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Math Notes

Complex Numbers in Cartesian Form (a + bi)

Introduction

Complex numbers are the sum of a real number (the kind we're used to working with) and an imaginary number.  We usually write complex numbers in the form a + bi, where a and b are both real numbers, and the letter i indicates which of these elements is the imaginary number .  Let's take a moment to introduce some terms and notation we are going to use in referring to complex numbers.

z is the variable we use to refer to complex numbers, rather than x.

Complex numbers are made up of a real part and an imaginary part.  Re(z) is the notation we use to refer to the real part of a complex number.  Im(z) is the notation we use to refer to the imaginary part of a complex number.  When a number is written in the form z = a + bi, Re(z) = a, and Im(z) = b.

The most common way we encounter complex numbers are through applying the quadratic formula to quadratic equations which have no real solution. [To refresh our memories, recall that a quadratic equation is of the form ax2 + bx + c = 0.  The quadratic formula helps us find the values of x that makes this equation true.   Remember that in our formula, the a, b, c refer back to the coefficients in our original quadratic equation.]

Consider the example x2 + x + 1 = 0.  This equation can't be factored to find a solution, so we can go the the quadratic formula to find a solution.  Unlike the equation x2 + 3x + 1 = 0, when we use the quadratic formula, we end up with a negative under the square root: .  Notice, that when we

simplify this expression as much as we can, we end up with a real part, -1/2 in this case, and an imaginary part, .  Just as when we end up with real solutions, we have two solutions whenever the number under the square root is not negative.  These two roots are known as complex conjugates of each other.

The complex conjugate, notated by z, is found by making the imaginary part of z the opposite sign.  So, for instance, if z = a + bi, then z = a - bi.  In our example above, one solution is and the other is .

Complex numbers allow us to make generalizations about polynomials that would not be possible using just real numbers.  The most important of these generalizations is the Fundamental Theorem of Algebra, which states that every polynomial of degree n has exactly n roots (counting real, complex and repeated roots).

One of the consequences of this theorem, is that every polynomial with real coefficients will have an even number of complex roots.  That even number may be zero, or it will be two, four, six, etc.  These roots will come in pairs of complex conjugates.

We want to be able to perform basic operations on complex numbers as well, such as addition, subtraction, multiplication and division.

When we are doing addition and subtraction, we are going to distribute signs and add like terms just as we would as if we are using variables.  For example: z1 = 3 - 4i  and z2 = 2 + 6i,
then z1 + z2 = (3 - 4i) + (2 + 6i) = (3 + 2) + (-4 + 6)i = 5 + 2i. 
And z1 - z2 = (3 - 4i) - (2 + 6i) = (3 -2) + (-4 - 6)i = 1 - 10i.

When we are doing multiplication, at first, we are also going to treat complex numbers as though we are dealing with a single variable, and FOIL.
So z1*z2 = (3 -4i)*(2 + 6i) = 6 + 18i - 8i - 24i2 = 6 + 10i - 24i2.  But we aren't done, because we have to simplify i2.  Remember, i2 = -1. So 6 + 10i - 24i2 = 6 + 10i - 24(-1) = 6 + 10i + 24 = 30 + 10i.

When we do division, we have to first remove any imaginary numbers from the denominator.  Consider a number like .  In order to remove the imaginary number from the denominator, we are going to multiply this fraction by where z2 is the complex number in the denominator.  One of the things we didn't talk about before when we dealt with complex conjugates is that when we multiply a complex number by it's conjugate the result is a purely real number.  It's just like multiplying the difference of squares formula, the middle terms drop out because they are of different signs.  So for this example z2 is 2 - 3i.  So to reduce this problem, we do the following calculation: .

This method of eliminating the imaginary number in the denominator is very similar to the process we use to rationalize denominators in expressions like .

Sometimes we want to look at complex numbers visually.  When we want to look at real numbers, we draw a one-dimensional number line and each point on that line represents a unique real number.  But when we want to look at complex numbers, because they have both a real and an imaginary part, we are going to need not one, but two dimensions.  This means that complex numbers are graphed not on a line, but on a plane.  We can treat complex numbers then like we would any other point in a plane, as a Cartesian coordinate, where the x-coordinate or first coordinate is the real part or Re(z), and the y-coordinate or second coordinate is the imaginary part or Im(z).  Now a + bi can be represented as simply (a, b).

The complex plane then looks like the picture at left.

The corners of the center square then are, starting in the first quadrant, 1 + i, second quadrant, -1 + i, third quadrant, -1 - i, and fourth quadrant, 1 - i.

We will do more with plotting complex numbers in the plane when we move on to Complex Numbers in Polar and Trigonometric Form.

 

 

 

Worked Examples

A.  State Re(z) and Im(z) for the following complex numbers.

     a. z = c + di              b. z = 3 - 4i             c. z = 5                  d. z = -4i

a. Re(z) = c, Im(z) = d;   b. Re(z) = 3, Im(z) = -4;  c. Re(z) = 5, Im(z) = 0;   d. Re(z) = 0, Im(z) = -4

We report Re(z) to be the part of a complex number lacking an i; we report Im(z) to be the part of the complex number which is attached to the i, but Im(z) is just the coefficient in front of the i, not the i itself.  Also note the either the real or imaginary part may be 0.

