# 3.6: Complex Zeros

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When finding the zeros of polynomials, at some point you’re faced with the problem \(x^{2} =-1\). While there are clearly no real numbers that are solutions to this equation, leaving things there has a certain feel of incompleteness. To address that, we will need utilize the imaginary unit, \(i\).

Imaginary number \(i\)

The most basic complex number is \(i\), defined to be \(i=\sqrt{-1}\), commonly called an **imaginary number**. Any real multiple of i is also an imaginary number.

Example 1

Simplify \(\sqrt{-9}\).

**Solution**

We can separate \(\sqrt{-9}\) as \(\sqrt{9} \sqrt{-1}\). We can take the square root of 9, and write the square root of -1 as \(i\).

\(\sqrt{-9} =\sqrt{9} \sqrt{-1} =3i\)

A complex number is the sum of a real number and an imaginary number.

complex number

A **complex ****number** is a number \(z=a+bi\), where \(a\) and \(b\) are real numbers

\(a\) is the real part of the complex number

\(b\) is the imaginary part of the complex number \[i=\sqrt{-1}\]

**Arithmetic on Complex Numbers**

Before we dive into the more complicated uses of complex numbers, let’s make sure we remember the basic arithmetic involved. To add or subtract complex numbers, we simply add the like terms, combining the real parts and combining the imaginary parts.

Example 3

Add \(3-4i\) and \(2+5i\).

**Solution**

Adding \((3-4i)+(2+5i)\), we add the real parts and the imaginary parts

\(3+2-4i+5i\)

\(5+i\)

try it now

Subtract \(2+5i\) from \(3-4i\).

**Answer**-
\(1. (3-4i)-(2+5i)=1-9i\)

We can also multiply and divide complex numbers.

Example 4

Multiply: \(4(2+5i)\).

**Solution**

To multiply the complex number by a real number, we simply distribute as we would when multiplying polynomials.

\(4(2+5i)\)

\(=4\cdot 2+4\cdot 5i\)

\(=8+20i\)

Example 5

Divide \(\frac{(2+5i)}{(4-i)}\).

**Solution**

To divide two complex numbers, we have to devise a way to write this as a complex number with a real part and an imaginary part.

We start this process by eliminating the complex number in the denominator. To do this, we multiply the numerator and denominator by a special complex number so that the result in the denominator is a real number. The number we need to multiply by is called the **complex conjugate**, in which the sign of the imaginary part is changed.

Here, 4+*i* is the complex conjugate of 4–*i*. Of course, obeying our algebraic rules, we must multiply by 4+*i* on both the top and bottom.

\(\frac{(2+5i)}{(4-i)} \, \cdot \frac{(4+i)}{(4+i)}\)

To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL – “first outer inner last”). In the numerator:

\((2+5i)(4+i)\) Expand

\(=8+20i+2i+5i^{2}\) Since \(i=\sqrt{-1}\), \(i^{2} =-1\)

\(=8+20i+2i+5(-1)\) Simplify

\(=3+22i\)

Following the same process to multiply the denominator

\((4-i)(4+i)\) Expand

\(=(16-4i+4i-i^{2} )\) Since \(i=\sqrt{-1}\), \(i^{2} =-1\)

\(=(16-(-1))\)

\(=17\)

Combining this we get \(\frac{3+22i}{17} =\frac{3}{17} +\frac{22i}{17}\)

try it now

Multiply \(3-4i\) and \(2+3i\).

**Answer**-
\((3-4i)(2+3i)=18+i\)

In the last example, we used the conjugate of a complex number

complex conjugate

The **conjugate** of a complex number \(a+bi\) is the number \(a-bi\).

The notation commonly used for conjugation is a bar: \(\overline{a+bi}=a-bi\)

**Complex Zeros of Polynomials**

Complex numbers allow us a way to write solutions to quadratic equations that do not have real solutions.

Example 6

Find the zeros of \(f(x)=x^{2} -2x+5\).

**Solution**

Using the quadratic formula,

\(x=\frac{2\pm \sqrt{(-2)^{2} -4(1)(5)} }{2(1)} =\frac{2\pm \sqrt{-16} }{2} =\frac{2\pm 4i}{2} =1\pm 2i.\)

try it now

Find the zeros of \(f(x)=2x^{2} +3x+4\).

**Answer**-
\(x=\frac{-3\pm \sqrt{(3)^{2} -4(2)(4)} }{2(2)} =\frac{-3\pm \sqrt{-23} }{4} =\frac{-3\pm i\sqrt{23} }{4} =\frac{-3}{4} \pm \frac{\sqrt{23} }{4} i\)

Two things are important to note. First, the zeros \(1+2i\) and \(1-2i\) are complex conjugates. This will always be the case when we find non-real zeros to a quadratic function with real coefficients.

Second, we could write \(f(x)=x^{2} -2x+5=\left(x-\left(1+2i\right)\right)\left(x-\left(1-2i\right)\right)\) if we really wanted to, so the Factor and Remainder Theorems hold.

How do we know if a general polynomial has any complex zeros? We have seen examples of polynomials with no real zeros; can there be polynomials with no zeros at all? The answer to that last question, which comes from the Fundamental Theorem of Algebra, is "No."

