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MATH 1113 PreCalculus
Day 37, Project #6
Complex Numbers
You have probably been introduced to complex numbers somewhere in your previous math classes as an algebraic idea: that is, the symbol "i" was invented to solve the equation, x2 = -1, which has no real number solution, since any real number squared is always non-negative. In this context, the historical (but very unfortunate) name "imaginary number" is used, as if other numbers are less imaginary, or these numbers are less "real."
In fact, in many (if not most) circumstances in mathematics, complex numbers are to be preferred to real numbers, because (as was mentioned in Chapter 4), the complex numbers are "algebraically complete:"
Every polynomial factors completely into a product of linear factors over the complex numbers.
This is often referred to as the "Fundamental Theorem of Algebra," although the most common statement is that "every polynomial equation of degree greater than or equal to one with complex coefficients has a solution over the complex numbers." (The link above has a small mistake, since it's missing the "or equal to," but since its explanation is otherwise so good, I've chosen it as the best link.) If you put this together with the "roots = factors" theorem, and apply a little induction, you get the more useful version I've stated.
The Algebra of Complex Numbers:
Considered formally (that is, just as a set of symbols), we can define the number "i" to be the principal square root of -1, and we can define the principal square root of any negative number to be the square root of the absolute value of the number, times "i:"
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Then a complex number is defined as any expression of the form: a + bi, where a and b are real numbers.
Consequently, the algebra of complex numbers basically boils down to treating "i" formally (like an "x"), then substituting "-1" for all occurrences of "i2":
(3 - 5i)(2 + 3i) = 6 -i - 15i2
6 - i - 15(-1) = -9 - i.
To perform division, you need to "rationalize" the denominator, using the "conjugate trick," just like you would for other square roots:

Most often, this is as far as you go, in an algebra class, in learning about complex numbers.
The Geometry of Complex Numbers:
Much of the power and usefulness (not to mention beauty) of complex numbers comes from thinking of them as ordered pairs of real numbers:
a+bi = (a, b)
so that the y-axis becomes the "imaginary" axis, and the x-axis becomes the "real" axis.
Thinking of them this way, we can then look at their polar form, by converting this point to polar coordinates:
x = a = r*cos(t), y = b = r*sin(t), so that a+bi = r*(cos(t) + i*sin(t))
Example: The complex number: 5 + 11i can be written in polar form:
r = sqrt(25 + 121) = sqrt(146), and t = arctan(5/11)
sqrt(146)*(cos(arctan(5/11)) + i*sin(arctan(5/11)).
Notice that, thanks to the sum of angles formula, we get a wonderful result about the product of two complex numbers:
(a + bi)*(c + di) = r1(cos(t1) + i*sin(t1))*r2(cos(t2) + i*sin(t2))
= r1*r2*(cos(t1)*cos(t2) + i*(sin(t1)cos(t2)+sin(t2)cos(t1)) + i2*sin(t1)*sin(t2))
= r1*r2*(cos(t1+t2) + i*sin(t1+t2))
That is, when you multiply complex numbers, the modulus (radius) is multiplied, while the angle is added.
So complex multiplications involve rotations! And, if you multiply by complex numbers with radius = 1, it is the same as a pure rotation.
Try this Geogebra Applet, to see what happens when you multiply two complex numbers together.
Complex Functions:
Complex functions are usually written as w = f(z),
where z = x + iy.
And the output is complex-valued, so w = u + iv.
With real-valued functions, we can simply input points along one axis, and output the result along the other. It's a bit more difficult to visualize complex functions, since, with a 2-dimensional domain (input: (x,y) ) and a 2-dimensional range (output (u,v) ), graphing the whole thing would require 4 dimensions. Instead, what we usually do is investigate what happens to particular geometric shapes in the domain, when the function is applied to them.
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Look at the function: f(z) = z^2. If you write it out as (x + iy)^2, you get: (x+iy)^2 = x^2 + 2*i*x*y + i^2y^2 = x^2 - y^2 + 2*i*x*y. Use the following Geogebra applets to answer the questions: 1. Using Applet #1, observe and describe what happens to points on the unit circle. 2. Using Applet #2, observe and describe (draw a picture!) what happens to points on the circle with radius = 1, centered at (0, 1). See if you recognize the object as a shape from polar functions! 3. Using Applet #3, observe and describe (draw pictures!) what happens to points on the two different squares. 4. Using Applet #4, investigate other shapes: A) Circles with the origin in their interior (but
not at the center)
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Page Last Modified: 21 March, 2007
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