Complex Numbers
Contents
Complex Numbers¶
Quadratic Formula¶
A quadratic formula has the form
\( \begin{align} f(x) =& ax^2 + bx + c \end{align} \)
For some contants \(a,b,c\) and variable \(x\).
We can set the constants to make a specific instance of the general quadratic formula.
\( \begin{align} f(x) =& 5x^2 + 2x - 3 \end{align} \)
This function creates a parabola. We may want to know the roots of the quadratic formula. These are the points at which \(f(x)=0\). We can plot the function and the line \(g(x)=0\) to see where they intersect.
This function intersects with the zero line at two points. This means there are two roots of the function.
In our case, the roots are \(x=-1\) and \(x=\frac{3}{5}\). If we plug either of these numbers in, the function evaluates to zero.
The first root is -1.
\( \begin{align} f(-1) =& 5(-1)^2 + 2(-1) - 3 \\ =& 5 -2 - 3 \\ =& 5-5 \\ =& 0 \end{align} \)
The second root is \(\frac{3}{5}\).
\( \begin{align} f\left(\frac{3}{5}\right) =& 5\left(\frac{3}{5}\right)^2 + 2\left(\frac{3}{5}\right) - 3 \\ =& 5\left( \frac{9}{25} \right) + \frac{6}{5} - 3 \\ =& \frac{9}{5} + \frac{6}{5} - 3 \\ =& \frac{15}{5} - 3 \\ =& 3 - 3 \\ =& 0 \end{align} \)
The roots of every quadratic expression can be solved using the following formula.
\( \begin{align} x =& \frac{ -b \pm \sqrt{b^2-4ac}}{2a} \end{align} \)
We can plug in the values from our example. We have \(a=5\), \(b=2\), and \(c=-3\).
\( \begin{align} x =& \frac{ -2 \pm \sqrt{2^2-4 * 5 * (-3)}}{2*5} \\ =& \frac{-2 \pm \sqrt{64}}{10} \\ =& \frac{-2 \pm 8}{10} \end{align} \)
We use use the plus version we get \(x=\frac{6}{10}=\frac{3}{5}\). If we use the minus version then we will get \(x=\frac{-10}{10}=-1\). The equation gives us both zeros.
This expression does work for every possible quadratic expression. We can plug in the expression for \(x\) and verify the answer is zero.
\( \begin{align} 0 =& ax^2+bx+c \\ =& a\left(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\right)^2+b\left(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\right)+c \\ =& \left( \frac{b^2 \pm -b\sqrt{b^2-4ac}-2ac }{2a} \right)+\left(\frac{-b^2 \pm b\sqrt{b^2-4ac}}{2a}\right)+c \\ =&\frac{(b^2-b^2)\pm(-b\sqrt{b^2-4ac}+b\sqrt{b^2-4ac})-2ac}{2a}+c \\ =&c-\frac{2ac}{2a} \\ =&0 \end{align} \)
What happens if we try different quadratic expressions?
\( \begin{align} h(x) = x^2 + 2x + 3 \end{align} \)
When we plot this function, we see that it never touches the zero line.
It appears that this function has no answers. We know \(a=1\), \(b=2\), and \(c=2\). What happens if we try to compute the answers?
\( \begin{align} x=&\frac{-2 \pm \sqrt{2^2-4*1*3}}{2*1} \\ =&\frac{-2 \pm \sqrt{-8}}{2} \end{align} \)
What is the square root of \(-8\)? We know that \(8=2*2*2\) To get negative eight, we need one of the following to happen:
\( \begin{align} -8 =& (-2) * 2 * 2 \\ -8 =& (-2) * (-2) * (-2) \end{align} \)
We can simplify \(\frac{-2 \pm \sqrt{-8}}{2}\) to \(-1 \pm \sqrt{-2}\). This as far as we can get without knowing what \(\sqrt{-1}\) is. We don’t know of any number that can be squared to get \(-1\). Regardless, we also have this \(\sqrt{-1}\) appear in our problem.
We introduce a new symbol, the imaginary number \(i=\sqrt{-1}\).
We know have our two solutions \(x=-1 \pm i \sqrt{2}\). The equation has two answers, but they are imaginary. They do not appear on the plot because we cannot plot \(\sqrt{-1}\) on this simple cartesian plane.
Imaginary Numbers¶
The Real Numbers are the set of numbers that do not include \(i\). If we combine all Real and Imaginary numbers, we get the set of Complex Numbers.
What happens when we compute the exponents of \(i\)?
