distribute the 4.
\[\begin{array}{l}4(2+5i)=(4⋅2)+(4⋅5i) \\ =8+20i\end{array}\]
Find the product \(-4(2+6i).\)

Jawaabta muuji

\(-8-24i\)
Multiply \((4+3i)(2-5i).\)

Jawaabta muuji

Use \((a+bi)(c+di)=(ac-bd)+(ad+bc)i\)
\[\begin{array}{l}(4+3i)(2-5i)=(4⋅2-3⋅(-5))+(4⋅(-5)+3⋅2)i \\ =(8+15)+(-20+6)i \\ =23-14i\end{array}\]
Multiply \((3-4i)(2+3i).\)

Jawaabta muuji

\(18+i\)
Find the complex conjugate of each number.
1. ⓐ \(2+i\sqrt{5}\)
2. ⓑ \(-\frac{1}{2}i\)
Jawaabta muuji
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 \(\frac{1}{2}i.\)
Divide \((2+5i)\) by \((4-i).\)

Jawaabta muuji

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 (the process commonly called FOIL).
\[\begin{array}{ll}\frac{(2+5i)}{(4-i)}⋅\frac{(4+i)}{(4+i)}=\frac{8+2i+20i+5{i}^{2}}{16+4i-4i-{i}^{2}} & \\ =\frac{8+2i+20i+5(-1)}{16+4i-4i-(-1)} & \text{Because}\ {i}^{2}=-1 \\ =\frac{3+22i}{17} & \\ =\frac{3}{17}+\frac{22}{17}i & \text{Separate real and imaginary parts}.\end{array}\]
Note that this expresses the quotient in standard form.
Let \(f(x)={x}^{2}-5x+2.\) Evaluate \(f(3+i).\)

Jawaabta muuji

Substitute \(x=3+i\) into the function \(f(x)={x}^{2}-5x+2\) and simplify.
Let \(f(x)=2{x}^{2}-3x.\) Evaluate \(f(8-i).\)

Jawaabta muuji

\(102-29i\)
Let \(f(x)=\frac{2+x}{x+3}.\) Evaluate \(f(10i).\)

Jawaabta muuji

Substitute \(x=10i\) and simplify.
\[\begin{array}{ll}\frac{2+10i}{10i+3} & \text{Substitute}\ 10i\text{ for}\ x. \\ \frac{2+10i}{3+10i} & \text{Rewrite the denominator in standard form}. \\ \frac{2+10i}{3+10i}⋅\frac{3-10i}{3-10i} & \text{Prepare to multiply the numerator and} \\ & \text{denominator by the complex conjugate} \\ & \text{of the denominator}. \\ \frac{6-20i+30i-100{i}^{2}}{9-30i+30i-100{i}^{2}} & \text{Multiply using the distributive property or the FOIL method}. \\ \frac{6-20i+30i-100(-1)}{9-30i+30i-100(-1)} & \text{Substitute -1 for}{i}^{2}. \\ \frac{106+10i}{109} & \text{Simplify}. \\ \frac{106}{109}+\frac{10}{109}i & \text{Separate the real and imaginary parts}.\end{array}\]
Let \(f(x)=\frac{x+1}{x-4}.\) Evaluate \(f(-i).\)

Jawaabta muuji

\(-\frac{3}{17}+\frac{5i}{17}\)
Evaluate \({i}^{35}.\)

Jawaabta muuji

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: \(35=4⋅8+3.\)
\[{i}^{35}={i}^{4⋅8+3}={i}^{4⋅8}⋅{i}^{3}={({i}^{4})}^{8}⋅{i}^{3}={1}^{8}⋅{i}^{3}={i}^{3}=-i\]
Explain how to add complex numbers.

Jawaabta muuji

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 the product of two imaginary numbers is not always imaginary.

Jawaabta muuji

\(i\) times \(i\) equals –1, which is not imaginary. (answers vary)
What is a characteristic of the plot of a real number in the complex plane?
\(\text{If}\ f(x)={x}^{2}+x-4,\) evaluate \(f(2i).\)

Jawaabta muuji

\(-8+2i\)
\(\text{If}\ f(x)={x}^{3}-2,\) evaluate \(f(i).\)
\(\text{If}\ f(x)={x}^{2}+3x+5,\) evaluate \(f(2+i).\)

Jawaabta muuji

\(14+7i\)
\(\text{If}\ f(x)=2{x}^{2}+x-3,\) evaluate \(f(2-3i).\)
\(\text{If}\ f(x)=\frac{x+1}{2-x},\) evaluate \(f(5i).\)

Jawaabta muuji

\(-\frac{23}{29}+\frac{15}{29}i\)
\(\text{If}\ f(x)=\frac{1+2x}{x+3},\) evaluate \(f(4i).\)
\((3+2i)+(5-3i)\)

Jawaabta muuji

\(8-i\)
\((-2-4i)+(1+6i)\)
\((-5+3i)-(6-i)\)

Jawaabta muuji

\(-11+4i\)
\((2-3i)-(3+2i)\)
\((-4+4i)-(-6+9i)\)

Jawaabta muuji

\(2-5i\)
\((2+3i)(4i)\)
\((5-2i)(3i)\)

Jawaabta muuji

\(6+15i\)
\((6-2i)(5)\)
\((-2+4i)(8)\)

Jawaabta muuji

\(-16+32i\)
\((2+3i)(4-i)\)
\((-1+2i)(-2+3i)\)

Jawaabta muuji

\(-4-7i\)
\((4-2i)(4+2i)\)
\((3+4i)(3-4i)\)

Jawaabta muuji

25

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Sida loo: Complex numbers

Express square roots of negative numbers as multiples of
Plot complex numbers on the complex plane.
Add and subtract complex numbers.
Multiply and divide complex numbers.
Write
Express

Su'aalaha dadku waydiiyaan

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Qaybaha ee boggaan waxaa la beddelay ka OpenStax Precalculus 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.

In ka badan Precalculus

Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

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	\({(-1)}^{17}⋅i\)	\(-{i}^{33}\)	\({i}^{31}\)	\({i}^{19}\)