| 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
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{, so same as }{{i}^{1}}} \right)\,\,\,\,\,\,\text{So, }\color{#800000}{{{{i}^{{77}}}}}=i\\\color{#800000}{{{{i}^{{110}}}}}&=\,\,\,?\,\,\,\,\,\,\left( {\frac{{110}}{4}=27R2\text{, so same as }{{i}^{2}}} \right)\,\,\,\,\,\text{So, }\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:
| Complex Operation | Notes |
| $ \begin{align}\color{#800000}{{\left( {4-3i} \right)-\left( {2+4i} \right)}}&=4-3i-2-4i\\&=\left( {4-2} \right)+\left( {-3i+-4i} \right)\\&=2+-7i\\&=2-7i\end{align}$ | Adding and Subtracting: Use distributive property and treat the $ i$ like a variable. Combine like terms – the reals versus the imaginaries (watch signs). If there are no reals or no imaginaries in the expression, don’t include it: you may just end up with a $ -7i$, for example. |
| $ \begin{align}\color{#800000}{{\left( {8i} \right)\left( {4+i} \right)}}&=\left( {8i} \right)\left( 4 \right)+\left( {8i} \right)\left( i \right)\\&=32i+8{{i}^{2}}\\&=32i+8\left( {-1} \right)\\&=32i-8\end{align}$ | Multiplying: Use the distributive property, but anywhere there is an $ {{i}^{2}}$, replace it with $ -1$. Anywhere there is an $ i$ raised to any power, simplify it. |
| $ \begin{align}\color{#800000}{{\left( {7i-1} \right)\left( {4i+3} \right)}}&=\left( {7i} \right)\left( {4i} \right)+\left( {7i} \right)\left( 3 \right)-\left( 1 \right)\left( {4i} \right)-\left( 1 \right)\left( 3 \right)\\&=\left( {28} \right)\left( {{{i}^{2}}} \right)+21i-4i-3\\&=-28+21i-4i-3\\&=-31+17i\end{align}$ | Multiplying Binomials: Multiply the binomials, like were did here in the Introduction to Multiplying Polynomials section, treating the “$ i$’s” like variables and replacing $ {{i}^{2}}$ with $ -1$. |
| $ \begin{align}\color{#800000}{{\frac{4}{{2+i}}}}&=\frac{4}{{2+i}}\times \frac{{2-i}}{{2-i}}\\&=\frac{{4\left( {2-i} \right)}}{{\left( {2+i} \right)\left( {2-i} \right)}}=\frac{{8-4i}}{{4-2i+2i-{{i}^{2}}}}\\&=\frac{{8-4i}}{{4-\left( {-1} \right)}}=\frac{{8-4i}}{5}\\&=\frac{8}{5}-\frac{4}{5}i=1.6-.8i\end{align}$ | Dividing: It’s bad “math grammar” to have any “$ i$’s” in a denominator. To fix this, multiply the top and bottom both (which makes it 1) by a complex conjugate: the real part of the complex number stays the same, but the $ i$ part is negated (plus turns to minus, minus turns to plus).
This magically gets rid of the $ i$ in the denominator! |
| Take reciprocal of $ 4-3i$: $ \begin{align}\color{#800000}{{\frac{1}{{4-3i}}}}&=\frac{1}{{4-3i}}\times \frac{{4+3i}}{{4+3i}}\\&=\frac{{1\left( {4+3i} \right)}}{{\left( {4-3i} \right)\left( {4+3i} \right)}}=\frac{{4+3i}}{{16-12i+12i-9{{i}^{2}}}}\\&=\frac{{4+3i}}{{16-\left( {-9} \right)}}=\frac{{4+3i}}{{25}}=\frac{4}{{25}}+\frac{3}{{25}}=.16-.12i\end{align}$ | Taking the reciprocal in this case means putting 1 on top and putting the complex number on the bottom.
Then use the complex conjugate method above to “fix” the complex number. The complex conjugate that we used to simplify a denominator with an imaginary number in it is similar to the radical conjugate we learned about here in the Introduction to Quadratics section. |
| $ \left| {4+3i} \right|$ | This is called the modulus, and, by definition, $ \left| {a+bi} \right|$ is defined as $ \sqrt{{{{a}^{2}}+{{b}^{2}}}}$. Thus, $ \left| {4+3i} \right|=\sqrt{{{{4}^{2}}+{{3}^{2}}}}=\sqrt{{25}}=5$. As we’ll see in the Trigonometry and the Complex Plane section, the modulus is defined as the distance between $ \left( {0,0} \right)$ and point $ \left( {a,b} \right)$ in the complex plane: |
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.
| Complex Operation | Notes |
| $ \begin{align}\color{#800000}{{\sqrt{{-3}}\cdot \sqrt{{-24}}}}&=\sqrt{3}i\cdot \sqrt{{24}}i\\&=\sqrt{3}i\cdot 2\sqrt{6}i\\&=2\sqrt{{18}}\cdot {{i}^{2}}=2\sqrt{2}\cdot \sqrt{9}\cdot {{i}^{2}}\\&=-2\sqrt{2}\cdot 3=-6\sqrt{2}\end{align}$ | Make sure to convert the radicals into imaginary (complex) numbers with “$ i$”’s first. Then, group the numbers and multiply, remembering that $ {{i}^{2}}=-1$.
