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\title{Complex Numbers and Complex Solutions}
\date{October 4, 2005}

\begin{document}
\maketitle
\section{Objectives}  
\begin{itemize}
\item Write the square of a negative number as a complex number and perform operations on complex numbers.
\item Determine the equality of two complex numbers.
\item Add, subtract and multiply complex numbers.
\item Use complex conjugates to divide complex numbers.
\end{itemize}
Many quadratic equations do not have real number solutions.    In order to find their solutions, we must extend our usual   understanding of numbers to include a more general notion of   numbers, called complex numbers. \\

\begin{itemize}
%---------------------------------------------------
\item \textbf{Recall 1}:  The equation $x^2 - x + 1 = 0$ has no real solutions.  If we attempt to use the quadratic   formula to find the solutions we get
$$x=\frac{1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}. $$
Simplifying, we get $x= \frac{1+ sqrt (-3)}{2}$.  But what does  that mean? There is no real number whose square is negative!  So, what do we do?  Extend the real numbers to a new category of   numbers called the complex numbers; these are numbers   of the form $a + ib$, where $a$ and $b$ are real numbers and $i$ represents $\sqrt{-1}$.  So, our solution is  two complex numbers:
$$\frac{1}{2} + i\frac{\sqrt{3}}{2}$$
and
$$\frac{1}{2} - i\frac{\sqrt{3}}{2}.$$ 
\end{itemize}
%---------------------------------------------------
\textbf{Note 1:} Notice that in Recall 1 $\frac{1}{2} + i\frac{\sqrt{3}}{2}$ is a complex number because it is of the form $a + bi$, where $a$ and $b$ are real numbers.  We call $a$ the \emph{real part} of the complex number and $b$ the \emph{imaginary part}.  Here, the real part is $\frac{1}{2}$, and the imaginary part is $\frac{\sqrt{3}}{2}$.

Moreover, you should notice that real numbers are a special case of complex numbers, since any real number $a$ may be written in the form $a +(0)i$. \\
%---------------------------------------------------
\textbf{Note 2:} By extending the notion of number to include complex number, we guarantee that any quadratic equation will have at least one solution, which may be real or complex.
%---------------------------------------------------
\begin{itemize}
\item Complex numbers may be \emph{added} together (or \emph{subtracted}) from one another by adding (or subtracting) real parts to real parts, and imaginary parts to imaginary parts:
$$(a + bi) \pm (c + di) = (a � c) \pm (b � d)i.$$
For example,
$$(3 + 2i) + (5 - 3i) = (3+5) + (2-3)i=8-i.$$ 
%---------------------------------------------------
\item Complex numbers may be \emph{multiplied} together to get another complex number.  Multiplying two complex numbers is the same as multiplying two binomials.  Keep in mind that $i*i=i^2=\sqrt{-1}^2=-1$, so:
$$(a + bi)(c + di) = (ac - bd) + (ad + bc)i.$$
For example,

$\begin{array}{ll}
(3 + 2i)(5 - 3i) &= (3\cdot5)+(3\cdot(-3i))+(2i\cdot5)+(2i\cdot(-3i))\\
\mbox{}&= 15-9i+10i-6i^2\\
\mbox{}&=15-6(-1)+i \\
\mbox{}&=15+6+i \\
\mbox{}&=21+i.
\end{array}$
%---------------------------------------------------
\item
\textbf{Recall 4}: Even though some quadratic equations may be split into a   product of linear terms as $(x - r_1)(x - r_2)$, it is not always obvious what the values of $r_1$ and $r_2$ should be.
Here is a way to find the values of $r_1$ and $r_2$.

  In splitting the quadratic $x^2 - 3x + 2 = 0$, we are looking  for an $r_1$ and $r_2$ that fit into

$$x^2 - 3x + 2 = (x - r_1)(x - r_2).$$

  By multiplying out the right-hand-side, we find that

$$x^2 - 3x + 2 = x^2 - (r_1 + r_2)x + r_1*r_2.$$

  By comparing the coefficients of the right side with those of the left side, we find that

