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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 in i-form and perform operations on numbers in i -form.
\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

(real number) + (another real number)$i$.

Moreover, you should notice that real numbers are a special case of complex numbers, since any real number may be written in the form

(real number) + (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 \textbf{Recall 2} Complex numbers may be \emph{added} together or \emph{subtracted} from one another, the result being a new complex number defined as:
$$(a + bi) \pm (c + di) = (a � c) \pm (b � d)i.$$
For example,
$$(3 + 2i) + (5 - 3i) = (3+5) + (2-3)i=8-i.$$ 
%---------------------------------------------------
\item \textbf{Recall 3}: Complex numbers may be \emph{multiplied} together to get another complex number.  Multiplying two complex numbers is the same as multiplying two binomials.  So:
$$(a + bi)(c + di) = (ac - bd) + (ad + bc)i.$$
For example,
$$(3 + 2i)(5 - 3i) = (3*5)+(3*(-3i))+(2i*5)+(2i*(-3i))= 21 + i.$$ 
%---------------------------------------------------
\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}