A polynomial in $z$z of degree $n$n has the form $P\left(z\right)=a_0+a_1z+a_2z^2+....+a_nz^n$P(z)=a0+a1z+a2z2+....+anzn, where the coefficients $a_0,a_1,a_2,...a_n$a0,a1,a2,...an are either purely real or complex.
A common convention is to use $z$z as the variable rather than $x$x whenever we are interested in solutions to the polynomial equation $P\left(z\right)=0$P(z)=0 over the complex number field.
In the specific instance where the coefficients are real, the complex solutions to $P\left(z\right)=0$P(z)=0, should there be any, will always occur in conjugate pairs.
This beautiful result is known as the Conjugate Root Theorem. Specifically, it states that if $a+bi$a+bi is a root of $P\left(z\right)=0$P(z)=0, then so also is its conjugate $a-bi$a−bi.
Without going into a full proof, we can demonstrate with an example why this must be case. To do this we first need to state a few results on conjugates.
With a little algebra, we can easily show that the conjugate of a sum of any two complex numbers is the same as the sum of the conjugates of each complex number.
That is to say, we can show that$\overline{\left(a+bi\right)+\left(c+di\right)}=\overline{a+bi}+\overline{c+di}$(a+bi)+(c+di)=a+bi+c+di with the following argument:
$LHS$LHS | $=$= | $\overline{\left(a+bi\right)+\left(c+di\right)}$(a+bi)+(c+di) |
$=$= | $\overline{\left(a+c\right)+\left(b+d\right)i}$(a+c)+(b+d)i | |
$=$= | $\left(a+c\right)-\left(b+d\right)i$(a+c)−(b+d)i | |
$=$= | $\left(a-bi\right)+\left(c-di\right)$(a−bi)+(c−di) | |
$=$= | $\overline{a+bi}+\overline{c+di}$a+bi+c+di | |
$=$= | $RHS$RHS |
With similar arguments, we can also show that:
So for example we know that:
$\overline{\left(1+3i\right)-\left(2-i\right)}=\overline{1+3i}-\overline{2-i}$(1+3i)−(2−i)=1+3i−2−i
and that:
$\overline{\left(1+3i\right)\times\left(2-i\right)}=\overline{1+3i}\times\overline{2-i}$(1+3i)×(2−i)=1+3i×2−i.
Note that this also means that $\overline{2z}=\overline{2}\times\overline{z}=2\overline{z}$2z=2×z=2z since the conjugate of the real number $2$2, like the conjugate of all reals, is itself.
It also means that, for integer powers $n$n, $\overline{z^n}=\left(\overline{z}\right)^n$zn=(z)n. So, for example $\overline{z^2}=\left(\overline{z}\right)^2$z2=(z)2.
Suppose we know that the complex number $2+2i$2+2i is a root of the polynomial $P\left(z\right)=z^2+4z+8$P(z)=z2+4z+8.
We ask ourselves is $z=\overline{2+2i}$z=2+2i also a root?
If we substitute $\overline{2+2i}$2+2i into $P\left(z\right)$P(z) we have:
$P\left(\overline{2+2i}\right)$P(2+2i) | $=$= | $\left(\overline{2+2i}\right)^2+4\left(\overline{2+2i}\right)+8$(2+2i)2+4(2+2i)+8 |
$=$= | $\overline{\left(2+2i\right)^2}+\overline{4\left(2+2i\right)}+\overline{8}$(2+2i)2+4(2+2i)+8 | |
$=$= | $\overline{\left(2+2i\right)^2+4\left(2+2i\right)+8}$(2+2i)2+4(2+2i)+8 | |
$=$= | $\overline{0}$0 | |
$=$= | $0$0 | |
In other words, the conjugate $2-2i$2−2i is also a root of $P\left(z\right)$P(z).
It's not hard to see how this result can be extended to any polynomial $P\left(z\right)$P(z) with real coefficients. Can you see why the argument will break down if any of the coefficients of $P\left(z\right)$P(z) are complex?
What this all means is, that if a polynomial function has real valued coefficients, then its complex roots, if there are any, exist as conjugate pairs.
The conjugate root theorem becomes useful even when finding the roots of polynomial functions with real domains.
$P\left(x\right)$P(x) of degree $n$n has either $n$n real roots, or else $n-2$n−2 or $n-4$n−4 or $n-6$n−6, etc real roots, because any complex roots it has will always exist in conjugate pairs.
We discuss a few examples to illustrate the point.
Consider the monic cubic function $y=f\left(x\right)$y=f(x) that has only one (and no other) real root at $x=a$x=a.
Then we know that it must have the form $y=\left(x-a\right)\left(x^2+qx+r\right)$y=(x−a)(x2+qx+r) where the contained quadratic polynomial expression is irreducible and hence has no real zeros itself.
However, if we admit complex numbers as candidates for zeros, we know that there are two other zeros that exist as a complex conjugate pair.
In addition, suppose we also know that this function has a complex root of $x=2+i$x=2+i and a particular function value given by $f\left(0\right)=35$f(0)=35.
With this new information, we immediately know two more things:
How many real roots might a degree $6$6 polynomial equation with real coefficients have if it is know to have the complex roots $1-i$1−i, $2-i$2−i, and $3-i$3−i?
The clear answer is none, because, by the conjugate root theorem, the other three roots are $1+i$1+i, $2+i$2+i and $3+i$3+i.
The polynomial $P\left(x\right)=x^3-x^2-20x+50$P(x)=x3−x2−20x+50 has a zero of $-5$−5. Find the other zeros.
Here is one way we can proceed:
We can immediately write that $x^3-x^2-20x+50=\left(x+5\right)\left(x^2+px+q\right)$x3−x2−20x+50=(x+5)(x2+px+q) and from equating the constant terms on both sides, we know that $5q=50$5q=50, from which $q=10$q=10.
Equating the terms in $x$x on both sides reveals that $q+5p=-20$q+5p=−20 and since $q=10$q=10, we have that $p=-6$p=−6.
Thus the quadratic factor becomes $x^2-6x+10$x2−6x+10. This could also have been deduced directly using polynomial division.
A quick calculation shows that the discriminant of the quadratic is $-4$−4, and so $P\left(x\right)$P(x) must only have exactly one real zero.
If we wish to find the complex roots, we could solve the equation $x^2-6x+10=0$x2−6x+10=0. We can do this by completing squares:
$x^2-6x+10$x2−6x+10 | $=$= | 0 |
$x^2-6x+9+1$x2−6x+9+1 | $=$= | 0 |
$\left(x-3\right)^2-i^2$(x−3)2−i2 | $=$= | 0 |
$\left(x-3-i\right)\left(x-3+i\right)$(x−3−i)(x−3+i) | $=$= | 0 |
$\therefore$∴ $x$x | $=$= | $3\pm i$3±i |
The polynomial $P\left(x\right)=x^3-7x^2+17x-15$P(x)=x3−7x2+17x−15 has a zero at $x=3$x=3. Solve for the other zeros of $P\left(x\right)$P(x).
The polynomial $P\left(x\right)=x^4+8x^2-9$P(x)=x4+8x2−9 has a zero at $x=3i$x=3i. Solve for the real zeros of $P\left(x\right)$P(x).
Manipulate complex numbers and present them graphically
Apply the algebra of complex numbers in solving problems