 Introduction to maginary Numbers
 Working with “\(i\)”
 Quadratic Formula with Complex Solutions
 Completing the Square with Complex Solutions
 More Practice
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 (because they are more easily used in advanced computations), but really don’t exist in life as we know it. Yet they are real in the sense that they do exist and can be explained quite easily in terms of math as the square root of a negative number.
Actually, imaginary numbers are used quite frequently in engineering and physics, such as an alternating current in electrical engineering, which is usually represented by a complex number. (Don’t worry; I don’t know too much about alternating currents either.)
Let’s 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” (nonreal) 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 nonreal 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 that has no real part. It can get a little confusing!
Now let’s 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:
\({{i}^{{\,2}}}\,\,=\,\,\,1\)
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).
And here’s something really cool: when we multiply \(i\)’s together, we notice a pattern:
Note that there’s a repeating pattern when raising “\(i\)” to an exponent. Notice that 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 PreCalculus, 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 
\(\begin{align}\color{#800000}{{\left( {43i} \right)\left( {2+4i} \right)}}&=43i24i\\&=\left( {42} \right)+\left( {3i+4i} \right)\\&=2+7i\\&=27i\end{align}\)  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. 
\(\begin{align}\color{#800000}{{\left( {8i} \right)\left( {4+i} \right)}}&=\left( {8i} \right)\left( 4 \right)+\left( {8i} \right)\left( i \right)\\&=32i+8{{i}^{2}}\\&=32i+8\left( {1} \right)\\&=32i8\end{align}\)  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. 
\(\begin{align}\color{#800000}{{\left( {7i1} \right)\left( {4i+3} \right)}}&=\left( {7i} \right)\left( {4i} \right)+\left( {7i} \right)\left( 3 \right)\left( 1 \right)\left( {4i} \right)\left( 1 \right)\left( 3 \right)\\&=\left( {28} \right)\left( {{{i}^{2}}} \right)+21i4i3\\&=28+21i4i3\\&=31+17i\end{align}\)  Multiplying Binomials: FOIL, treating the “\(i\)’s” like variables and replacing \({{i}^{2}}\) with \(1\). 
\(\begin{align}\color{#800000}{{\frac{4}{{2+i}}}}&=\frac{4}{{2+i}}\times \frac{{2i}}{{2i}}\\&=\frac{{4\left( {2i} \right)}}{{\left( {2+i} \right)\left( {2i} \right)}}=\frac{{84i}}{{42i+2i{{i}^{2}}}}\\&=\frac{{84i}}{{4\left( {1} \right)}}=\frac{{84i}}{5}\\&=\frac{8}{5}\frac{4}{5}i=1.6.8i\end{align}\)  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 \(43i\): \(\begin{align}\color{#800000}{{\frac{1}{{43i}}}}&=\frac{1}{{43i}}\times \frac{{4+3i}}{{4+3i}}\\&=\frac{{1\left( {4+3i} \right)}}{{\left( {43i} \right)\left( {4+3i} \right)}}=\frac{{4+3i}}{{1612i+12i9{{i}^{2}}}}\\&=\frac{{4+3i}}{{16\left( {9} \right)}}=\frac{{4+3i}}{{25}}=\frac{4}{{25}}+\frac{3}{{25}}i\end{align}\) 
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.

Here are a few more without the “\(i\)”s; be careful to do the operations in the correct order:
Complex Operation  Notes 
\(\begin{align}\color{#800000}{{\sqrt{{3}}\cdot \sqrt{{24}}}}=\sqrt{3}i\cdot \sqrt{{24}}i\\=\sqrt{3}i\cdot 2\sqrt{6}i\\=2\sqrt{{18}}\cdot {{i}^{2}}\\=2\cdot 3\sqrt{2}\\=6\sqrt{2}\end{align}\)  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 \(\displaystyle \frac{i}{i}\). 
\(\displaystyle \begin{align}\color{#800000}{{\sqrt{{8}}+3\sqrt{{4}}}}&=\sqrt{8}i+3\sqrt{4}i\\&=2\sqrt{2}i+3\sqrt{4}i\\&=2\sqrt{2}i+3\cdot 2i=\left( {2\sqrt{2}+6} \right)i\end{align}\)  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!
You first have to 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 you can use “2^{nd} .” (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.
You can even turn the final answers into fractions using MATH ENTER (for Frac) after you get your final answer.

