Lesson: Complex Number Review

Question 2

1 point

Transcript — Introduction

We begin by reviewing the properties of complex numbers that will be required for what we do in the rest of the course.

Definition: We define i to be a number such that i-squared = -1, and define the set of complex numbers to be the set C equals the set of all (a + bi) such that a and b are real numbers. For a complex number z = a + bi, we define the real part of z by the real part of z is equal to a, and the imaginary part of z by the imaginary part of z equals b. If b = 0, then we say that z is real. If a = 0 and b is not equal to 0, then we say that z is imaginary.

We now look at the common operations on complex numbers. We define addition of two complex numbers (a + bi) and (c + di) by (a + bi) + (c + di) = (a + c) + (b + d)i. Remember, it is important in linear algebra to look for similarities and make connections. We notice that this definition of addition looks a lot like standard addition of vectors in a real vector space. That is, we are just adding the vectors component-wise. So, if we define real scalar multiplication by t(a + bi) = ta + (tb)i for any real scalar t, then we can show that the set C with these operations satisfies ten familiar properties. Yes, that is right—we have that it is a real vector space. As usual, we should find a basis and the dimension of this real vector space. It is easy to see that every vector in C is a linear combination of 1 and i. Moreover, the set {1, i} is linearly independent since no real number times 1 can give i. Thus, the standard basis for C as a real vector space is {1, i}, and thus, it is a 2-dimensional real vector space.

Hmm, what do we know about 2-dimensional real vector spaces? They are isomorphic to every other 2-dimensional real vector space. In particular, this implies that C is isomorphic to the plane, and so we often talk about the complex plane. Normally, we relate the complex number (a + bi) to the point (a, b) in R2.

It is really neat that C is a 2-dimensional real vector space, but there is a drawback to this. We had to restrict to multiplying by a real scalar. That is, we have no way of multiplying two complex numbers—or do we? Consider a simple example. We want i(a + bi) to be equal to (-b + ai). Notice that this looks like a linear mapping named i. The mapping i takes the vector (a + bi) and outputs the vector (-b + ai). As usual, we can find the matrix of the linear mapping with respect to the standard basis {1, i}. We get that the standard matrix of this linear mapping is [0, -1; 1, 0]. Whoa. We recognize this matrix as the matrix of a rotation by pi/2, so multiplying by i gives a rotation of 90 degrees.

Some students will ask why this is useful. Well, I have recently heard that the phone company is going to make use of this. Whenever someone dials a number that is not in service, they will now hear the message “You have dialed an imaginary number. Please rotate your phone by 90 degrees and try again.”

Similarly, we can show that multiplying by any complex number alpha corresponds to a rotation and a stretch. This is an excellent exercise.

Viewing complex multiplication as a linear mapping is cool, but not practical. So instead, we define multiplication of two complex numbers (a + bi) and (c + di) by (a + bi)(c + di) = ac – bd + (ad + bc)i.

Let’s practice some of the basic operations with a couple easy examples. Example: Evaluate the following. (a) (2 – 3i) + (3 + i). Using the definition of addition, we get (2 – 3i) + (3 + i) = 5 – 2i. (b) (2 – i)(3 + 2i). Evaluating, we find that this is 8 + i. And finally, (c) (2 – i)(2 + i) gives 5.

Complex Properties and Additional Definitions

We get the following familiar properties. Theorem 11.1.1: If z1, z2, and z3 are complex numbers, then

  1. z1 + z2 = z2 + z1
  2. z1z2 = z2z1
  3. z1 + (z2 + z3) = (z1 + z2) + z3
  4. z1(z2z3) = (z1z2)z3, and
  5. z1(z2 + z3) = z1z2 + z1z3

We also need to define division of complex numbers, but before we do this, it is helpful to define another operation.

Definition: Let z = a + bi be a complex number. The complex conjugate of z is the conjugate of z is a – bi.

Example: We have the conjugate of (3 – 4i) is 3 + 4i, the conjugate of 2i is -2i, and the conjugate of -5 is -5.

As usual, when looking at a new operation, we should prove some properties of that operation. Theorem 11.1.2: If z = a + bi, z1, and z2 are all complex numbers, then

  1. The complex conjugate of (the complex conjugate of z) is z.
  2. z is real if and only if the conjugate of z is equal to z.
  3. If z is not equal to 0, then z is imaginary if and only if the complex conjugate of z equals –z.
  4. The conjugate of (z1 + z2) is equal to (the conjugate of z1) plus (the conjugate of z2).
  5. The conjugate of (z1z2) is equal to (the conjugate of z1)(the conjugate of z2).
  6. z plus (the conjugate of z) is 2 times the real part of z.
  7. z minus (the conjugate of z) is 2 times i times the imaginary part of z. And finally,
  8. z(the conjugate of z) is equal to a-squared + b-squared.

We observe that a-squared + b-squared is the square of the Euclidean length of the point (a, b) from the origin in R2. Thus, we make the following definition. Definition: Let z = a + bi be a complex number. We define the absolute value of z to be the absolute value of z is the square root of (z(the conjugate of z)), which is equal to the square root of (a-squared + b-squared).

Theorem 11.1.3: If w and z are complex numbers, then

  1. The absolute value of z is a real number, and the absolute value of z is always greater than or equal to 0.
  2. The absolute value of z equals 0 if and only if z = 0.
  3. The absolute value of (wz) is equal to (the absolute value of w)(the absolute value of z). And
  4. The absolute value of (w + z) is less than or equal to ((the absolute value of w) + (the absolute value of z)).

Finally, we can define division of two complex numbers. For any complex numbers w and z, with z not equal to 0, we have w/z = (w(z-conjugate))/(z(z-conjugate)), which is equal to (w(z-conjugate))/((the absolute value of z)-squared).

Example: Calculate (4 + 3i)/(2 – 3i). Solution: To calculate (4 + 3i)/(2 – 3i), we multiply top and bottom by the conjugate of (2 – 3i). Then simplifying, we get –(1/13) + (18/13)i.

Complex Systems of Equations

Since we can now add, subtract, multiply, and divide complex numbers, this means that we can also row reduce matrices with complex entries. That is, we can solve systems of linear equations with complex entries. This is, of course, exactly the same as in the real case, except the calculations are a little more complex, and that any free variable takes on any complex value.

Example: Find the solution set of the homogeneous system of linear equations z1 + (2i)z2 + (1 + 3i)z3 = 0, (i)z1 + (-2 + i)z2 + (-3 + 3i)z3 = 0, and -2z1 + (3i)z2 + (-2 + 8i)z3 = 0. Solution: We row reduce the coefficient matrix of the system to reduced row echelon form as normal. Doing this carefully, we get the reduced row echelon form [1, 0, 1 – i; 0, 1, 2; 0, 0, 0]. Thus, z3 is a free variable. Again, since we are solving this system over C, this means that z3 can take any complex value. Hence, we let z3 equals any complex number alpha, and so a vector equation for the solution set is z = (alpha)[-1 + i; -2; 1] for any complex number alpha.

When row reducing complex matrices, it is highly recommended that you check your answer whenever possible. In this case, we can verify that if we take alpha = 1, then z1 = -1 + i, z2 = -2, and z3 = 1 does satisfy the original system.

Note that you will likely find, at first, that row reducing complex matrices can be very slow and error-prone. However, it is very likely that you will need to row reduce at least one matrix with complex entries on the final exam. Therefore, it is very, very important that you practice these enough so that you can quickly row reduce complex matrices without making errors.

In the next lecture, we will extend the idea of a real vector space to a complex vector space. This ends this lecture.

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