23.2 Conjugates and Polynomial Equations
AHL 1.12 AHL 1.14
- what the complex conjugate \(z^*\) is, and the results \(z + z^* = 2\operatorname{Re}(z)\), \(z - z^* = 2i\operatorname{Im}(z)\) and \(zz^* = x^2 + y^2\)
- the main properties of conjugates
- why the non-real roots of a polynomial equation with real coefficients come in conjugate pairs
- how to solve cubic and quartic equations with complex roots, and how to form a polynomial from its roots
The complex conjugate
Complex conjugate Not in the formula booklet
The conjugate of \(z = x + iy\) is
\[ z^* = x - iy . \]
- \(z\): a complex number
- \(x\): the real part of \(z\)
- \(y\): the imaginary part of \(z\)
- \(z^*\): the conjugate of \(z\)
We change the sign of the imaginary part only. For example, \((5 + 2i)^* = 5 - 2i\), \(\;(-3 - i)^* = -3 + i\), \(\;(4i)^* = -4i\) and \(7^* = 7\).
Conjugates are very useful for division, as we saw in Section 23.1. Three simple results explain why.
Three results for \(z = x + iy\) Not in the formula booklet
\[ z + z^* = 2\operatorname{Re}(z), \qquad z - z^* = 2i\operatorname{Im}(z), \qquad zz^* = x^2 + y^2 . \]
- \(z\): a complex number
- \(z^*\): the conjugate of \(z\)
- \(x\): the real part of \(z\)
- \(y\): the imaginary part of \(z\)
Proof. \(z + z^* = (x + iy) + (x - iy) = 2x\), \(\quad z - z^* = (x + iy) - (x - iy) = 2iy\),
\[ zz^* = (x + iy)(x - iy) = x^2 - i^2y^2 = x^2 + y^2 . \]
So \(z + z^*\) and \(zz^*\) are always real, and \(z - z^*\) is always purely imaginary (or \(0\)).
| \(z\) | \(z^*\) | \(z + z^*\) | \(z - z^*\) | \(zz^*\) |
|---|---|---|---|---|
| \(5 + 2i\) | \(5 - 2i\) | \(10\) | \(4i\) | \(29\) |
| \(1 - 4i\) | \(1 + 4i\) | \(2\) | \(-8i\) | \(17\) |
| \(-2 + 3i\) | \(-2 - 3i\) | \(-4\) | \(6i\) | \(13\) |
When an equation contains both \(z\) and \(z^*\), put \(z = x + iy\) and equate real and imaginary parts.
Example 1 Find all complex numbers \(z\) such that \(zz^* + 4z = 16 - 8i\).
Let \(z = x + iy\), \(x, y \in \mathbb{R}\). Then \(zz^* = x^2 + y^2\), so
\[ x^2 + y^2 + 4x + 4iy = 16 - 8i . \]
Im: \(4y = -8\), so \(y = -2\).
Re: \(x^2 + 4 + 4x = 16\), so \(x^2 + 4x - 12 = 0\), \((x + 6)(x - 2) = 0\), \(x = 2\) or \(x = -6\).
So \(z = 2 - 2i\) or \(z = -6 - 2i\).
Properties of conjugates
For any complex numbers \(z_1\), \(z_2\):
Properties of conjugates Not in the formula booklet
- \((z_1 + z_2)^* = z_1^* + z_2^*\)
- \((z_1 - z_2)^* = z_1^* - z_2^*\)
- \((z_1z_2)^* = z_1^*\,z_2^*\)
- \(\left(\dfrac{z_1}{z_2}\right)^* = \dfrac{z_1^*}{z_2^*}\) \(\;(z_2 \neq 0)\)
- \((z^*)^* = z\)
- \((z^n)^* = (z^*)^n\) for \(n \in \mathbb{Z}^+\)
- \((kz^n)^* = k(z^*)^n\) for \(k \in \mathbb{R}\)
- \(z_1, z_2\): complex numbers
- \(z_1^*, z_2^*\): the conjugates of \(z_1\) and \(z_2\)
- \(z\): a complex number, with conjugate \(z^*\)
In words: the conjugate of a sum, difference, product, quotient or power is the same combination of the conjugates. Property 3 can be proved by expanding both sides with \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\). Property 6 follows from property 3 (use it again and again). Property 7 is true because a real number \(k\) is its own conjugate.
