### Complex numbers: Complex polynomials

### Zeros of complex polynomials

We will now examine the zeros of a complex polynomial in general.

Linear factors of polynomials correspond with zeros

If #p(z)# is a *complex polynomial* in #z# of *degree* #n# and #a# is a complex number which is a *zero* to #p(z)=0#, then #p(z)# has a factor #z-a #; that is to say: there is polynomial #q(z)# of degree #n-1# that satisfies \[p(z)=(za)\cdot q(z)\tiny.\]

*Proof:* For each choice of #a# we can divide #p(z)# by #z-a#. We find a *quotient* #q(z)# and a *remainder* #r(z)#: \[

p(z)=(z-a )\cdot q(z)+r(z)\tiny.

\] Because the remainder has degree less than #1#, it is a complex number #r#. Now #p(a)=0#, such that

\[

0=p(a)=(a-a)\cdot q(a)+r=0\cdot q(a)+r=r\tiny,

\]

hence, #r=0#. We conclude that \[

p(z)=(za)\cdot q(z)\tiny,

\] just as we needed to prove.

If #p(a)=0#, then #p(z)# can be written as #(za)\cdot q(z)#. If #q(a)=0#, then #q(z)# also contains a factor #(z-a)#, and therefore \[p(z)=(za)^2\cdot r(z)\,,\ \ \text{and so on}\]

A complex number #a# is called a zero of **multiplicity** #m# of a complex polynomial #p(z)# if there exists a polynomial #t(z)# with #t(a)\neq 0#, such that

\[

p(z)=(za)^m\cdot t(z)\tiny .

\]

The multiplicity of a zero #a# hence is the number of factors #(z-a)# of #p(z)#.

The number #0# is a zero #z^n# with multiplicity #n#.

The other extreme, #n# zeros of a complex polynomial of degree #n# each with multiplicity #1#, also occurs. First we will deal with the case #n=2#. Let #w# be a complex number that is not equal to #0# and consider the equation \[

z^2=w\tiny.\] According to the statement *Higher power roots *this equation has two solutions, both with absolute value #\sqrt{|w|}# and respectively with #\frac12\arg (w)# and # \frac12 \arg (w)+\pi# as argument. This means that the polynomial #z^2-w# can be decomposed into two linear factors: \[ \begin{array}{rcl}z^2-w&=&\left(z-\sqrt{|w|}\cdot\left(\cos\left(\frac12\arg (w)\right)+\sin\left(\frac12\arg (w)\right)\cdot\ii\right)\right)\\ &&\cdot\left(z+\sqrt{|w|}\cdot\left(\cos\left(\frac12\arg (w)\right)+\sin\left(\frac12\arg (w)\right)\cdot\ii\right)\right)\end {array}\] The multiplicities of the zeros of the complex polynomial #z^2-w# are thus all equal to #1#.

In general, this is also the case for the polynomial #z^nw# as long as #w\ne0#.

If #p(z)# is a polynomial of degree #n#, from the previous follows immediately that the number of zeros of #p(z)#, each counted with its multiplicity, is no more than #n#. In fact, there is a much stronger property, which we will treat *later*.

The polynomial #p(z)# can be factored as follows:

\[p(z)=\left(z+\ii\right) ^{2} \cdot \left(z^3+\ii\cdot z+3\right)\tiny.\] The value of the second factor at #-\ii# is equal to #\ii+4\ne0#, so no higher power of #z+\ii# divides the polynomial #p(z)#.

This result is found by

- equating the number of factors #z+\ii# of #p(z)# we are looking for to #k=0# ;
- verifying whether #p(z)# is divisible by #z+\ii# by determining whether #p(-\ii)=0#;
- if so, adding #1# to #k# and treating the quotient #q(z)# of the division by #z+\ii# like #p(z)# in step 2 and repeating step 2 until #q(z)# is no longer divisble by #z+\ii#.

\[ \begin{array}{rcl} p(-\ii) &=& 0\\

&&\phantom{xyzuvw}\color{blue}{\text{substitution }z=-\ii}\\

p(z) &=& \left(z+\ii\right)\cdot p_1(z)\\

&&\phantom{x}\text{ where }\\

p_1(z) &=& z^4+\ii\cdot z^3+\ii\cdot z^2+2\cdot z+3\cdot \ii\\

&&\phantom{xyzuvw}\color{blue}{\text{division of }p(z)\text{ by }z+\ii}\\

p_1(-\ii)&=& 0\\ &&\phantom{xyzuvw}\color{blue}{\text{substitution }z=-\ii}\\ p(z) &=& \left(z+\ii\right)^2\cdot p_2(z)\\ &&\phantom{x}\text{ where }\\ p_2(z)&=& z^3+\ii\cdot z+3\\ &&\phantom{xyzuvw}\color{blue}{\text{division of } p_1(z)\text{ by }z+\ii}\\

p_{2}(-\ii) & = & \ii+4\ne0\\

\end {array}\] The conclusion is that #p(z)=\left(z+\ii\right) ^{2} \cdot p_{2}(z)# and that #-\ii# is not a zero of #p_{2}(z)#. In particular, #-\ii# is a zero #p(z)# of multiplicity #2#.

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