# Complex Numbers

## Introduction to Complex Numbers

A complex number has the form [latex]a+bi[/latex], where [latex]a[/latex] and [latex]b[/latex] are real numbers and [latex]i[/latex] is the imaginary unit.### Learning Objectives

Describe the properties of complex numbers and the complex plane### Key Takeaways

#### Key Points

- A complex number is a number that can be expressed in the form [latex]a+bi[/latex], where [latex]a[/latex] and [latex]b[/latex] are real numbers and [latex]i[/latex] is the imaginary unit.
- The real number [latex]a[/latex] is called the real part of the complex number [latex]z=a+bi[/latex] and is denoted [latex]\text{Re}\{a+bi\}=a[/latex]. The real number [latex]b[/latex] is called the imaginary part of [latex]z=a+bi[/latex] and is denoted [latex]\text{Im}\{a+bi\}=b[/latex].

#### Key Terms

**real number**: An element of the set of real numbers. The set of real numbers include the rational numbers and the irrational numbers, but not all complex numbers.**imaginary number**: a number of the form [latex]ai[/latex], where [latex]a[/latex] is a real number and [latex]i[/latex] the imaginary unit**complex**: a number, of the form [latex]a+bi[/latex], where [latex]a[/latex] and [latex]b[/latex] are real numbers and [latex]i[/latex] is the square root of [latex]-1[/latex].

### The Complex Number System

A complex number is a number that can be put in the form [latex]a+bi[/latex] where [latex]a[/latex] and [latex]b[/latex] are real numbers and [latex]i[/latex] is called the imaginary unit, where [latex]i^2=-1[/latex]. In this expression, [latex]a[/latex] is called the real part and [latex]b[/latex] the imaginary part of the complex number. We will write [latex]\text{Re}\{a+bi\}=a[/latex] to indicate the real part of the complex number, and [latex]\text{Im}\{a+bi\}=b[/latex] to indicate the imaginary part. For example, to indicate that the real part of the number [latex]2+3i[/latex] is [latex]2[/latex], we would write [latex]\text{Re}\{2+3i\}=2[/latex]. To indicate that the imaginary part of [latex]4-5i[/latex] is [latex]-5[/latex], we would write [latex]\text{Im}\{4-5i\} = -5[/latex]. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. The complex number [latex]a+bi[/latex] can be identified with the point [latex](a,b)[/latex]. Thus, for example, complex number [latex]-2+3i[/latex] would be associated with the point [latex](-2,3)[/latex] and would be plotted in the complex plane as shown below.**The complex point [latex]-2+3i[/latex]:**The complex number [latex]-2+3i[/latex] is plotted in the complex plane, [latex]2[/latex] to the left on the real axis, and [latex]3[/latex] up on the imaginary axis.

## Addition and Subtraction of Complex Numbers

Complex numbers can be added and subtracted by adding the real parts and imaginary parts separately.### Learning Objectives

Calculate the sums and differences of complex numbers by adding the real parts and the imaginary parts separately### Key Takeaways

#### Key Points

- Complex numbers can be added and subtracted to produce other complex numbers. This is done by adding the corresponding real parts and the corresponding imaginary parts.
- It is possible for two non-real complex numbers to add to a real number. However, two real numbers can never add to be a non-real complex number.

### Sums of Complex Numbers

Complex numbers can be added and subtracted to produce other complex numbers. This is done by adding the corresponding real parts and the corresponding imaginary parts. For example, the sum of [latex]2+3i[/latex] and [latex]5+6i[/latex] can be calculated by adding the two real parts [latex](2+5)[/latex] and the two imaginary parts [latex](3+6)[/latex] to produce the complex number [latex]7+9i[/latex]. Note that this is always possible since the real and imaginary parts are real numbers, and real number addition is defined and understood. As another example, consider the sum of [latex]1-3i[/latex] and [latex]4+2i[/latex]. In this case, we would add [latex]1[/latex] and [latex]4[/latex] to produce [latex]5[/latex] and also would add [latex]-3[/latex] and [latex]2[/latex] to produce [latex]-1[/latex]. Thus we would write: [latex-display](1-3i)+(4+2i)=5-i[/latex-display]### Differences (Subtraction) of Complex Numbers

