A strange type of complex root

Question

This function is related to the zeta function for $-7.$

Here is a function which has $2$ complex roots at $\frac{1}{2}$ which I find very strange. https://www.wolframalpha.com/input/?i=Roots+n%5E2%2F12+-+(7+n%5E4)%2F24+%2B+(7+n%5E6)%2F12+%2B+n%5E7%2F2+%2B+n%5E8%2F8

My intuition with complex roots is that they help solve equations like $x^2 +1=0$ which don't cross the $x$-axis. Thus we can shift this function by a constant so that the roots become real.

It seems like there will only be one value where these roots become real. But it is not like a root with odd multiplicity where the function clearly has zero derivative near the imaginary points.

My question is, what is a simple example of a function with a root with these properties. And what shift to the function makes this root real?

Answer 1

I figured out that if I write the equation in the form of products of it's roots. You can then remove the real roots, and the equation becomes 1/24 n^2 (n + 1)^2 (3 n^4 + 6 n^3 - n^2 - 4 n + 2). This let's me see where the complex roots come from.

The roots are actually quite regular, but looked strange

Answer 2

I'm having a hard time understanding your question.

My intuition with complex roots is that they help solve equations like x^2 +1=0 which don't cross the x axis.

The way complex numbers are defined is as numbers of the form $a+bi$ for $a$ (the "real" part) and $b$ (the "imaginary" part) are real, and $i$ is such that $i^2=-1$. This definition leads to complex numbers which do help with solving equations like $x^2+1=0$, but complex numbers are interesting for other reasons than just helping with solving polynomial equations. This is why we have a more general definition of complex numbers.

It's important to note that you can have a function with real and non-real roots. So the graph of a polynomial on the real line might cross the x-axis any number of times, and still have non-real roots. The function you provided is such an example.

It seems like there will only be one value where these roots become real. But it is not like a root with odd multiplicity where the function clearly has zero derivative near the imaginary points.

I'm not quite sure what you mean here. The sense of the "multiplicity" of a root is the power of the binomial in the factored form. For example $x=2$ is a root of multiplicity of 3 and $x=4$ is a root of miltiplicity of 1 in the polynomial $f(x)=12(x-2)^3(x-4)$. What I suspect you might be observing is the fact that non-real roots of polynomials with real coefficients come in "complex-conjugate pairs". Meaning that if $a+bi$ (where $a$ and $b$ are real) is a root $f(a+bi)=0$, then $a-bi$ is a root $f(a-bi)=0$. This is true for any polynomial with real coefficients. Is this what you are referring to?

And what shift to the function makes this root real?

I'm not sure that there is a general guidline on how shifting a polynomial by a constant will make a root with non-zero imaginary part into a real root. Shifting a polynomial by a constant would, in many cases, turn real roots into non-real roots.

One very noteworthy fact about complex numbers is that any polynomial with complex coefficients (which includes any polynomial with real coefficients) with degree $n>0$ has a complex root. A consequence is that any polynomial with complex or real coefficients $p(x)$ with degree $n>0$ can be written in the form $p(x)=c(x-a_1)(x-a_2)...(x-a_n)$ (where some of these roots $a_1,...,a_n$ might equal one another).

Answer 3

Multiply the equation by $24$ to get whole numbers $$2 n^2-7 n^4+14 n^6+12 n^7+3 n^8=0$$we can factor $n^2$ and write $$n^2(2 -7 n^2+14 n^4+12 n^5+3 n^6)=0$$ By inspection $n=-1$ is a root so $$n^2(n+1)(2-2 n-5 n^2+5 n^3+9 n^4+3 n^5)=0$$ By inspection again $n=1$ is a root so $$n^2(n+1)^2(2-4 n-n^2+6 n^3+3 n^4)=0$$

Now, consider $$f(n)=2-4 n-n^2+6 n^3+3 n^4$$ $$f'(n)=-4-2n+18n^2+12n^3$$ By inspection, $n=-\frac 12$ is a root making $$f'(n)=\left(n+\frac 12 \right)(-8+12 n+12 n^2)$$ So, the zeros of the derivative are $$n_1=-\frac 12 \qquad n_2=\frac{1}{6} \left(-3-\sqrt{33}\right) \qquad n_3=\frac{1}{6} \left(-3+\sqrt{33}\right)$$ $$f(n_1)=\frac{51}{16}\qquad f(n_2)=\frac{2}{3} \qquad f(n_3)=\frac{2}{3}$$ So the quartic never intersects the $x$ axis and then no real roots to it. You will then have four complex roots conjugate by pairs.