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For an equation $f(z) = z^5 - 6z^4 + 15z^3 - 34z^2 +36z -48$ show that roots $f(z) = 0$ of this equation include 2 purely imaginary roots, and find them.

I thought to substitute in $z=x+iy$ to show that you can only get a solution for $y$s, but that seems like a really long and complicated process. I then thought to use $z=re^{i\theta}$ but I'm not sure how I'd use this to get the roots from the resultant equation.

Mark Bennet
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inya
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  • This is false - the polynomial is of odd degree with real coefficients and therefore has at least one real root. Unless you want to show that it has two imaginary roots (but may have other roots too)? – Jack M Jan 01 '16 at 17:32
  • I have edited to indicate what is true - that there is a pair of imaginary roots. – Mark Bennet Jan 01 '16 at 17:34

4 Answers4

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Your polynomial $f$ has real coefficients. Therefore, if $r$ is one root of $f$, $\overline r$ will be another. If $r$ is also imaginary, then $\overline r = -r$. Thus if there is an imaginary root $r$ of $f$, then we must have $f(r)=f(-r)=0$, in other words, the polynomials $f(x)$ and $f(-x)$ have at least one common root, namely $r$.

Therefore, calculate the GCD of $f(x)$ and $f(-x)$, since it must contain all of these common roots. This turns out to be $x^2+3$, whose roots are $i\sqrt 3$ and $-i\sqrt 3$. Therefore these are roots of $f$.

Jack M
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    Jack, how would i be able to calculate the greatest common divisor - I can't seem to find any good resources online for calculating the greatest common divisor for two polynomials. Thanks. – inya Jan 01 '16 at 22:55
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    @inya Have you tried using Euclid's algorithm? – kasperd Jan 01 '16 at 23:22
  • I'm researching it now you've mentioned it - but resources on it I find are poorly explaining it - I'm not sure how you find this d(x) which is meant to divide into both gcd's and therefore provide a greatest common denominator. Can you recommend some resources? – inya Jan 01 '16 at 23:44
  • http://sites.millersville.edu/bikenaga/number-theory/exteuc/exteuc.html

    Here's a great resource.

     – user795305 Jan 02 '16 at 06:42
  • @inya Do you understand basic polynomial algebra concepts like polynomial long division, the remainder theorem, etc? Khan Academy's "Polynomial Arithmetic" course covers it, or you could look for a textbook on (modern) algebra and skip to the chapter on polynomials. If you understand all these concepts then I think the Wiki article on the Polynomial GCD should be sufficient if you soldier through it. – Jack M Jan 03 '16 at 09:06
  • Yes, I understand the basics of polynomial long division. Will have a look at Wikipedia and Benjamin's article. – inya Jan 04 '16 at 03:42
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Note that if you have a pair of conjugate imaginary roots you have a factor of the form $(z+ai)(z-ai)=z^2+a^2$. Here $a$ need not be an integer, of course, so we look for a factor of the form $z^2+b$ with $b$ a positive real number by substituting $z^2=-b$ in the original equation to obtain $$b^2z-6b^2-15bz+34b+36z-48=0$$

We then note that this splits as $$z(b^2-15b+36)-(6b^2-34b+48)=0$$

If we are looking for real $b$ we know that $z$ is imaginary for our roots so both expressions in $b$ must vanish. We find a common factor $b-3$ - the common factor here shows that there is a pair of purely imaginary roots. This gives us a factor $z^2+3$.

Mark Bennet
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  • Mark, is this common factor b-3 found by dividing the second quadratic by the first quadratic in your second equation? not too sure how you found it. – inya Jan 01 '16 at 18:26
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    @inya Well I factorised the first quadratic as $(b-3)(b-12)$ and tried $b=3$ in the second. But I could just have solved the second too, or used Euclid's algorithm for polynomials to find the common factor. – Mark Bennet Jan 01 '16 at 18:29
  • So I think I'm right in thinking that in essence, we have b=3 so thats where $z^2=3$ comes from, since $z^2=-b$. However, what 'happens' to b = 12? Why isn't that in the second factor and how come it doesn't contribute to another root of the equation? Thanks. – inya Jan 01 '16 at 23:05
  • @inya $b=12$ is an extraneous root as is $b=\frac 83$ from the second equation - substituting them in gives a false equation. If there were no common root, the assertion of purely imaginary roots would be false. – Mark Bennet Jan 01 '16 at 23:10
  • I understood your answer intuitively best, although the other answers also provided great insight. Thanks. – inya Jan 01 '16 at 23:51
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This fifth degree polynomial has 5 roots (counting multiplicities), one of them real an the others complex conjugates. To rule out multiple roots, compute the greatest common divisor with its derivative, which is 1. So it has 5 different roots.

By Descartes' rule of signs it has at most 5 positive roots (i.e, it could have 1, 3, or 5 positive roots), and at most 0 negative ones.

The rational root test tells you possible rational roots are divisors of 48. Sure enough, $p(4) = 0$.

My tame computer algebra system (maxima) tells me $p(z) = (z - 4) (z^2 + 3) (z^2 - 2 z + 4)$.

Yes, it's got two pure imaginary roots.

You could have found this out by trial division by $z^2 + a$, adjusting $a$ so the division goes through...

vonbrand
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  • What do you mean by positive root or negative root? I.e does it simply mean z=c is positive and z=-c is a negative root? – inya Jan 01 '16 at 23:01
  • @inya, positive and negative real roots. Complex numbers are neither. – vonbrand Jan 02 '16 at 00:19
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Let's consider the function $$g(z) = f(iz) = iz^5 - 6z^4 -15iz^3 + 34 z^2 + 36iz - 48$$ Then, $f$ has a pure imaginary root if (and only if) $g$ has a real root. But, if $g$ has a real root, then the polynomials $$R(z) = \operatorname{Re}(g(z)) = -6z^4 + 34 z^2 - 48$$ $$I(z) = \operatorname{Im}(g(z)) = z^5 - 15 z^3 + 36z = z(z^4 - 15 z^2 + 36)$$ must have this same real root. Thus, we look for common roots of $R$ and $I$.

Since $R(z)$ is quadratic in $z^2$, any root $z$ satisfies $$z^2 = \frac{-34 \pm \sqrt{34^2 - 4(-6)(-48)}}{2(-6)} = \frac{34 \pm 2}{12} = 3, \frac{5}{2}.$$ so, the roots of $I(z)$ are $$z = \pm \sqrt{3}, \pm \sqrt{\frac{5}{2}}.$$ Putting these values in for $z$ in $I(z)$, we find that $z= \pm\sqrt{3}$ are also roots of $I(z)$ (and $\pm \sqrt{\frac{5}{2}}$ are not roots), so $$f(\pm i\sqrt{3}) = g(\pm\sqrt{3}) = R(z) + iI(z) = 0$$ and $f$ has two purely imaginary roots.