Note that you can’t work this problem this way: $ \color{#800000}{{\sqrt{{-3}}\cdot \sqrt{{-24}}}}=\sqrt{{-3\cdot -24}}=\sqrt{{72}}=6\sqrt{2}$ . |
| $ \displaystyle \begin{align}\color{#800000}{{\frac{{3\sqrt{2}}}{{\sqrt{{-8}}}}}}&=\frac{{3\sqrt{2}}}{{2\sqrt{2}i}}\\&=\frac{3}{{2i}}\cdot \frac{i}{i}\\&=\frac{{3i}}{{-2}}=-\frac{3}{2}i\end{align}$ | Make sure to convert the radicals into imaginary (complex) numbers with “$ i$”’s first.
Since we typically don’t want “$ i$”’s in the denominator, multiply by $ \displaystyle \frac{i}{i}$ ($ 1$). |
| $ \displaystyle \begin{align}\color{#800000}{{\sqrt{{-8}}+3\sqrt{{-4}}}}&=\sqrt{8}i+3\sqrt{4}i\\&=2\sqrt{2}i+3\sqrt{4}i\\&=2\sqrt{2}i+3\cdot 2i=\left( {2\sqrt{2}+6} \right)i\end{align}$ | Make sure to convert the radicals into imaginary (complex) numbers with “$ i$”’s first. Then, group the numbers and multiply. |
You can also put complex expressions in the graphing calculator:
| Graphing Calculator Screens | Notes |
 | 
You can put complex expressions in the graphing calculator!
First push MODE and scroll down to REAL and scroll over to a+bi and hit ENTER. It’s OK to leave your calculator like this.
Then, use “2nd .” (to the right of the “0” button) for “$ \boldsymbol {i}$” in any expression. After you do the computations, the answer will be in a+bi form.
If needed, turn the final answers into fractions using MATH, ENTER (for Frac), ENTER after you get your final answer. |
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.
| Quadratic with Complex Solutions | Solve Using Quadratic Formula |
| $ {{x}^{2}}-2x+2=0$
$ \begin{array}{l}a=1\\b=-2\\c=2\end{array}$ | $ \require{cancel} \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 1 \right)\left( 2 \right)}}}}{{2\left( 1 \right)}}\\&=\frac{{2\pm \sqrt{{-4}}}}{2}=\frac{{2\pm 2i}}{2}\\&=\frac{{\cancel{2}(1\pm i)}}{{\cancel{2}}}=1\pm i\end{align}$ |
| $ 3{{x}^{2}}-2x+2=0$
$ \begin{array}{l}a=3\\b=-2\\c=2\end{array}$ | $ \displaystyle \begin{align}\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}&=\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 3 \right)\left( 2 \right)}}}}{{2\left( 3 \right)}}\\&=\frac{{2\pm \sqrt{{-20}}}}{6}=\frac{{2\pm 2\sqrt{5}\,i}}{6}\\&=\frac{{{}^{1}\cancel{2}(1\pm \sqrt{5}\,i)}}{{{{{\cancel{6}}}^{3}}}}=\frac{1}{3}\pm \frac{{\sqrt{5}i}}{3}\text{ }\text{ or }\text{ }\frac{1}{3}\pm \frac{{\sqrt{5}}}{3}i\text{ }\end{align}$ |
(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:
| Completing the Square with Complex (Imaginary) Solutions | |
| $ \displaystyle {{x}^{2}}-2x+2=0$
$ \displaystyle \begin{align}{{x}^{2}}-2x&=-2\\\,\,\,\,\,\,{{x}^{2}}-2x+\,\,\underline{{\,\,\,\,\,\,}}&=-2+\,\,\underline{{\,\,\,\,\,\,}}\\{{x}^{2}}-2x+\underline{{\color{#03A89E}{{{{{\left( {\frac{2}{2}} \right)}}^{2}}}}}}&=-2+\underline{{\color{#03A89E}{{{{{\left( {\frac{2}{2}} \right)}}^{2}}}}}}\\\,\,\,{{x}^{2}}-2x+\underline{{\color{#03A89E}{{{{{\left( 1 \right)}}^{2}}}}}}&=-2+\underline{{\color{#03A89E}{{{{{\left( 1 \right)}}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {x-1} \right)}^{2}}&=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\left( {x-1} \right)}}^{2}}}}&=\sqrt{{-1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-1&=\pm \sqrt{{-1}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=\pm \,i+1=1\pm \,\,i\end{align}$ | $ \displaystyle 3{{x}^{2}}-2x+2=0$
$ \require{cancel} \displaystyle \begin{align}3\left( {{{x}^{2}}-\frac{2}{3}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=-2+\,\,\underline{{\,\,\,\,\,\,}}\\\frac{{\cancel{3}}}{{\cancel{3}}}\left( {{{x}^{2}}-\frac{2}{3}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=\frac{{-2}}{3}+\,\,\underline{{\,\,\,\,\,\,}}\\\,\,\,\,{{x}^{2}}-\frac{2}{3}x+\underline{{\color{#03A89E}{{{{{\left( {\frac{{\cancel{{{}^{1}2}}}}{{\cancel{{{}^{3}6}}}}} \right)}}^{2}}}}}}&=\frac{{-2}}{3}+\underline{{\color{#03A89E}{{{{{\left( {\frac{1}{3}} \right)}}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {x-\frac{1}{3}} \right)}^{2}}&=\frac{{{}^{{-6}}\cancel{{-2}}}}{{\cancel{{{}^{9}3}}}}+\frac{1}{9}=\frac{{-5}}{9}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\left( {x-\frac{1}{3}} \right)}}^{2}}}}&=\pm \sqrt{{-\frac{5}{9}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-\frac{1}{3}&=\pm \frac{{\sqrt{5}}}{3}\,\,i\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=\,\frac{1}{3}\,\,\pm \frac{{\sqrt{5}}}{3}\,\,i\end{align}$ |
Yeah! We got the same answers as when we solved with the Quadratic Equation!
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