The Conjugate Root Theorem states that if the complex number a + ib is a root of a polynomial in one variable with real coefficients, then the complex conjugate a - bi also a root of that polynomial. This is a useful theorem for solving polynomials with real coefficients. Since the coefficients of the polynomial are real numbers, complex roots must always come in pairs and more than that they must be conjugate pairs - this way two complex numbers can multiply to give a real number.
On this page, you should learn to
- Understand that complex roots occur in conjugate pairs
- Solve polynomial equations
- Make links with this topic and Factor and Remainder Theorem and Sums and Products of Roots
The following video will help you understand all the concepts from this page
In the following video we will look at an example involving complex roots of a polynomial equation. Since the coefficients of the polynomial are real numbers, complex roots must always come in pairs and more than that they must be conjugate pairs - this way two complex numbers can multiply to give a real number.
The conjugate root theorem states that if the complex number \(a+ib\) is a root of a polynomial f(x) in one variable with real coefficients, then the complex conjugate \(a-ib\) also a root of that polynomial.
One root of the equation \(4z^4-4z^3-25z^2+55z-42=0\) is \(1+\frac {\sqrt{3}}{2}i\)
Find the other roots of the equation.
Notes from the video
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Here is a quiz that practises the skills from this page
START QUIZ!
2 + 3i is a root of a polynomial.
a + bi is another root.
Find a and b
a =
b =
According to the conjugate root theorem, complex roots of polynomials come in conjugate pairs.
If a + bi is a root, then a - bi is a root
\(\frac{1}{1-2i} \) is a root of a cubic equation. Which of the following is another root?
\(\frac{1}{1-2i} =\frac{1+2i}{(1-2i)(1+2i)} \)
\(\frac{1+2i}{1-4i²} =\frac{1+2i}{5} \)
Another root is the complex conjugate of this = \(\frac{1-2i}{5}\)
If (z - 2 - i)(z - 2 + i) \(\equiv\) z² + az + b
work out a and b
a =
b =
(z - 2 - i)(z - 2 + i)
= z² - 2z + zi
- 2z + 4 - 2i
- zi + 2i - i²
= z² - 4z + 5
If (z - 1 - 3i)(z - 1 + 3i) \(\equiv\) z² + az + b
work out a and b
a =
b =
(z - 1 - 3i)(z - 1 + 3i)
= z² - 2z + 3zi
- z + 1 - 3i
- 3zi + 3i - 9i²
= z² - 2z + 10
If (2z - 3)(z² - 4z + 8) \(\equiv\) 2z3 + az² + bz - 24
find a and b
a =
b =
2z(z² - 4z + 8) - 3(z² - 4z + 8)
= 2z3 - 8z² + 16z - 3z² + 12z - 24
= 2z3 - 11z² + 28z - 24
If (z - p)(z² + 2z - 1) \(\equiv\) z3 - 2z² + az + b
Work out p
p =
Consider the z² term in the expansion of (z - p)(z² + 2z - 1)
= \(z\times2z-p\times z^2=(2-p)z^2\)
z² term in z3 - 2z² + az + b is equal to -2z²
2 - p = -2
p = 4
If (z - p)(z² - 4z + 5) \(\equiv\) z3 + az² - 3z + b
Work out p
p =
Consider the z term in the expansion of (z - p)(z² - 4z + 5)
\(=z\times5-p\times(-4z)=(5+4p)z\)
z term in z3 + az² - 3z + b is equal to -3z
5 + 4p = - 3
p = - 2
How many roots does the following cubic have?
real root(s)
complex root(s)
The graph cuts the x axis 3 times.
Therefore, it has 3 real roots.
How many roots does the following cubic have?
real root(s)
complex root(s)
The graph cuts the x axis 1 time.
Therefore, it has 1 real root.
There are 3 roots altogether, so it must have 2 complex roots.
How many roots does the following quartic have?
real root(s)
complex root(s)
The graph cuts the x axis 2 times.
Therefore, it has 2 real roots.
There are 4 roots altogether, so it must have 2 complex roots.
One root of the equation z² + bz + c = 0 is 2+3i where \(b,c\in\mathbb{Z}\).
Find the value of b and the value of c.
Hint
Full Solution
\(\frac{2}{1+i}\) is a root to the quadratic equation z² + px + q = 0
a) Find the other root
b) Hence find the values of p and q.
Hint
Full Solution
2 - 3i is a root of the equation \(z^3-7z^2+az+b=0\quad ,a, b\in\ \mathbb{R}\)
Work out a and b and the other roots of the equation.
Hint
Full Solution
The quartic equation \(z^4+az^3+bz^2+cz+d\) has roots 2 + i and 2i
a) Work out the other roots of the equation
b) Find the values of a , b , c and d
Hint
Full Solution
The equation \(2z^{ 4 }−9z^{ 3 }+pz^{ 2 }+qz−174=0 \quad,\quad p,q\in\mathbb{Z}\) has two real roots \(\alpha\) and \(\beta\) and two complex roots \(\gamma\) and \(\delta\) where \(\gamma=2-5i\).
a. Show that \(\alpha+\beta=\frac{1}{2}.\)
b. Find \(\alpha\beta\).
c. Hence find the two real roots α and β.
d. Find the values of p and q.
Hint
Full Solution
How much of Complex Numbers - Roots of Polynomials have you understood?
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