$3 = r_1 + r_2$ and $2 = r_1*r_2$

  If we want $r_1$ to be an integer, then it must be $\pm 2$ or $\pm 1$, for the only two integers that multiply together to give 2 are 2   and 1.  Test $\pm 2$ and $\pm 1$ to find the right combination that   gives $r_1 + r_2 = 3$ and $r_1*r_2 = 2$. You can see that $r_1 = 1$ and   $r_2 = 2$ is the right choice.  Hence, we know that

$$x^2 - 3x + 2 = (x - 1)(x - 2)$$

  and therefore that $x = 1$ and $x = 2$ are solutions to the   equation $x^2 - 3x + 2 = 0$. 
%---------------------------------------------------
\end{itemize}
\textbf{Note:} It is important to check your solutions by substituting them into the original equations.  In the case of the solutions $x = 1$ and $x = 2$ above, we check that
$$(1)^2 - 3(1) + 2 = 0.$$ and that $$(2)^2 - 3*(2) + 2 = 0.$$
In general, checking is easy and worth the little extra time. 
%---------------------------------------------------
\begin{itemize}
\item \textbf{Recall 5}:   If the factors of a quadratic equation are not immediately   apparent, you may use the \emph{quadratic formula} to find the solutions to

$$ax^2 + bx + c = 0.$$

The quadratic formula says that
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

For example, the equation $$x^2 - x - 1 = 0$$ has $a=1$, $b=-1$, and $c=-1$.  So the solutions are
$$x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(-1)},$$
which reduce to 
$$x= \frac{1 + \sqrt{5}}{2}$$  and  $$x= \frac{1 - \sqrt{5}}{2}.$$ 
This means that there are two solutions to the equation $x^2 - x - 1 = 0,$ namely,   
$$x= \frac{1 + \sqrt{5}}{2}$$  and  $$x= \frac{1 - \sqrt{5}}{2}.$$ 

\end{itemize}

%---------------------------------------------------
\section{Examples}
\begin{itemize} 
\item \textbf{Example 1.} Find the solutions to the equation $x^2 + 5x + 6 = 0.$

\emph{Solution}:   Look for two integers whose product is 6 and whose sum is -5.  The factors of 6 are $\pm 2$,$\pm 3$, $\pm 1$ and $\pm6$. So, candidates for $r_1$ and $r_2$ are

$r_1 = 2$ and $r_2 = 3$,$r_1 = -2$ and $r_2 = -3$, $r_1 = 1$ and $r_2 = 6$, or $r_1 = -1$ and $r_2 = -6.$

Next, we test the product
$$(x + 2)(x + 3),$$
to see if it is equal to $x^2 + 5x + 6$.   It is.  Therefore, $x = -2$   and $x = -3$ are the solutions to our equation. 
%---------------------------------------------------
\item \textbf{Example 2.}Find the solutions to the equation $2x^2 + 10x + 12 = 0$.

\emph{Solution}:    After factoring out the 2, the equation becomes
$$2(x^2 + 5x + 6) = 0$$
 Divide both sides by 2 to get the equivalent equation
$$x^2 + 5x + 6 = 0.$$
This is the equation from Example 1 and, therefore, it   has the same solutions $x = -2$ and $x = -3$.
%---------------------------------------------------
\item \textbf{Example 3.} Find the solutions to the equation $3x^2 + 7x - 6 = 0$.

\emph{Solution}:    Notice that leading term is a 3, not a 1, and that not all of the other terms are divisible by 3, so we cannot factor it out. We begin by noticing that the only way to get a 3 as the leading term is if one of the factors of $3x^2 + 7x - 6$ has a factor of 3 as its leading term.  The only factors of 3 are 3 and 1; so, we look for factors of the form
$$(3x - r_1)(x - r_2).$$
The factors of 6 are $\pm 2$,$\pm 3$, $\pm 1$ and $\pm6$. So candidates for $r_1$ and $r_2$ are 

$r_1 = 2$ and $r_2 = 3$,$r_1 = -2$ and $r_2 = -3$, $r_1 = 1$ and $r_2 = 6$, or $r_1 = -1$ and $r_2 = -6.$

 Next, we test the product with one of our candidates, say $r_1 = -1$ and $r_2 = 6$:
$$(3x - (-1))(x - 6)$$
 to see if it is equal to $3x^2 + 7x - 6$.   It is not.  Therefore, we must test  another possible set of values for $r_1$ and $r_2$ , say $r_1 = 2$ and $r_2 = -3$.  This  means that our factors have the form
$$(3x - 2)(x - (-3)),$$  or
$$(3x - 2)(x + 3).$$
Test this last product to see that it multiplies out to
$$3x^2 + 7x - 6.$$
 This means that $3x - 2 = 0$ and $x + 3 = 0$.  We may conclude that the solutions are
$x =\frac{2}{3}$  and $x = -3$. 