(Note that 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.)
Quadratic Formula with Complex Solutions
Now let’s solve a quadratic equation that has complex (imaginary) solutions.
Let’s take the equation \({{x}^{2}}2x+2\). 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.)
Let’s use the quadratic equation to find this solution, and one that’s a little more complicated:
Quadratic with Complex Solutions  Solve Using Quadratic Formula 
\({{x}^{2}}2x+2=0\)
\(\begin{array}{l}a=1\\b=2\\c=2\end{array}\) 
\(\require{cancel} \displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{\left( {2} \right)\pm \sqrt{{{{{\left( {2} \right)}}^{2}}4\left( 1 \right)\left( 2 \right)}}}}{{2\left( 1 \right)}}\\&=\frac{{2\pm \sqrt{{4}}}}{2}=\frac{{2\pm 2i}}{2}\\&=\frac{{\cancel{2}(1\pm i)}}{{\cancel{2}}}=1\pm i\end{align}\) 
\(3{{x}^{2}}2x+2=0\)
\(\begin{array}{l}a=3\\b=2\\c=2\end{array}\) 
\(\displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{\left( {2} \right)\pm \sqrt{{{{{\left( {2} \right)}}^{2}}4\left( 3 \right)\left( 2 \right)}}}}{{2\left( 3 \right)}}\\&=\frac{{2\pm \sqrt{{20}}}}{6}=\frac{{2\pm 2\sqrt{5}\,i}}{6}\\&=\frac{{{}^{1}\cancel{2}(1\pm \sqrt{5}\,i)}}{{{{{\cancel{6}}}^{3}}}}=\frac{1}{3}\pm \frac{{\sqrt{5}i}}{3}\text{ }\text{ or }\text{ }\frac{1}{3}\pm \frac{{\sqrt{5}}}{3}i\text{ }\end{align}\)

Completing the Square with Complex Solutions
Let’s try completing the square with a quadratic with complex solutions:
Completing the Square with Complex (Imaginary) Solutions  
\(\displaystyle {{x}^{2}}2x+2=0\)
\(\displaystyle \begin{align}{{x}^{2}}2x&=2\\\,\,\,\,\,\,{{x}^{2}}2x+\,\,\underline{{\,\,\,\,\,\,}}&=2+\,\,\underline{{\,\,\,\,\,\,}}\\{{x}^{2}}2x+\underline{{\color{#03A89E}{{{{{\left( {\frac{2}{2}} \right)}}^{2}}}}}}&=2+\underline{{\color{#03A89E}{{{{{\left( {\frac{2}{2}} \right)}}^{2}}}}}}\\\,\,\,{{x}^{2}}2x+\underline{{\color{#03A89E}{{{{{\left( 1 \right)}}^{2}}}}}}&=2+\underline{{\color{#03A89E}{{{{{\left( 1 \right)}}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {x1} \right)}^{2}}&=1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\left( {x1} \right)}}^{2}}}}&=\sqrt{{1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x1&=\pm \sqrt{{1}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=\pm \,i+1=1\pm \,\,i\end{align}\) 
\(\displaystyle 3{{x}^{2}}2x+2=0\)
\(\require{cancel} \displaystyle \begin{align}3\left( {{{x}^{2}}\frac{2}{3}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=2+\,\,\underline{{\,\,\,\,\,\,}}\\\frac{{\cancel{3}}}{{\cancel{3}}}\left( {{{x}^{2}}\frac{2}{3}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=\frac{{2}}{3}+\,\,\underline{{\,\,\,\,\,\,}}\\\,\,\,\,{{x}^{2}}\frac{2}{3}x+\underline{{\color{#03A89E}{{{{{\left( {\frac{{\cancel{{{}^{1}2}}}}{{\cancel{{{}^{3}6}}}}} \right)}}^{2}}}}}}&=\frac{{2}}{3}+\underline{{\color{#03A89E}{{{{{\left( {\frac{1}{3}} \right)}}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {x\frac{1}{3}} \right)}^{2}}&=\frac{{{}^{{6}}\cancel{{2}}}}{{\cancel{{{}^{9}3}}}}+\frac{1}{9}=\frac{{5}}{9}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\left( {x\frac{1}{3}} \right)}}^{2}}}}&=\pm \sqrt{{\frac{5}{9}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\frac{1}{3}&=\pm \frac{{\sqrt{5}}}{3}\,\,i\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=\,\frac{1}{3}\,\,\pm \frac{{\sqrt{5}}}{3}\,\,i\end{align}\) 
Yeah! We got the same answers as when we solved with the Quadratic Equation!
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 \(3i\) is also a root (the conjugate is always the second imaginary solution). Interesting!
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