### Complex numbers: Complex polynomials

### Real polynomials

Now, we now look back to polynomials whose coefficients are real. Such a polynomial is also called **real.**

*Proof:* Write #p(z)=a_nz^n+a_{n-1}z^{n-1}+ \cdots +a_1z+a_0#, in which #a_n,a_{n-1},\ldots,a_1,a_0# are real. Consider \[a_nw ^n+a_{n-1}w ^{n-1}+ \cdots + a_1w + a_0=0\tiny .\] Complex conjugating gives \[\overline{a_nw ^n+a_{n-1}w ^{n-1}+ \cdots + a_1w +a_0}=\overline{0}\tiny.\] Application of the *rules of calculation for complex conjugation* gives \[a_n\overline{w }^n+a_{n-1}\overline{w }^{n-1}+ \cdots + a_1\overline{w }+ a_0=0\tiny,\] hence, #p(\overline{w })=0#.

The abc-formula for complex solutions of a real quadratic equation

The quadratic equation

\[az^2+bz+c=0\] with real coefficients #a#, #b#, #c# for which #a\ne0#, has as complex solution \[z=\frac{-b\pm\sqrt{D}}{2a}\tiny,\] in which #D= b^2-4\cdot a\cdot c# is the *discriminant*.

- If #D\gt0#, then both solutions are real and different.
- If #D = 0#, there is one solution that is real and with multiplicity #2#.
- If #D\lt0#, then both solutions are not real: #z=\frac{-b\pm\sqrt{-D}\cdot\ii}{2a}#.

The cases #D\gt0# and #D=0# are known from the real *abc-formula*.

If #D\lt0#, then #\sqrt{D}# is the *square root of a negative real number*, from which is known it can be written as #\sqrt{-D}\cdot\ii#. The same derivation as for the real case (for example by completing the square), gives that the complex solutions are #z=\frac{-b\pm\sqrt{D}}{2a}#, which can be rewritte as #z=\frac{-b\pm\sqrt{-D}\cdot\ii}{2a}#.

Real version of the fundamental theorem of algebra

Each polynomial not equal to the zeropolynomial with real coefficients can be factored into factors of degree 1 or 2 with real coefficients.

*Proof:* Let #p(z)# be a polynomial with real coefficients. If the degree of #p(z)# is equal to #0# , then #p(z)# is a constant polynomial and we need to prove nothing. Hence, assume that the degree of #p(z)# is at least #1#. We apply the *fundamental theorem of algebra* on #p(z)#, treating it as a complex polynomial. This has as result that #p(z)# has a zero #w#, which on its turn is a complex number.

If #w# is real, then #p(z)# can be written as \[p(z)=(zw )\cdot q(z)\tiny,\]

in which #q(z)# is also a polynomial with real coefficients.

If #w# is a non-real zero of #p(z)#, then #\overline{w}# is also a zero, different from #w#, and

\[ \begin{array}{rcl}p(z) & =& (zw )\cdot(z-\overline{w})\cdot r(z)\\

& = &(z^2-(w +\overline{w })\cdot z+w \cdot \overline{w})\cdot r(z)\\

& = &(z^2-2{\Re} (w )\cdot z+|w |^2)\cdot r(z) \end {array}

\]

The first factor has real coefficients, hence, #r(z)# as well.

Since #q(z)# and #r(z)# have a lower degree than #p(z)#, we can repeat this construction, just as long until we found a polynomial of degree #0#, hene, a real constant, for #q(z)# or #r(z)#.

We give three derivations:

1. From the theory of

*Complex conjugate*we know the equation has the form #z^2-2\,a\cdot z+r=0#, in which #a=\Re(-4-\ii)=-4# and #r = \left(-4-\ii\right)\cdot\left(\overline{-4-\ii}\right)=-4^2-1^2 =17#. Hence, the answer is #z^2+8\cdot z+17#.

2. The requested polynomial has as zero #-4-1\cdot\ii#, but also has, because the polynomial has to be real,

*complex conjugate*#\overline{-4-\ii}=-4+1\cdot\ii#. These two numbers are different, hence the product

\[\left(z+4+1\cdot\ii\right)\cdot\left(z+4-1\cdot\ii\right)=z^2+8\cdot z+17\]is a divisor of the requested polynomial. Because the polynomial is of degree #2# and the leading coefficient is#1#, this is the requested polynomial.

3. Given is that the requested polynomial is of the form #z^2+b\cdot z + c# with a non-real solution. According to the theory, the zeros are \[\frac{-b\pm\sqrt{4c-b^2}\cdot\ii}{2}\tiny.\]Equating the solutions with a plus in front of the root sign to #-4-\ii# gives the equations

\[\eqs{-4&=&-\frac{b}{2}\cr -1&=&\frac{\sqrt{4c-b^2}}{2}\cr}\] The first equation leads to #b=8# and entering this value for #b# in the second equation leads to #c=17#, such that the requested polynomial is #z^2+8\cdot z+17#.

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