Lesson: The Complex Conjugate

Question 2

1 point


Everything we study comes with its own special operations. With matrices, we have the determinant. With polynomials, we get the idea of factoring polynomials, and of evaluating polynomials. The special operation for complex numbers is called conjugation.

The complex conjugate of the complex number z = x + yi is z = x – yi, and is denoted overline-z.

So for example, the conjugate of (1 + 2i) equals 1 – 2i. The conjugate of 3 – 4i is 3 + 4i. The conjugate of 5 is simply 5.

Let’s look at the following theorem outlining properties of the complex conjugate. So for complex numbers z1 = x1 + y1i and z2, we have that

  • The conjugate of (z1-conjugate) is z1.
  • z1 is purely real if and only if z1-conjugate = z1.
  • z1 is purely imaginary if and only if z1-conjugate = -z1.
  • The conjugate of (z1 + z2) equals (the conjugate of z1) + (the conjugate of z2).
  • The conjugate of the product (z1z2) equals (the conjugate of z1)(the conjugate of z2).
  • The conjugate of (z1-to-the-n) equals (the conjugate of z1)-quantity-to-the-n.
  • z1 + z1-conjugate will equal 2 times the real part of z1, which is 2x1, while z1 – z1-conjugate will equal i times 2 times the imaginary part of z1, which is i times 2y1.
  • And lastly, z1(z1-conjugate) = x1-squared + y1-squared.

The proofs of these properties are easy consequences of the definition of conjugation, which makes them great practice. I’ll prove properties 1, 2, and 4 now.

Proof of property 1: Let z1 = x1 + y1i be a complex number. Then z1-conjugate = (x1 + y1i)-conjugate, which equals x1 – y1i. And so we have that (z1-conjugate)-conjugate will equal (x1 – y1i)-conjugate, which equals x1 + y1i, which equals z1 as desired.

Let’s look at the proof of property 2. Suppose that z1 = x1 + y1i is a complex number that is purely real. Then y1 = 0, and z1 = x1. Moreover, we see that the conjugate of z1, which equals x1 – y1i, will equal x1 – 0i, which equals x1, which equals z1. So if z1 is purely real, then the conjugate of z1 equals z1. Now let’s prove the other direction of our if-and-only-if sentence. So let’s suppose that z1 is a complex number such that the conjugate of z1 equals z1. Well, then this means that x1 + y1i = x1 – y1i, so (x1 + y1i) – (x1 – y1i) = 0. Well, this means that 2y1i = 0, which can only happen if y1 = 0. So we see that if z1-conjugate = z1, then y1 = 0, which means that z1 is purely real.

Last, we’ll look at a proof of property 4. Let z1 = x1 + y1i and z2 = x2 + y2i be complex numbers. Then we see that the conjugate of (z1 + z2) will equal the conjugate of ((x1 + y1i) + (x2 + y2i)). Well, this is the conjugate of ((x1 + x2) + (y1 + y2)i), which will equal (x1 + x2) – (y1 + y2)i. Well, this equals (x1 + x2) + (-y1 – y2i), and we can rewrite this as the sum (x1 – y1i) + (x2 – y2i), which is (x1 + y1i)-conjugate + (x2 + y2i)-conjugate. That is to say, it equals z1-conjugate + z2-conjugate.

Now, we have already noticed that complex conjugation does not affect real numbers. This will become of great importance during our study of the complex numbers, as it turns out that when we generalize many of the things we did in the real numbers, we will need to introduce complex conjugation to the mix in order to have our formulas turn out correctly. I like to think that, secretly, we’ve been conjugating these numbers all the time, and just didn’t realize it because it doesn’t make a difference in the real numbers. But it will make a difference now, so be on the lookout for conjugation throughout the rest of the course.

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