In a similar fashion, complex numbers can be subtracted. The key again is to combine the real parts together and the imaginary parts together, this time by subtracting them. Thus to compute: [latex-display](4-3i)-(2+4i)[/latex-display] we would compute [latex]4-2[/latex] obtaining [latex]2[/latex] for the real part, and calculate [latex]-3-4=-7[/latex] for the imaginary part. We would thus write [latex](4-3i)-(2+4i) = 2-7i[/latex]. Note that the same thing can be accomplished by imagining that you are distributing the subtraction sign over the sum [latex]2+4i[/latex] and then adding as defined above. Thus you could write: [latex-display](4-3i)-(2+4i) = (4-3i)+(-2-4i) = 2-7i.[/latex-display] Note that it is possible for two non-real complex numbers to add to a real number. For example, [latex](1-3i)+(1+3i)=2+0i=2[/latex]. However, two real numbers can never add to be a non-real complex number.## Multiplication of Complex Numbers

Complex numbers can be multiplied using the FOIL algorithm.### Learning Objectives

Calculate the product of complex numbers using FOIL and the properties of [latex]i[/latex]### Key Takeaways

#### Key Points

- The imaginary unit [latex]i[/latex] has the property that [latex]i^2=-1[/latex]
- Complex numbers can be multiplied using the FOIL algorithm

### The Square of the Imaginary Unit [latex]i[/latex]

In the following calculations, it is important to remember that [latex]i^2=-1[/latex]. It is best to not let this fact confuse you, but to just remember it as a fact. Any time an [latex]i^2[/latex] appears in a calculation, it can be replaced by the real number [latex]-1.[/latex]### Multiplying Complex Numbers

Two complex numbers can be multiplied to become another complex number. The key to performing the multiplication is to remember the acronym FOIL, which stands for First, Outer, Inner, Last. Thus, we multiply [latex]a+bi[/latex] and [latex]c+di[/latex] by writing [latex](a+bi)(c+di)[/latex] and multiplying the First terms [latex]a[/latex] and [latex]c[/latex], and then the Outer terms [latex]a[/latex] and [latex]di[/latex] and then the Inner terms [latex]bi[/latex] and [latex]c[/latex] and then the Last terms [latex]bi[/latex] and [latex]di[/latex]. Note that this last multiplication yields a real number, since: [latex-display]bi\cdot di = bd \cdot i^2=bd\cdot (-1) =-bd[/latex-display] Note that the FOIL algorithm produces two real terms (from the First and Last multiplications) and two imaginary terms (from the Outer and Inner multiplications). We then combine these to write our complex number in standard form. Thus we have: [latex](a+bi)(c+di)=ac+adi+bci-bd=(ac-bd)+(ad+bc)i[/latex]. For example, consider the product [latex](2+3i)(4+5i)[/latex]. We would compute: [latex]\begin {align}(2+3i)(4+5i)&=8+10i+12i-15 \\&=(8-15)+(10+12)i \\&=-7+22i \end {align}[/latex] As another example, consider the product: [latex]\begin {align}(1-i)(2+4i)&=1\cdot 2 +1\cdot 4i -2 i -(4i^2) \\&= 2+2i-(-4) \\&=6+2i \end {align}[/latex] Note that if a number has a real part of [latex]0[/latex], then the FOIL method is not necessary. For example: [latex]\begin {align}(0+5i)(2+5i)&=5i(2+5i) \\&= 10i+25i^2 \\&=-25+10i \end {align}[/latex] Similarly, a number with an imaginary part of [latex]0[/latex] is easily multiplied as this example shows: [latex](2+0i)(4-3i)=2(4-3i)=8-6i.[/latex] Note that it is possible for two nonreal complex numbers to multiply together to be a real number. For example: [latex]\begin {align} (2-3i)(2+3i)&=4+6i-6i-9i^2 \\&=4+9 \\&=13 \end {align}[/latex]## Complex Numbers and the Binomial Theorem