Example 2 Let \(z = 3 - 2i\). Show that \((z^2)^* = (z^*)^2\).
\(z^2 = (3 - 2i)^2 = 9 - 12i + 4i^2 = 5 - 12i\), so \((z^2)^* = 5 + 12i\).
\(z^* = 3 + 2i\), so \((z^*)^2 = 9 + 12i + 4i^2 = 5 + 12i\).
The two results are equal, so \((z^2)^* = (z^*)^2\).
Polynomial equations with complex roots
Look at the polynomial \(P(z) = z^3 - 2z^2 + 7\). Its coefficients are real. Let \(w\) be any complex number. Using properties 7 and 1, and the fact that \(7^* = 7\):
\[ \begin{aligned} P(w^*) &= (w^*)^3 - 2(w^*)^2 + 7 \\ &= (w^3)^* - (2w^2)^* + 7^* \\ &= \left(w^3 - 2w^2 + 7\right)^* \\ &= \big(P(w)\big)^* . \end{aligned} \]
The same argument works for any polynomial \(P(z) = a_nz^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0\) whose coefficients \(a_n, \ldots, a_0\) are real:
\[ P(z^*) = a_n(z^*)^n + \cdots + a_1z^* + a_0 = \left(a_nz^n + \cdots + a_1z + a_0\right)^* = \big(P(z)\big)^* . \]
So if \(P(w) = 0\), then \(P(w^*) = 0^* = 0\).
Complex conjugate root theorem Not in the formula booklet
If a polynomial equation \(P(z) = 0\) has real coefficients, and \(w\) is a non-real root, then \(w^*\) is also a root.
The non-real roots of a polynomial equation with real coefficients occur in conjugate pairs.
Two useful consequences:
- A conjugate pair \(w, w^*\) gives a real quadratic factor: \[ (z - w)(z - w^*) = z^2 - (w + w^*)z + ww^* = z^2 - 2\operatorname{Re}(w)\,z + ww^* . \] For example, \(w = 1 + 2i\) gives \(z^2 - 2z + 5\).
- A polynomial of odd degree with real coefficients has at least one real root, because the non-real roots pair up.
The condition “real coefficients” is essential. The equation \(z^2 - iz + 2 = 0\) has roots \(2i\) and \(-i\), which are not conjugates.
Example 3 Without a calculator, solve \(z^3 + z + 10 = 0\).
A cubic with real coefficients has a real root. Try factors of \(10\): \(P(-2) = -8 - 2 + 10 = 0\), so \(z + 2\) is a factor (factor theorem, Section 11.2).
\[ z^3 + z + 10 = (z + 2)(z^2 - 2z + 5) . \]
Solve \(z^2 - 2z + 5 = 0\): \(\;z = \dfrac{2 \pm \sqrt{4 - 20}}{2} = \dfrac{2 \pm 4i}{2} = 1 \pm 2i\).
The roots are \(z = -2\), \(z = 1 + 2i\) and \(z = 1 - 2i\).
Example 4 Show that \(z = 2 + i\) is a root of \(z^3 - 3z^2 + z + 5 = 0\). Write down a second root, and hence solve the equation completely.
\((2 + i)^2 = 4 + 4i - 1 = 3 + 4i\) and \((2 + i)^3 = (3 + 4i)(2 + i) = 6 + 3i + 8i - 4 = 2 + 11i\).
\[ (2 + 11i) - 3(3 + 4i) + (2 + i) + 5 = (2 - 9 + 2 + 5) + (11 - 12 + 1)i = 0 \;✓ \]
The coefficients are real, so \(2 - i\) is also a root. The real quadratic factor is
\[ \big(z - (2 + i)\big)\big(z - (2 - i)\big) = z^2 - 4z + 5 . \]
Compare: \(z^3 - 3z^2 + z + 5 = (z^2 - 4z + 5)(z + 1)\) (check the constant: \(5 \times 1 = 5\) ✓).