%---------------------------------------------------
\item \textbf{Example 4.} Find the solutions to the equation $2x^2 + 3x - 3 = 0.$

\emph{Solution}:      If you search for factors of the form
$$(2x - r_1)(x - r_2)$$
where $r_1$ and $r_2$ are rational numbers, you will not find them.  There are none.  In this case, the best you can do is use the quadratic formula by comparing $2x^2 + 3x - 3 = 0$ with $ax^2 + bx + c = 0$.  Notice that these last two quadratic equations are the same when $a = 2$, $b = 3$ and $c = -3$. (Note that  $c$ must be -3, not +3.)  Then, the solutions are
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
where $a = 2$, $b = 3$ and $c = -3$.  Thus, x becomes
$$x=\frac{-(3) \pm \sqrt{(3)^2-4(2)(-3)}}{2(-3)}$$
Notice that these solutions are not rational. 

%---------------------------------------------------
\end{itemize}


\textbf{Note:} How do you know when to look for factors of the quadratic and when to use the quadratic formula?  You may \emph{always} use the quadratic formula.  Even in the cases where the equation factors.  However, to check if the formula is necessary, try computing the quantity $b^2 - 4ac$.  If this quantity is a perfect square, then the factors will be rational numbers; otherwise, the factors will not be rational.  The next example shows what happens if you use the quadratic formula on an equation that has simple integer factors.
%---------------------------------------------------
\begin{itemize}
\item \textbf{Example 5.} Find the solutions to the equation $2x^2 + 3x - 2 = 0.$

 \emph{Solution}:     If you compute the value of $b^2 - 4ac$, as suggested above, you find that it is equal to 25, since $b = 3$, $a = 2$ and $c = -2$.  This tells us that $b^2 - 4ac$ is a perfect square, since $25 = 5^2$.  So the quadratic equation factors.  Indeed, it factors as

$$(2x - 1)(x + 2) = 0,$$

from which we may conclude that $x =\frac{1}{2}$  or $x = -2$.

But, what if we used the \emph{quadratic formula} to find the solutions to $2x^2 + 3x - 2 = 0$?

Since $a = 2$, $b = 3$ and $c = -2$, 
$$x=\frac{-(3) \pm \sqrt{(-3)^2-4(2)(-2)}}{2(-2)},$$
which reduces to $x = \frac{1}{2}$ or $x = -2$.  Notice that these are the same solutions that we computed from the factoring of the equation.    This means that you are always welcome to use the quadratic formula. 
\end{itemize}
%---------------------------------------------------

\section{Exercises}
Solve the following quadratic equations for the appropriate variable:\\

1. $x^2 - x - 2 = 0$\\
 
 

2. $x^2 - 7x + 12 = 0$\\
 
 

3  $x^2 + 2x - 8 = 0$\\
 
 

4 $2x^2 + 6x + 4 = 0$\\
 
 

5 $2x^2 - x - 1 = 0$\\
 
 

6 $3x^2 + 7x + 2 = 0$\\
 
 

7.  $x^2 + 3x - 5 = x + 3$ Hint:  Subtract $(x + 3)$ from both sides.\\
 
 

8.  $3x^2 - 1 = x^2 + x$              Hint:  Collect all terms to the left side.\\
 
 

9.  $x^2 + x - 1 = 0$\\
 
 

10. $3x^2 - 2x - 4 = 0$\\
 
 

11.  $5x^2 = 5 - 2x$\\
 
 

12.  $3x^2 + 4x + 1 = 0$\\
 
 

13.  $6x^2 - 5x - 6 = 0$\\
 
 

14.  $15x^2 - 16x - 15 = 0$\\
 
 

15.  $x^2 + 0.9x - 2.52 = 0$\\
 
 

16.  $25x^2 -  4 = 0$\\
 
 

17. $4x^2 - 20x + 25 = 0$\\
\end{document}