Powers of complex numbers can be computed with the the help of the binomial theorem.### Learning Objectives

Connect complex numbers raised to a power to the binomial theorem### Key Takeaways

#### Key Points

- The binomial theorem can be used to compute powers of complex numbers. To compute [latex](a+bi)^n[/latex] we consider the expression [latex](x+y)^n [/latex] where [latex]x=a[/latex] and [latex]y=bi[/latex].
- The powers of [latex]i[/latex] are [latex]i^2=-1[/latex], [latex]i^3=-i[/latex], [latex]i^4=1[/latex], and [latex]i^5=i[/latex], etc.

#### The Powers of [latex]i[/latex]

In what follows, it is useful to keep in mind the powers of the imaginary unit [latex]i[/latex]: [latex]\begin {align}i^1&=i \\i^2&=-1 \\i^3&=-i \\i^4&=1 \end {align}[/latex] After that, they repeat, since [latex]i^5=i^4\cdot i = i[/latex].### Computing Powers of Complex Numbers

Powers of complex numbers can be computed with the the help of the binomial theorem. Recall the binomial theorem, which tells how to compute powers of a binomial like [latex]x+y[/latex]. It says: [latex-display]{ (x+y) }^{ n }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ n-k }{ y }^{ k }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ k }{ y }^{ n-k }[/latex-display] For example, consider the case [latex]n=4.[/latex] We have: [latex-display](x+y)^4 = x^4 +4x^3y+6x^2y^2 + 4xy^3+y^4[/latex-display] We can use this to compute the fourth power of a complex number [latex]a+bi[/latex] by letting [latex]x=a[/latex] and [latex]y=bi[/latex]. Then we have: [latex-display](a+bi)^4=a^4+4a^3bi+6a^2b^2i^2+4ab^3i^3+b^4i^4[/latex-display] Now recalling the powers of [latex]i[/latex], we have: [latex-display](a+bi)^4 = a^4+4a^3bi-6a^2b^2+-4ab^3i+b^4[/latex-display] If we gather the real terms and the imaginary terms, we have the complex number: [latex-display](a^4-6a^2b^2+b^4)+(4a^3b-4ab^3)i[/latex-display]### Example 1

Suppose you wanted to compute [latex](2+3i)^4[/latex]. Using the previous example as a guide, we have: [latex-display](2+3i)^4=2^4+4\cdot 2^3\cdot 3i -6\cdot 2^2\cdot 3^2 -4\cdot 2 \cdot 3^3 i +3^4 [/latex-display] which can be written as: [latex-display]16 + 96i-216-216i+81 = -119-120i[/latex-display]### Example 2

Suppose you wanted to compute [latex](1+i)^3[/latex]. Using the binomial theorem directly, this can be written as: [latex-display]1^3+3\cdot 1^2 \cdot i + 3\cdot 1 \cdot i^2+i^3[/latex-display] which can be simplified to: [latex-display]1+3i-3-i=-2+2i[/latex-display]### Example 3

Suppose you wanted to compute [latex](2+i)^5[/latex]. Recall that the binomial coefficients (from the 5th row of Pascal's triangle) are [latex]1, 5, 10, 10, 5, \text{and}\, 1.[/latex] Using the binomial theorem directly, we have: [latex-display](2+i)^5 =2^5 + 5\cdot 2^4 i + 10\cdot 2^3 i^2 + 10\cdot 2^2 i^3 + 5\cdot 2 \cdot i^4 + i^5[/latex-display] This can in turn be written as: [latex-display]32+80i-80-40i+10+i = -38+41i[/latex-display]## Complex Conjugates

The complex conjugate of the number [latex]a+bi[/latex] is [latex]a-bi[/latex]. Two complex conjugates of each other multiply to be a real number with geometric significance.### Learning Objectives

Explain how to find a complex number's conjugate and what it is used for### Key Takeaways

#### Key Points

- The complex conjugate of [latex]a+bi[/latex] is [latex]a-bi[/latex], and vice versa.
- Two complex conjugates multiply together to be the square of the length of the complex number.
- If a complex number is a root of a polynomial equation, then its complex conjugate is a root as well.

