The Conjugate, and the Three Things It Is For
Chapter Nine
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 19 to 21 of 303
In one line
The conjugate of x + iy is x minus iy, and it is the tool that makes a complex denominator real, pulls out the two parts, and gives the modulus squared.
The definition
The conjugate of z = x + iy is the complex number with the sign of the imaginary part reversed. It is written with a bar over the z, and in this book, where a bar cannot be typeset in the text, as conjugate(z).
z = x + i y
conjugate(z) = x - i y
| z | conjugate(z) |
|---|---|
| 3 + 4i | 3 - 4i |
| 5 - 2i | 5 + 2i |
| 7 | 7 |
| 6i | -6i |
Two of those rows are the special cases. A real number is its own conjugate, because there is no imaginary part to change. A purely imaginary number has its own negative as its conjugate.
The three things it is for
This is the reason the conjugate gets a chapter rather than a line.
One: it turns a complex denominator real. Multiply any complex number by its conjugate and the answer is real.
(x + i y)(x - i y) = x^2 - i^2 y^2 = x^2 + y^2
There is no i left. That single fact is the whole method of dividing complex numbers, which is the next chapter.
Two: it extracts the real and imaginary parts. Adding z to its conjugate cancels the imaginary part; subtracting cancels the real part.
z = x + i y
z + conjugate(z) = 2x
z - conjugate(z) = 2 i y
So the two parts can be written without ever mentioning x and y.
z = x + i y
(z + conjugate(z))/2 = x
(z - conjugate(z))/(2i) = y
Those two formulae are used in proofs and in locus questions, where you are given a condition on z and have to turn it into a condition on x and y.
Three: it gives the modulus squared. From the first fact, z times its conjugate is x squared plus y squared, which is the square of the distance of the point from the origin.
z = x + i y
z conjugate(z) = x^2 + y^2
The chapter on the modulus makes that the definition of the modulus, and this identity is how the modulus is computed in practice.
The rules the conjugate obeys
All of these are proved by writing out both sides, and all of them are set as short theory questions.
conjugate(conjugate(x + i y)) = x + i y
conjugate((a + i b) + (c + i d)) = conjugate(a + i b) + conjugate(c + i d)
conjugate((a + i b)(c + i d)) = conjugate(a + i b) conjugate(c + i d)
The Conjugate, and the Three Things It Is For
In words: conjugating twice gets you back where you started; the conjugate of a sum is the sum of the conjugates; and the conjugate of a product is the product of the conjugates. The same holds for a quotient.
The product rule is the useful one, and it is worth seeing why it is not obvious. On the left you multiply first and flip the sign of the answer's imaginary part. On the right you flip both signs first and then multiply. Those are different procedures and they give the same result, which is a genuine fact about how the multiplication rule is built.
Here it is on numbers.
conjugate((2 + 3i)(1 - i)) = conjugate(5 + i) = 5 - i
conjugate(2 + 3i) conjugate(1 - i) = (2 - 3i)(1 + i) = 5 - i
What it means on the diagram
Conjugating a complex number reflects its point in the real axis. The real part is unchanged, the imaginary part changes sign, so the point flips from above the axis to below it or the other way.
That makes several facts obvious that are tedious to prove algebraically. The conjugate has the same modulus, because a reflection does not change distance from the origin. Its argument is the negative of the original argument, because the angle has flipped to the other side of the axis. And a real number sits on the axis, so reflecting it does nothing, which is why a real number is its own conjugate.
Conjugates and real coefficients
Here is the fact that makes conjugates matter beyond this module.
If a polynomial has real coefficients and a complex number is a root of it, then the conjugate of that number is also a root. Complex roots of a real polynomial always come in conjugate pairs.
You have already seen this without being told. In the chapter on why i had to be invented, Cardano's cubic threw up 2 + i and 2 minus i, a conjugate pair, and their sum was real. It is also why the quadratic formula, applied to a real quadratic with negative discriminant, gives two answers differing only in the sign of the i.
x = (-2 + sqrt(2^2 - 415))/2
x = -1 + 2i
That is one root of x squared plus 2x plus 5, and the other is minus one minus 2i. Both parts of the pair are needed, and their sum, minus two, is real, as is their product, five.
This fact is what makes Module 2 work. When the auxiliary equation of a differential equation has complex roots, they arrive as a conjugate pair, and the pair combines into a real answer in sines and cosines. Without conjugate pairs, a real equation would have a complex answer, which would be absurd.
The Conjugate, and the Three Things It Is For
Check yourself
| Question | Answer |
|---|---|
| conjugate(4 - 7i) | conjugate(4 - 7i) = 4 + 7i |
| (5 + 2i) times its conjugate | (5 + 2i)(5 - 2i) = 29 |
| Add 3 - i to its conjugate | (3 - i) + (3 + i) = 6 |
| Subtract the conjugate of 3 - i from it | (3 - i) - (3 + i) = -2i |
| conjugate(i^3) | conjugate(i^3) = i |
| The other root when 1 + 3i is a root of a real quadratic | conjugate(1 + 3i) = 1 - 3i |
The fifth one is worth a moment: i cubed is minus i, and the conjugate of minus i is i.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.