# 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:

\(f(x)=3x^4+5x^3+25x^2=45x-18 = 3(x+3i)(x-3i)(x+2)(x-1/3)\)