#### Key Terms

**modulus**: The length of a complex number, [latex]\sqrt{a^2+b^2}[/latex]**complex conjugate**: For the number [latex]a + bi[/latex], this is [latex]a-bi[/latex].

### Complex Conjugates

The complex conjugate (sometimes just called the*conjugate*) of a complex number [latex]a+bi[/latex] is the complex number [latex]a-bi[/latex]. Thus, for example, the conjugate of [latex]2+3i[/latex] is [latex]2-3i[/latex] and the conjugate of [latex]1-5i[/latex] is [latex]1+5i[/latex]. Since the conjugate of a conjugate is the original complex number, we say that the two numbers are conjugates of each other. The symbol for the complex conjugate of [latex]z[/latex] is [latex]\overline{z}[/latex]. So, we might write: [latex-display]\overline{3-6i} = 3+6i.[/latex-display]

### The Product of Two Conjugates

The product of two conjugates is always a real number. Note that: [latex-display](a+bi)(a-bi)=a^2-abi+abi-b^2i^2=a^2+b^2[/latex-display] This number has a geometric significance.**The length of a complex number:**The length of the line segment from the origin to the point [latex]a+bi[/latex] is [latex]\sqrt{a^2+b^2}[/latex]. This comes from the Pythagorean Theorem.

*modulus*of the complex number [latex]z=a+bi[/latex]. The symbol for the modulus of [latex]z[/latex] is [latex]|z|[/latex]. Thus we can write [latex-display]\begin{align} z\overline{z} &= (a+bi)(a-bi)\\&=a^2+b^2\\&=|z|^2 \end{align}[/latex-display] Or in other words: [latex-display]|z|=\sqrt{z\overline{z}}[/latex-display] The modulus symbol looks just like the absolute value symbol, which is okay because whenever [latex]b=0[/latex] so that [latex]z=a+bi=a[/latex] is a real number, we have that the conjugate is [latex]a-bi=a[/latex]. In this case we have [latex-display]\begin{align} \sqrt{z\overline{z}} &= \sqrt{a^2}\\&=|a|\\&=|z| \end{align}[/latex-display] So the symbol is consistent with the use of the absolute value symbol.

### Complex Roots Come in Conjugate Pairs

One important fact about conjugates is that whenever a complex number is a root of polynomial, its complex conjugate is a root as well. This can be seen in the quadratic formula whenever the discriminant [latex]b^2-4ac[/latex] is negative. For example, consider the equation: [latex-display]x^2+x+1=0[/latex-display] By the quadratic formula, the roots are [latex-display]x=\dfrac{-1 \pm \sqrt{1-4\cdot 1 \cdot 1}} {2} [/latex-display] Simplifying gives the two complex numbers [latex]-1/2+(\sqrt{3}/2)i[/latex] and [latex]-1/2-(\sqrt{3}/2)i[/latex], which are complex conjugates of each other.## Division of Complex Numbers

Division of complex numbers is accomplished by multiplying by the multiplicative inverse of the denominator. The multiplicative inverse of [latex]z[/latex] is [latex]\frac{\overline{z}}{|z|^2}.[/latex]### Learning Objectives

Use the complex conjugate to divide complex numbers### Key Takeaways

#### Key Points

- Division of complex numbers is accomplished by multiplying by the multiplicative inverse of the denominator.
- The multiplicative inverse of the complex number [latex]z[/latex] is [latex]\frac{\overline{z}}{|z|^2}[/latex]
- If [latex]w[/latex] and [latex]z[/latex] are complex numbers, we have [latex]\frac{w}{z} = w \cdot \frac{\overline{z}}{|z|^2}[/latex]

