Introduction to Imaginary Numbers | Completing the Square with Complex Solutions |

Working with “$ i$” | More Practice |

Quadratic Formula with Complex Solutions |

Note: There’s an example of expanding a complex binomial here in the **Binomial Expansion** section.

## Introduction to Imaginary Numbers

Think of **imaginary numbers** as numbers that are typically used in mathematical computations to get to/from “real” numbers, sometimes since they are easier to use in advanced computations. They are quite frequently used in engineering and physics, such as an alternating current in electrical engineering, Imaginary numbers really don’t exist as we know it, they are real in the sense that they can be explained quite easily in terms of math as the **square root of a negative number**. See how they don’t seem “real”: how can we square a number and get a negative number?

To see where they fit in in the grand scheme of things, look again at the Venn Diagram from the **Types of Numbers and Algebraic Properties** section:

A **complex number** consists of a “real” part and an “imaginary” (non-real) part, and typically looks like $ a+bi$, where “$ a$” is the real part, and “$ b$” is the imaginary part, following by “$ i$”, to indicate the “imaginary” unit. Note that complex numbers consist of both real numbers ($ a+0i$, such as **3**) and non-real numbers ($ a+bi,\,\,\,b\ne 0$, such as $ 3+i$); thus, all real numbers are also complex.

An imaginary number is the “$ i$” part of a real number, and exists when we have to take the square root of a negative number. So technically, an imaginary number is only the “$ i$” part of a complex number, and a pure imaginary number is a complex number whose “real part” is $ 0$. It can get a little confusing!

(Note: The definition of a **“non-real” number** is any number that does not lie on the real number line in the complex plane. This includes imaginary numbers, and complex numbers which have both a real, even $ 0$, **and** an imaginary part.)

As an example, look at the following graph and notice that the parabola **never** touches the $ x$-axis, so there aren’t any $ x$-intercepts, although the “roots”, “zeros”, “solutions”, and “values” are “complex” or “imaginary”.

As we learned here in the **Introduction to Quadratics** section, the **discriminant** can be used to determine what type of solutions a quadratic has. For imaginary solutions, since the graph has no roots, it has a discriminant $ \left( {{{b}^{2}}-4ac} \right)$ that is **less than** **0** (if it were equal to **0**, there’d be one solution, and if it were greater than zero, there’d be two solutions). The graph will never touch the $ x$-axis, yet we can still find imaginary roots, and the roots will have “$ i$”s in them, as we see later. But we can’t find these roots with a graphing calculator!

## Working with “$ i$”, the $ \sqrt{{-1}}$

Before working problems that have imaginary solutions, we need to learn about the value of a special “number” called “$ i$”. $ i$** **is simply $ \sqrt{{-1}}$, which can’t exist in our “real” system, since we can never take two “real” numbers multiplied together to get $ –1$. Since $ i$ equals $ \sqrt{{-1}}$, then it follows that $ \boldsymbol {{{i}^{{\,2}}}=-1}$. Thus, $ \displaystyle \color{#800000}{{\sqrt{{-16}}}}=4i$, since $ \displaystyle \sqrt{{-16}}=\sqrt{{\left( {16} \right)\times \left( {-1} \right)}}=\left( {\sqrt{{16}}} \right)\left( {\sqrt{{-1}}} \right)=4i$.

Similarly, $ \color{#800000}{{\sqrt{{-3}}}}=i\sqrt{3}$, or $ \sqrt{3}\,i$ (if you put the $ i$ at the end, make sure it is clearly outside of the square root sign in this case).

When we multiply $ i$’s together, we notice a pattern:

Note that there’s a repeating pattern when raising “$ i$” to an exponent. Every **fourth **exponent number repeats, so $ {{i}^{1}}={{i}^{5}}={{i}^{9}}=i$, and so on. Because of this, we can easily compute “$ i$” raised to any exponent by dividing that exponent by **four**, and examining the **remainder**, as shown in the examples:

$ \displaystyle \begin{align}\color{#800000}{{{{i}^{{77}}}}}&=?\,\,\,\left( {\frac{{77}}{4}=19R1\text{, same as }{{i}^{1}}} \right);\,\,\,\color{#800000}{{{{i}^{{77}}}}}=i\\\color{#800000}{{{{i}^{{110}}}}}&=?\,\,\,\left( {\frac{{110}}{4}=27R2\text{, same as }{{i}^{2}}} \right);\,\,\,\color{#800000}{{{{i}^{{110}}}}}=-1\end{align}$

Again, when dealing with complex numbers, expressions contain a **real** part and an **imaginary** part. Together they form a **complex** number that typically looks like $ a+bi$, where “$ a$” is the real part, and “$ b$” is the imaginary part, following by “$ i$”, to indicate the “imaginary” unit. (Later, in the **Trigonometry and the Complex Plane **section, we’ll see how these can be graphed on a coordinate system, where the “$ x$” is the real part and the “$ y$” is the imaginary part.)

For example, “$ 4+5i$” indicates the number $ \displaystyle 4+\left( 5 \right)\left( {\sqrt{{-1}}} \right)$, and we **cannot mix the real parts with the imaginary parts **when adding or subtracting, so that the “$ i$’s” are treated somewhat like variables (like radicals were thought as variables, back in the **Exponents and Radicals in Algebra **section).

Thus, when we perform operations on $ i$, we pretty much treat it like a variable, except when we’re multiplying the “$ i$’s” together – and then we can simplify. **Note that for good “math grammar” we want our final answer to be in the form **$ \boldsymbol{a+bi}$. Here are some examples:

Here are a few more without the “$ i$”s; be careful to do the operations in the correct order. The rule of thumb is to **convert square roots of negative numbers to imaginary numbers first**.

You can also put complex expressions in the **graphing calculator**:

## Quadratic Formula with Complex Solutions

As we saw in an example above, many quadratic equations have complex (imaginary) solutions; here is the graph for $ {{x}^{2}}-2x+2$.

We know that since the discriminant $ \left( {{{b}^{2}}-4ac} \right)$ for $ a,\,b,$ and $ c$ in $ a{{x}^{2}}+bx+c=0$ is negative ( $ -4$), there are no real solutions to the equation, but there are two imaginary solutions. (Note that if there are imaginary solutions, there are always **two** of them.)

Use the **quadratic equation** to find this solution, and one that’s a little more complicated.

(Note: We learned earlier here in the **Introduction to Quadratics** section that when we have an irrational value for a root, the **conjugate** is also a root. (For example, if $ \displaystyle 3+\sqrt{{17}}$ is a root, then $ \displaystyle 3-\sqrt{{17}}$ is also a root). Similarly, if we have a complex root, the **complex conjugate** is also a root; this is called the **Complex Conjugate Root Theorem**, or **Complex** **Conjugate Zeros Theorem**. For example, if $ 3+i$** **is a root, then $ 3-i$ is also a root (the conjugate is always the second imaginary solution). Interesting!)

## Completing the Square with Complex Solutions

We learned how to complete the square of a quadratic here in the **Solving Quadratics by Factoring and Completing the Square** section. Now we’ll work with a quadratic with complex solutions:

Yeah! We got the same answers as when we solved with the Quadratic Equation!

**Learn these rules and practice, practice, practice!**

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the **Mathway** site, where you can register for the **full version** (steps included) of the software. You can even get math worksheets.

You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

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On to **Solving Absolute Value Equations and Inequalities**– you are ready!