\( \begin{align} i^0 =& 1 \\ i^1 =& \sqrt{-1} \\ i^2 =& -1 \\ i^3 =& -i \\ i^4 =& 1 \end{align} \)
A Complex Number is any number that has for format \(a+bi\) where \(a\) and \(b\) are real numbers.
We can work with Complex Numbers.
We can add two complex numbers together.
\( \begin{align} (a+bi) + (x+yi) =& (a+x) + (b+y) i \end{align} \)
We can subtract two complex numbers.
\( \begin{align} (a+bi) - (x+yi) =& (a-x) + (b-y)i \end{align} \)
We can multiply two complex numbers.
\( \begin{align} (a+bi) * (x+yi) =& (ax-by) + (ay+bx)i \end{align} \)
Using the multiplication formula we can also square a complex number. This can be extended for any exponent.
\( \begin{align} (a+bi)^2 =& (a^2-b^2)+2abi \end{align} \)
Next, we introduce some additional operations on Complex Numbers.
The Absolute Value of a complex number is given by the following expression. The absolute value is the distance from the origin (0,0) to the point (a,b) on the cartesian plane.
\( \begin{align} \left|(a+bi)\right| = \sqrt{a^2+b^2} \end{align} \)
We can find the reciprocal of a complex number.
\( \begin{align} \frac{1}{a+bi} = \frac{a}{a^2+b^2}+\frac{-b}{a^2+b^2}i \end{align} \)
We can also find the complex conjugate.
\( \begin{align} (a+bi)^{*} = a-bi \end{align} \)
What happens when we multiply a number by it’s complex conjugate?
\( \begin{align} (a+bi)(a+bi)^{*} =& (a+bi)(a-bi) \\ =& a^2-abi+abi+b^2 \\ =& a^2+b^2 \\ =& \left| a+bi \right|^2 \end{align} \)
The complex conjugate is a mirror of the imaginary number across the x-axis. Some areas us \(\overline{a+bi}\) as the complex conjugate symbol. This book will use the \((a+bi)^{*}\) notation.
Euler’s Number¶
Euler’s Number, also called the natural number, is the solution to the following summation.
\( \begin{align} e =& \sum_{n=0}^{\infty} \left( \frac{1}{n!} \right) \\ =& 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \cdots \\ \approx& 2.71828 \end{align} \)
Euler’s number appears in many areas. One of the most common is the computation of compound interest.
Euler’s number also has a geometric definition. It is the point \(p\) at which the area under the curve \(\frac{1}{x}\) from \(1\) to \(p\) is equal to exactly 1.
We can write this using an integeral. We want to know \(p\) such that
\( \begin{align} \int_{1}^{p} \frac{1}{x} = 1 \end{align} \)
The answer to this question is \(p=e\).
It is also the base of the natural log.
\( \begin{align} \ln e^k = k \end{align} \)
It also appears in Euler’s identity which is ludicrous and shouldn’t be true.
\( \begin{align} e^{i \pi} + 1 = 0 \end{align} \)
There is a relationship between \(e\) and Complex Numbers.
\( \begin{align} e^{ix} =& \cos x + i \sin x \end{align} \)
The formula for \(e^{ix}\) creates a circle! We can take a point \(a+bi\) and convert it to a value \(re^{ix}\) to get a point on the circle.
Every Complex Number can be represented by a point on the circle. Let \(z=a+bi\) be a complex number, then we can fine a matching radius \(r\) and angle \(\theta\) on the circle.
\( \begin{align} r=& \left| z \right| \\ \theta =& \tan^{-1}(b/a) \end{align} \)
We can figure out where on the circle the complex number \(9+5i\) is.
\( \begin{align} r = & |9+5i| = \sqrt{9^2+5^2} = 56.62155 \\ \theta =& \tan^{-1} \frac{5}{9} = 0.50798 \text{ Radians} \end{align} \)
Representing complex numbers as points on a circle gives us a geometric realization of the numbers. This can provide advantages when working with complex numbers. For example, the complex conjugate is easily visualized.
The complex conjugate of \(9+5i\) is \(9-5i\). The points on the circle are computed as
\( \begin{align} r = & |9-5i| = \sqrt{9^2+(-5)^2} = 56.62155 \\ \theta =& \tan^{-1} \frac{-5}{9} = -0.50798 \text{ Radians} \end{align} \)
C++ Class¶
As a running example, we will simulate the mathematics behind each example using C++. The first component we need is a Complex Number class. There is one included in the standard library but for educational purpose we will build up our own.