#### Multiplicative Inverses of Complex Numbers

We have seen how to add, subtract, and multiply complex numbers, but it remains to learn how to divide them. The key is to think of division by a number [latex]z[/latex] as multiplying by the multiplicative inverse of [latex]z[/latex]. You are probably already familiar with this concept for ordinary real numbers: dividing by [latex]2[/latex] is the same as multiplying by [latex]\frac12[/latex], dividing by 3 is the same as multiplying by [latex]\frac13[/latex], and so on. Algebraically, we write [latex]\frac{x}{k} = x \cdot \frac1k[/latex]. For complex numbers, the multiplicative inverse can be deduced using the complex conjugate. We have already seen that multiplying a complex number [latex]z=a+bi[/latex] with its complex conjugate [latex]\overline{z}=a-bi[/latex] gives the real number [latex]a^2+b^2[/latex]. Thus, we have [latex]z\overline{z}=a^2+b^2[/latex], and dividing through by [latex]a^2+b^2[/latex] gives: [latex-display]z \cdot \frac{\overline{z}}{a^2+b^2}=1[/latex-display] So the multiplicative inverse of [latex]z[/latex] must be the complex conjugate of [latex]z[/latex] divided by its modulus squared. We can write [latex]\frac{1}{z}= \frac{\overline{z}}{|z|^2}[/latex]### Example 1

The multiplicative inverse of [latex]1+2i[/latex] is: [latex-display]\frac{1-2i}{1+4}=(1/5)-(2/5)i[/latex-display] To see that this is correct, we can multiply these numbers to see if we get the multiplicative identity number [latex]1[/latex]. Using FOIL, we have: [latex]\begin {align}(1+2i)((1/5)-(2/5)i) &= 1/5 -(2/5)i+(2/5)i+4/5 \\&=1/5+4/5 \\&=1 \end {align}[/latex]### Example 2

The multiplicative inverse of [latex]3-4i[/latex] is: [latex-display]\frac{3+4i}{9+16} = (3/25)+(4/25)i[/latex-display] Again, checking through multiplication, we have: [latex]\begin {align}(3-4i)((3/25)+(4/25)i&=9/25 -(12/25)i+(12/25)i+16/25 \\&=\frac{25}{25} \\&=1 \end {align}[/latex]### Division of Complex Numbers

Suppose you wanted to divide the complex number [latex]z=2+3i[/latex] by the number [latex]1+2i[/latex]. Since dividing by [latex]1+2i[/latex] is the same as multiplying by the multiplicative inverse (which we have seen above is [latex](1/5)-(2/5)i[/latex]), we have: [latex-display]\frac{2+3i}{1+2i}=(2+3i)((1/5)-(2/5)i)[/latex-display] If we multiply out (FOIL) this last expression we obtain: [latex-display]2/5 -(4/5)i+(3/5)i+(6/5)=(8/5)-(1/5)i[/latex-display] In general, for complex numbers [latex]w[/latex] and [latex]z[/latex], we have [latex]\frac{w}{z} =w \cdot \frac{1}{z} = w\cdot \frac{\overline{z}}{|z|^2}[/latex] As an example, let's use this formula directly to compute [latex]\frac{1-i}{3-4i}[/latex] [latex]\begin {align}\frac{1-i}{3-4i} &= (1-i)\cdot\frac{3+4i}{9+16} \\&=(1-i)((3/25)+(4/25)i) \end {align}[/latex] This multiplies out to be [latex]\begin {align}(1-i)((3/25)+(4/25)i) &= 3/25 +(4/25)i-(3/25)i+4/25 \\&= 7/25 + (1/25)i \end {align}[/latex]## Complex Numbers in Polar Coordinates

Complex numbers can be represented in polar coordinates using the formula [latex]a+bi=re^{i\theta}[/latex]. This leads to a way to visualize multiplying and dividing complex numbers geometrically.### Learning Objectives

