Complex Numbers and DeMoivre's Theorem

Complex Numbers- In previous math classes you have probably come upon an equation of the form x2+4 = 0 When you attempted to solve this equation you got x = Square root(-4) and your instructor probably said that is a non-real solution and we will not deal with that is this class. Well, this is the class that will deal with that problem. We will define the square root of -1 to equal i. If we have a quantity such as 3i we will call that an imaginary number (an unfortunate name that comes to us from history). If one of these special numbers has a real part and an imaginary part we will call it a compex number. In cartesian form we will write compex numbers a + bi, where a is the "real" part of the number and "bi" is the imaginary part of the number. Examples of complex numbers are shown below:

  1. 3 + 2i
  2. -4 + i
  3. -2 - 3i
  4. 5 -2i
  5. 0 + 3i
  6. 5 - 0i

Graphs/Plots - If we want to graph complex numbers we can use a set of cartesian coordinates (x, y) that has the real part on the horizontal axis and the imaginary part of the number on the vertical axis. The numbers shown above are plotted on the graph to the right. You plot complex numbers in a manner very similar to that used to plot x,y coordinates. The real part goes on the x axis and the imaginary part goes on the y axis. So 3 + 2i means 3 units to the right of the origin and then 2 units up. Now that we can graph complex numbers we will next look at the ways we can manipulate them. Specifically, we will look at addition, subtraction, multiplication and division.

Addition and Subtraction - When adding or subtracting complex numbers you will group real with real and imaginary with imaginary. This is very similar to the concept of grouping "like" terms. Real terms are "like" other real terms and imaginary terms are "like" other imaginary terms. Here are some examples:

Multiplication - When multiplying complex numbers the FOIL (First, Outer, Inner, Last) technique that you (probably) learned in previous math classes will do the job. So to multiply (6 - 7i) times (2 + 4i) we would proceed as follows:

Division - Suppose we want to divide 3 - 6i by 2 + 4i. To do this we will:

    Now we will use numbers
  1. [(3 - 5i)/(2 + 4i)] * [(2 - 4i)/(2 - 4i)]
  2. Multiplying numerator terms gives: 6 - 12i - 10i - 20i2 = 26 - 22i
  3. Multiplying denominator terms gives: 4 - 8i + 8i -16i2 = 20 + 0i (no imaginary component)
  4. Final result: 26/20 - 22i/20 = 13/10 - (11/10)i = 1.3 - 1.1i

Forms of Complex Numbers At this point we have looked at the basics of complex numbers. The next step is to look at the various forms of the complex number and then to do some mathematics with the various forms. Your text books (Connally and Durbin) cover three forms of the complex number. These are:

  1. The cartesian form: a + bi
  2. The trigonometric form: r [cos(q) + i sin(q)] - where r is the absolute value of the complex number and q is an angle
  3. The Polar form: reiq - where r is the absolute value of the complex number and q is an angle but must be in radians

Trigonometric FormWe have already looked at the cartesian form. The next form we will look at is the trigonometric form. To the right you see a plot of the point a + bi. The length along the horizontal axis is a, the length in the vertical plane is b. The angle to the point, measured counterclockwise from the horizontal axis is theta. The trigonometric functions of the angle are shown in the figure the sine is equal to b/r and the cosine to a/r, where r is the absolute value of the complex number. When we solve the two trigonometric equations for a and b we get b = r sin(q) and a = r cos(q). It we next substitute these values into a + bi we get r cos(theta) + i * r sin(theta). Factoring r gives, r[ cos(q) + i sin(q)] which is the trigonometric form of the complex number. So to convert the a + bi form to trigonometric form you:

  1. Calculate the absolute value of the complex number, r (square root of a2 + b2)
  2. substitue r cos(q) for a and r sin(q) for b
  3. factor the r to get r[ cos(q) + i sin(q)]

Polar FormYour text book tells you (I hope!) that Euler developed the formula eiq = cos(q) + i sin(q). If you multiply both sides of this equation by r you get the expression reiq = r[cos(q) + i sin(q)]. This is the Polar form of the a complex number. To convert from either of the other forms we need only find r (the absolute value of the complex number) and q the angle (in radians).

Converting from one Form to Another - Examples

Cartesian to Trigonometric and Polar - Assume we have a complex number of the form 3 + 4i.

  1. Find the absolute value - square root [32 + 42] = square root[ 9 + 16] = 5, so r = 5
  2. Find q - the point 3 + 4i is a first quadrant point so the angle we want will be a first quadrant angle. tan(q) = 4/3 and tan-1(4/3) = 0.9273 radians.
  3. Trigonometric form is: 5 [cos(q) + i sin(q)]
  4. Polar form is : 5 e0.9273i

Trigonometric to Cartesian and Polar - Assume we have a complex number of the form 8 [cos(2.2166) + i sin(2.2166)]

  1. The absolute value is given as 8
  2. The angle q is given as 2.2166 radians. 2.2166 radians places this angle is the second quadrant.
  3. a = 8 cos(2.2166) = -0.6018 and b = 8 sin(2.2166) = 0.7986, the complex number is cartesian form is -0.6018 + 0.7986 i
  4. In Polar form the number is 8 e2.2166i

Polar to Cartesian and Trigonometric - Assume we are given a complex number of the form 2.5 e5.9721i

  • The absolute value is given as 2.5
  • The angle q is given as 5.9721 radians
  • a = 2.5 cos(5.9721) = 2.3800, b = 2.5 sin(5.9721) = -0.7652 and the number in cartesian form is 2.3800 - 0.7652 i
  • r = 2.5 and q = 5.9721 so the Trigonometric form is 2.5 [cos(5.9721) + i sin(5.9721)]

    DeMoivre's Theorem - To this point we have covered the basics of complex numbers. The next question we will address is taking a complex number to a power followed by finding the roots of a complex number. Let's start with the question of taking a complex number to a power by first looking at multiplying two complex numbers.

    An example - Suppose we have the complex number -6 + 8i and we want to take it to the third power. First we will convert it to trigonometric form and then we will use DeMoivre's Theorem to take the trigonomeric form to the third power.

    Find the roots of a Complex Number To end this section let's look at the problem of finding the roots of a complex number. When we took the number to a power we converted to trigonometric form and then used the relationship:

    (a + bi)n = [r (cosq + i sinq]n

    Which is equal to rn [ cos(n * q) + i sin(n * q)]

    If we have the above result and go want to go backwards it seems obvious that we will have to take the nth root of rn and divide the arguments of cos(n*q) and sin(n*q) by n. This is almost correct. There is one adjustment that we must make. In the example worked just before this our final answer was 380.8 degrees but we wrote it as 380.8o - 360sup>o = 20.8sup>o. It made no difference to our answer and simplified the equation. But, it says that when we have a solution and are going for a root we don't really know the actual solution. So we will write (q + k360o) and divide that by n. If we are working in radians we will add k2p instead of 360o. Then we will let k vary from 0 to 1 to 2 and calculate results only stopping when our angle exceeds the usual limit of 360o. In actual practice if it is a third root there will be three answers, if it is an nth root there will be n answers and we will quit when k = n - 1.

    An Example - Let's find the cube roots of -i.