Fundamental theorem of algebra

A non-constant polynomial \(f\) with real or complex coefficients will have at least one real or complex zero.

This theorem is an example of an "existence" theorem in mathematics. It guarantees the existence of at least one zero, but provides no algorithm to use for finding it.

Now suppose we have a polynomial \(f(x)\) of degree \(n\). The Fundamental Theorem of Algebra guarantees at least one zero \(z_{1}\), then the Factor Theorem guarantees that \(f\) can be factored as \(f(x)=\left(x - z_{1} \right)q_{1} (x)\), where the quotient \(q_{1} (x)\) will be of degree \(n - 1\).

If this function is non-constant, than the Fundamental Theorem of Algebra applies to it, and we can find another zero. This can be repeated \(n\) times.

complex factorization theorem

If \(f\) is a polynomial \(f\) with real or complex coefficients with degree \(n \ge 1\), then \(f\) has exactly \(n\) real or complex zeros, counting multiplicities.

If \(z_{1} ,z_{2} ,\ldots ,z_{k}\) are the distinct zero of \(f\) with multiplicities \(m_{1} ,m_{2} ,\ldots ,m_{k}\) respectively, then \(f(x)=a\left(x-z_{1} \right)^{m_{1} } \left(x-z_{2} \right)^{m_{2} } \cdots \left(x-z_{k} \right)^{m_{k} }\)

Example 7

Find all the real and complex zeros of \(f(x)=12x^{5} -20x^{4} +19x^{3} -6x^{2} -2x+1\).

**Solution**

Using the Rational Roots Theorem, the possible real rational roots are

\(\left\{\pm \frac{1}{1} ,\pm \frac{1}{2} ,\pm \frac{1}{3} ,\pm \frac{1}{4} ,\pm \frac{1}{6} ,\pm \frac{1}{12} \right\}\)

Testing \(\frac{1}{2}\),

Success! Because the graph bounces at this intercept, it is likely that this zero has multiplicity 2. We can try synthetic division again to test that.

The other real root appears to be \(-\frac{1}{3}\) or \(-\frac{1}{4}\). Testing \(-\frac{1}{3}\),

Excellent! So far, we have factored the polynomial to

\(f(x)=\left(x-\frac{1}{2} \right)^{2} \left(x+\frac{1}{3} \right)\left(12x^{2} -12x+12\right)=12\left(x-\frac{1}{2} \right)^{2} \left(x+\frac{1}{3} \right)\left(x^{2} -x+1\right)\)

We can use the quadratic formula to find the two remaining zeros by setting \(x^{2} -x+1=0\), which are likely complex zeros.

\(x=\frac{1\pm \sqrt{(-1)^{2} -4(1)(1)} }{2(1)} =\frac{1\pm \sqrt{-3} }{2} =\frac{1\pm i\sqrt{3} }{2} .\)

The zeros of the function are \(x=\frac{1}{2} ,-\frac{1}{3} ,\frac{1+i\sqrt{3} }{2} ,\frac{1-i\sqrt{3} }{2}\). We could write the function fully factored as \(f(x)=12\left(x-\frac{1}{2} \right)^{2} \left(x+\frac{1}{3} \right)\left(x-\frac{1+i\sqrt{3} }{2} \right)\left(x-\frac{1-i\sqrt{3} }{2} \right)\).

When factoring a polynomial like we did at the end of the last example, we say that it is **factored completely over the complex ****numbers**, meaning it is impossible to factor the polynomial any further using complex numbers. If we wanted to factor the function over the **real numbers**, we would have stopped at \(f(x)=12\left(x-\frac{1}{2} \right)^{2} \left(x+\frac{1}{3} \right)\left(x^{2} -x+1\right)\). Since the zeros of \(x^{2} -x+1\) are nonreal, we call \(x^{2} -x+1\) an **irreducible ****quadratic** meaning it is impossible to break it down any further using real numbers.

It turns out that a polynomial with real number coefficients can be factored into a product of linear factors corresponding to the real zeros of the function and irreducible quadratic factors which give the nonreal zeros of the function. Consequently, any nonreal zeros will come in conjugate pairs, so if \(z\) is a zero of the polynomial, so is \(\bar{z}\).

try it now

Find the real and complex zeros of \(f(x)=x^{3} -4x^{2} +9x-10\).

**Answer**-
Cauchy’s Bound limits us to the interval [-11, 11]. The rational roots theorem gives a list of potential zeros: \(\left\{\pm 1,\pm 2,\pm 5,\pm 10\right\}\). A quick graph shows that the likely rational root is \(x = 2\).

Verifying this,

So \(f(x)=(x-2)(x^{2} -2x+5)\)

Using quadratic formula, we can find the complex roots from the irreducible quadratic.

\(x=\frac{-(-2)\pm \sqrt{(-2)^{2} -4(1)(5)} }{2(1)} =\frac{2\pm \sqrt{-16} }{2} =\frac{-2\pm 4i}{2} =-1\pm 2i\).

The zeros of this polynomial are \(x=2,\; -1+2i,\; -1-2i\)

**Important Topics of This Section**

Complex and Imaginary numbers

Finding Complex zeros of polynomials