Explain how to represent complex numbers in polar coordinates and why it is useful to do so### Key Takeaways

#### Key Points

- The complex number [latex]z=a+bi[/latex] can be written [latex]z=re^{i\phi}[/latex] where [latex]r=\sqrt{a^2+b^2}[/latex] is the modulus of [latex]z[/latex], and [latex]\phi[/latex] is the angle that the line segment from the origin to [latex]z[/latex] makes with the horizontal.
- Using polar coordinates, it becomes easier to multiply and divide complex numbers, using the geometric interpretation as a guide.

**The parameters for polar coordinates:**A point in the first quadrant with Cartesian coordinates [latex](x, y)[/latex] and polar coordinates [latex](r, \phi)[/latex].

### Multiplying Complex Numbers in Polar Coordinates

If [latex]z=re^{i\phi}[/latex] and [latex]w=se^{i\theta}[/latex] are complex numbers, then the product of [latex]z[/latex] and [latex]w[/latex] is [latex]zw=rse^{i(\phi+\theta)}[/latex], which comes from simply multiplying as usual for exponential functions. We can then see that the product of [latex]z[/latex] and [latex]w[/latex] is the complex numbers whose distance from the origin is the product of the distances from the origin of [latex]z[/latex] and [latex]w[/latex], and whose angle with the horizontal is the sum of the angles of [latex]z[/latex] and [latex]w[/latex] with the origin. For example, consider the complex numbers [latex]z=\sqrt2e^{i\pi/4}=1+i[/latex] and [latex]w=\sqrt2e^{3i\pi/4} = -1+i[/latex]. So [latex]z[/latex] is the complex number which is [latex]\sqrt2[/latex] units from the origin and whose angle with the horizontal is [latex]\pi/4[/latex] radians, which is [latex]45 [/latex] degrees. Then [latex]w[/latex] is the number whose distance from the origin is [latex]\sqrt2[/latex] and whose angle with the origin is [latex]3\pi/4[/latex] radians which is [latex]135[/latex] degrees. When we multiply [latex](1+i)(-1+i)[/latex] by FOILing, we obtain [latex]-1+i-i-1=-2[/latex]. Perhaps more easily we could multiply [latex-display]\begin{align} zw&=\sqrt2 e^{i\pi/4}\cdot\sqrt2 e^{3i\pi/4} \\&= 2e^{i\pi} \\&= -2 \end{align}[/latex-display] Realizing that we are getting the number whose distance from the origin is [latex]2[/latex] and whose angle with the horizontal is [latex]\frac{\pi}{4}+\frac{3\pi}{4}=\pi,[/latex] or [latex]180[/latex] degrees.### Dividing Complex Numbers in Polar Coordinates

Similarly, if [latex]z=re^{i\phi}[/latex] and [latex]w=se^{i\theta}[/latex] then [latex]\frac{z}{w}[/latex] is the result of dividing [latex]\frac{re^{i\phi}}{se^{i\theta}} = \frac{r}{s} e^{i(\phi-\theta)}[/latex] In other words, when dividing by a complex number, the result is a number whose distance from the origin is the quotient of the distances of the two numbers from the origin, and whose angle with the horizontal is the difference of the angles with the horizontal of the two numbers. For example, If you were to divide [latex]z=\sqrt2e^{i\pi/4} = 1+i[/latex] by [latex]w=\sqrt{2}e^{3i\pi/4}=-1+i[/latex], the result would be: [latex-display]\frac{\sqrt2}{\sqrt2} e^{i(\pi/4 -3\pi/4)} = e^{-i\pi/2} = -i[/latex-display] The result is one unit from the origin and at an angle of [latex]-\pi/2[/latex] (or [latex]-90[/latex] degrees) with the horizontal. This way of thinking about multiplying and dividing complex numbers gives a geometric way of thinking about those operations.## Licenses & Attributions

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