The roots are \(2 + i\), \(2 - i\) and \(-1\).
The results for the sum and product of the roots from Section 11.3 are true for complex roots too. In Example 4 the sum of the roots is \((2 + i) + (2 - i) + (-1) = 3 = -\dfrac{-3}{1}\) ✓. They are often the quickest way to find a missing root.
Example 5 Given that \(1 + 2i\) is a root of \(z^3 + az^2 + bz - 15 = 0\), where \(a, b \in \mathbb{R}\), find \(a\) and \(b\). Hence solve the equation.
The coefficients are real, so \(1 - 2i\) is also a root. Let the third root be \(c\).
Product of the roots (cubic, so it is \(-\dfrac{a_0}{a_3}\)):
\[ (1 + 2i)(1 - 2i)\,c = -\frac{-15}{1} \;\Longrightarrow\; 5c = 15 \;\Longrightarrow\; c = 3 . \]
So
\[ z^3 + az^2 + bz - 15 = (z^2 - 2z + 5)(z - 3) = z^3 - 5z^2 + 11z - 15 . \]
Therefore \(a = -5\) and \(b = 11\), and the roots are \(1 + 2i\), \(1 - 2i\) and \(3\).
Another method: substitute \(z = 1 + 2i\), using \((1 + 2i)^2 = -3 + 4i\) and \((1 + 2i)^3 = -11 - 2i\): \(\;(-11 - 2i) + a(-3 + 4i) + b(1 + 2i) - 15 = 0\). Equating parts gives \(-3a + b = 26\) and \(4a + 2b = 2\), so \(a = -5\), \(b = 11\).
Forming a polynomial from its roots
If we know the roots, we can build the polynomial. Remember to include the conjugate of each non-real root when the coefficients must be real.
Example 6 Find a quartic polynomial with real coefficients and leading coefficient \(1\) that has \(2i\) and \(3 - i\) as two of its zeros.
The coefficients are real, so \(-2i\) and \(3 + i\) are zeros too.
\[ (z - 2i)(z + 2i) = z^2 + 4, \qquad \big(z - (3 - i)\big)\big(z - (3 + i)\big) = z^2 - 6z + 10 . \]
\[ P(z) = (z^2 + 4)(z^2 - 6z + 10) = z^4 - 6z^3 + 14z^2 - 24z + 40 . \]
On a Calculator page: Menu → Algebra → Polynomial Tools → Find Roots of Polynomial. Choose the degree and set Roots to Complex (with Real, the non-real roots are not shown). Enter the coefficients. For \(z^3 - 3z^2 + z + 5\) the calculator gives \(-1\), \(2 - i\) and \(2 + i\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \((3 - 2i)^* = -3 + 2i\) | \((3 - 2i)^* = 3 + 2i\) | Only the sign of the imaginary part changes. |
| \(zz^* = x^2 - y^2\) | \(zz^* = x^2 + y^2\) | \((x + iy)(x - iy) = x^2 - i^2y^2\) and \(-i^2 = +1\). |
| “\(2 + i\) is a root, so \(-2 - i\) is a root” | \(2 - i\) is a root | The partner is the conjugate, not the negative. |
| Using conjugate pairs for \(z^2 - iz + 2 = 0\) | Check the coefficients are real first | The theorem needs real coefficients. |
| \((z - (2 + i))(z - (2 - i)) = z^2 - 4z + 3\) | \(z^2 - 4z + 5\) | The constant term is \(ww^* = 2^2 + 1^2 = 5\). |
Exercises
1 Let \(z = 4 - 3i\). Find \(z^*\), \(z + z^*\), \(z - z^*\) and \(zz^*\).
\(z^* = 4 + 3i\), \(\;z + z^* = 8\), \(\;z - z^* = -6i\), \(\;zz^* = 16 + 9 = 25\).
2 Let \(z_1 = 1 + 2i\) and \(z_2 = 3 - i\). Show that \((z_1z_2)^* = z_1^*\,z_2^*\).
\(z_1z_2 = 3 - i + 6i - 2i^2 = 5 + 5i\), so \((z_1z_2)^* = 5 - 5i\).
