Complex Zeros in Math
These are my notes on complex zeros in math.
Complex Zeros
A variable in the complex number system is referred to as a complex variable. A
complex number r is called a complex zero of f if f(r)=0.
We have learned that some quadratic equations have no real solutions, but that
in the complex number system every quadratic equation has a solution, either
real or complex.
Every complex polynomial function f(x) of degree \(n \geq 1\) has at least one
complex zero.
Let \(f(x)=a_n^n+a_{n-1}x^{n-1}+...+a_1x+a_0\)
By the fundamental theorem of algebra, f has at least one zero, r1. Then, by the
factor theorem, x-r1 is a factor and \(f(x)=(x-r_1)q_1(x)\) where where q1(x) is
a complex polynomial of degree n-1 whose leading coefficient is an. Repeating
this argument n times, we arrive at \(f(x)=(x-r_1)(x-r_2)...(x-r_n)q_n(x)\)
where \(q_n(x)\) is a complex polynomial of degree n-n=0 whose leading
coefficient is \(a_n\). That is, \(q_n(x)=a_nx^n=a_n\), and so
\(f(x)=a_n(x-r_1)(x-r_2)...(x-r_n)\). We conclude that every complex polynomial
function f(x) of degree \(n \geq 1\) has exactly n zeros.
Conjugate Pairs Theorem
We can use the fundamental theorem of algebra to obtain valuable information
about the complex zeros of polynomials whose coefficients are real numbers.
In other words, for polynomials whose coefficients are real numbers, the complex
zeros occur in conjugate pairs. This result should not be all that surprising
since the complex zeros of a quadratic function occurred in conjugate pairs.
The importance of this result should be clear. Once we know that, say, \(3+4i\)
is a zero of a polynomial with real coefficients, then we know that \(3-4i\) is
also a zero. This result has an important corollary. A polynomial f of odd
degree with real coefficients has at least one real zero.
For example, the polynomial \(f(x)=x^5-3x^4+4x^3-5\) has at least one zero that
is a real number, since f is of degree 5 (odd) and has real coefficients.
Example 1
Using the conjugate pairs theorem
A polynomial f of degree 5 whose coefficients are real numbers has the zeros 1,
5i, and 1+i. Find the remaining two zeros.
Solution:
Since f has coefficients that are real numbers, complex zeros appear as
conjugate pairs. It follows that -5i, the conjugate of 5i, and 1-i, the
conjugate of 1+i, are the two remaining zeros.
Example 2
Finding a polynomial function whose zeros are given
Find a polynomial function f of degree 4 whose coefficients are real numbers
that has the zeros 1, 1, and -4+i.
Solution:
since -4+i is a zero, by the conjugate pairs theorem, -4-i must also be a zero
of f. Because of the factor theorem, if f(c)=0, then x-c is a factor of of f(x).
So we can now write f as:
\(f(x)=a(x-1)(x-1)[x-(-4+i)][x-(-4-i)]\) where a is any real number.
\(f(x)=a(x^2-2x+1)[x^2-(-4+i)x-(-4-i)x+(-4+i)(-4-i)]\)
\(a(x^2-2x+1)(x^2+4x-ix+4x+ix+16+4i-4i-i^2)\)
\(a(x^2-2x+1)(x^2+8x+17)\)
\(a(x^4+8x^3+17x^2-2x^3-16x^2-34x+x^2+8x+17)\)
\(a(x^4+6x^3+2x62-26x+17)\)
Every polynomial function with real coefficients can be uniquely factored over
the real numbers into a product of linear factors and/or irreducible quadratic
factors.
This second degree polynomial has real coefficients and is irreducible over the
real numbers. So, the factors of f are either linear or irreducible quadratic
factors.
Example 3
Finding the complex zeros of a polynomial
\(f(x)=3x^4+5x^3+25x^2+45x-18\)
Solution:
1. The degree of f is 4, so f will have four complex zeros.
2. The rule of signs provides information about the real zeros. For this
polynomial, there is one positive real zero. There are three or one negative
real zeros because \(f(-x)=3x^4-5x^3+25x^2-45x-18\) has three variations in
sign.
3. The rational zeros theorem provides information about the potential rational
zeros of polynomials with integer coefficients. For this polynomial, which has
integer coefficients, the potential rational zeros are:
\(\pm 1/3, \pm 2/3, \pm 1, \pm 2, \pm 6, \pm 9, \pm 18\)
Since f(-2)=0, then -2 is a zero and x+2 is a factor of f. the depressed
equation is:
\(3x^3-x^2+27x-9=0\)
Factor by grouping
\(x^2(3x-1)+9(3x-1)=0\)
Factor out the common factor 3x-1.
\((x^2+9)(3x-1)=0\)
\(x^2+9=0\)
\(3x-1=0\)
\(x^2=-9\)
\(x=1/3\)
\(x=-3i,3i,1/3\)
So the complex zeros are \(-3i.3i.-2,1/3\)
Factored form of f is:
\(f(x)=3x^4+5x^3+25x^2=45x-18 = 3(x+3i)(x-3i)(x+2)(x-1/3)\)