B. Give  the real and imaginary parts of the solutions to the quadratic equation 2x2 - x + 3 = 0.

Use the quadratic formula to get .Re(x) = 1/4, Im(x) = .

C. Give the complex conjugate of a. z = 4 - 6i, b. z = 5, c. z = i

To find the complex conjugate, just change the sign of the imaginary part.  For a. z = 4 + 6i; for b. z = 5 because there is no imaginary part, it's just like 5 + 0i, and 5 - 0i is exactly the same thing; for c. z = -i.

D. For the equation x4 + 3x3 + 6x2 + x - 4 = 0, how many roots does this polynomial have?

Since the highest degree term is x4, this is a degree-4 polynomial, so it has to have 4 roots.

E. Add  and subtract the following complex numbers: z1 = 2 + 4i, z2 = -7 + 3i.

When we add we get 2 + 4i + (-7 + 3i) = (2 - 7) + (4 + 3)i = -5 + 7i.
When we subtract, we get 2 + 4i - (-7 + 3i) = 2 + 4i + 7 - 3i = (2 + 7) + (4 - 3)i = 9 + i.

F. Multiply and divide the following complex numbers: z1 = 1 + i, z2 = 3 - 4i

z1*z2 = (1 + i)(3 - 4i) = 3 - 4i + 3i - 4i2 = 3 - i - 4i2 = 3 - i - 4(-1) = 3 - i + 4 = 7 - i.
z1/z2 =

Problem Solving Tips

When dealing with complex numbers, there are some strategies you can use:

  • If you are having trouble getting your head around what a complex number is exactly, you can think of the i term just like a variable,  Complex numbers like 2 + 9i will add, subtract and multiply exactly as 2 + 9x does.  It may help you to think of complex numbers in terms of having certain rules to follow.  You'll find, after working with them for a while, that they begin to seem like they make more sense when they seem less unfamiliar.  However, you can also try some links at the bottom of the page for additional explanations.
  • Remember to remove complex numbers from a denominator using the complex conjugate before attempting to simplify further.
  • Don't forget to replace i2 with (-1).
  • When you are asked for Im(z), the answer never has an i in it.

Additional Problems with Answers

1. State Re(z) and Im(z) of the following complex numbers:

    a. 4 + 6i                 b.           c. -i                 d. 1/4

2. Find the real and imaginary parts of the solutions to the following polynomials:

    a. 5x2 - 12x -3 = 0      b. -3x2  + 2x - 3 = 0         c. x3 - 1 = 0

3.  Find the complex conjugates of:

    a. z = 1 + 0.4i           b. z = 3 - 5i         c. -.0006i        d. i2

4. How many real or complex roots do the following polynomials have?

    a. x3 - 1 = 0              b. x5 + 6x -2 = 0           c. x6 + 5x4 - 3x2 - x + 4 = 0

5. Given the following complex numbers: z1 = 2 - i, z2 = 3 + 2i, z3 = 4 - 5i, z4 = 2 + 4i, z5 = -2 - 3i, simplify the following expressions:

   a. z1 + z2            b. z2 + z3         c. z3 - z4         d. z4 - z5         e. z5*z1        f. z1*z3         g. z2/z4        h. z3/z5

6. Plot the following complex numbers in the plane:

    a. 2 - i                b. i                 c. 1 + 3i           d. 4                e. 0

Answers:

1. a. Re(z) = 4, Im(z)=6; b. Re(z) = 1/2, Im(z) = -3/2; Re(z) = 0, Im(z) = -1; Re(z) = 1/4, Im(z) = 0
2. a. Re(x) =
, Im(x) = 0; b. Re(x) = -1/3, Im(x) = ; c. Re(x) = 1, Im(x) = 0 AND Re(x) = -1/2, Im(x) =

3. a. 1 - 0.4i, b. 3 + 5i, c. 0.0006i, d. -1 (this number is real when it's reduced, so it's the same)
4. a. 3, b. 5, c. 6
5. a. 5 + i, b. 7 - 3i, c. 2 -9i, d. 4 + 7i, e. -7 - 4i, f. 3 - 14i, g. 7/10 - (2/5)i, h. 7/13 + (22/13)i
6. Plot the points with the following coordinates (2, -1), (0, 1), (1, 3), (4, 0), (0, 0).

Links to Outside Sources

An introduction to complex numbers
Complex Numbers
Complex numbers - Wikibooks
College Algebra Tutorial on Complex Numbers
Complex numbers - A complete course in algebra
Problems about Complex numbers
Complex Numbers
Using the Graphing Calculator with Complex Numbers

Links to Supporting Topics

Imaginary Numbers
Complex Numbers in Polar or Trigonometric Form and Exponential Form (rcos(t) + risin(t)) & reit
Math Forum: Ask Dr. Math FAQ: Imaginary Numbers
Answers and Explanations -- Do "Imaginary Numbers" Really Exist?
Imaginary Number -- from MathWorld
Imaginary Numbers
Imaginary Numbers
Imaginary Numbers Are Not Imaginary.

History

Question Corner -- The Origin of Complex Numbers
A History of Hypercomplex Numbers
Re: so, who invented the IMAGINARY NUMBER?
Wikipedia article

 

 
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