\(z_1^*\,z_2^* = (1 - 2i)(3 + i) = 3 + i - 6i - 2i^2 = 5 - 5i\). The two are equal.
3 Find the complex number \(z\) such that \(2z + z^* = 6 - 5i\).
Let \(z = x + iy\): \(\;2x + 2iy + x - iy = 3x + iy = 6 - 5i\). So \(x = 2\), \(y = -5\), and \(z = 2 - 5i\).
4 Find all complex numbers \(z\) such that \(zz^* + 2z = 19 + 8i\).
With \(z = x + iy\): \(\;x^2 + y^2 + 2x + 2iy = 19 + 8i\).
Im: \(2y = 8\), \(y = 4\). Re: \(x^2 + 16 + 2x = 19\), \(x^2 + 2x - 3 = 0\), \((x + 3)(x - 1) = 0\).
So \(z = 1 + 4i\) or \(z = -3 + 4i\).
5 Let \(w = 3 - 2i\). Write \((z - w)(z - w^*)\) as a quadratic in \(z\), and explain why its coefficients are real.
\((z - w)(z - w^*) = z^2 - (w + w^*)z + ww^* = z^2 - 6z + 13\).
The coefficients are \(-(w + w^*) = -2\operatorname{Re}(w)\) and \(ww^* = 3^2 + 2^2\), which are always real.
6 Without a calculator, solve \(z^3 - 3z^2 + 9z + 13 = 0\).
\(P(-1) = -1 - 3 - 9 + 13 = 0\), so \(z + 1\) is a factor: \(\;z^3 - 3z^2 + 9z + 13 = (z + 1)(z^2 - 4z + 13)\).
\(z = \dfrac{4 \pm \sqrt{16 - 52}}{2} = \dfrac{4 \pm 6i}{2} = 2 \pm 3i\).
The roots are \(-1\), \(2 + 3i\), \(2 - 3i\).
7 Given that \(3 - i\) is a root of \(z^3 - 5z^2 + 4z + 10 = 0\), find the other two roots.
Real coefficients, so \(3 + i\) is a root. Quadratic factor: \(z^2 - 6z + 10\).
\(z^3 - 5z^2 + 4z + 10 = (z^2 - 6z + 10)(z + 1)\), so the third root is \(-1\).
(Or: sum of roots \(= 5\), so \(6 + c = 5\), \(c = -1\).)
8 Given that \(2i\) is a root of \(z^4 - 2z^3 + 6z^2 - 8z + 8 = 0\), solve the equation.
\(-2i\) is also a root, so \((z - 2i)(z + 2i) = z^2 + 4\) is a factor.
\(z^4 - 2z^3 + 6z^2 - 8z + 8 = (z^2 + 4)(z^2 - 2z + 2)\).
\(z^2 - 2z + 2 = 0\) gives \(z = \dfrac{2 \pm \sqrt{-4}}{2} = 1 \pm i\).
The roots are \(2i\), \(-2i\), \(1 + i\), \(1 - i\).
9 The equation \(z^3 + pz + q = 0\), where \(p, q \in \mathbb{R}\), has a root \(1 - 3i\). Use the sum of the roots to find the real root, and hence find \(p\) and \(q\).
\(1 + 3i\) is also a root. There is no \(z^2\) term, so the sum of the roots is \(0\): \((1 - 3i) + (1 + 3i) + c = 0\), so \(c = -2\).
\[ (z^2 - 2z + 10)(z + 2) = z^3 + 2z^2 - 2z^2 - 4z + 10z + 20 = z^3 + 6z + 20 . \]
So \(p = 6\) and \(q = 20\).
10 The cubic \(z^3 + az^2 + bz + c\) has real coefficients, and two of its zeros are \(-3\) and \(2 - 5i\). Find \(a\), \(b\) and \(c\).
The third zero is \(2 + 5i\), giving the factor \(z^2 - 4z + 29\).
\[ (z + 3)(z^2 - 4z + 29) = z^3 - 4z^2 + 29z + 3z^2 - 12z + 87 = z^3 - z^2 + 17z + 87 . \]
So \(a = -1\), \(b = 17\), \(c = 87\).