# Assume $p(z)$ is a polynomial with $\deg p(z)=9$… how many real roots can $p(z)$ have?

Assume $p(z)$ is a polynomial with $\deg(p(z))=9$ with real coefficients. Assume that $p(1+i)=0$ and that $p(z)$ has at least one pure imaginary root (i.e on the form $ai, \ a\in\mathbb{R}$).

1. How many real roots can $p(z)$ have the most?
2. How many real roots can $p(z)$ have the least?

1) Since the coefficients $\in\mathbb{R}$, then $\overline{1+i}=1-i$ and $\overline{ai}=-ai$ are also roots. So the maxmum number of roots for $a\neq 0$ that $p(z)$ can have is $9-4=5$.

If however $a=0$, is the root classified real or imaginary? Depending on the answer here,the number $5$ will change.

2) Since it's a degree 9 polynomial with real coefficients, all komplex roots come in pairs. So the last root has to be real, thus the least number of real roots has to be 1 regardless of the sign of $a$.

• Yes, both look good, and $a \neq 0$ seems implied as $0$ can't be called pure imaginary. – Macavity Oct 22 '17 at 11:26

Complex roots are pairwise conjugate. Hence there can be $0, 2,4 ,6,8$ complex roots, whence $...$ real roots.
I suppose that whoever created this exercise had in his or her mind that $a\neq0$. However, you cannot deduce automatically that the maximal number of real roots of $5$. You should provide an example, such as$$(x-1-i)(x-1+i)(x^2+a^2)x(x-1)(x-2)(x-3)(x-4).$$
And, yes, the minimal number of real roots is $1$ whenever all coefficients are real and the degree is odd, althou I do not understand what is that you mean when you mention tha “last” root.
HINT.-$p(z)=(z^2-2z+2)(z^2+a^2)(b_5z^5+b_4z^4+b_3z^3+b_2z^2+b_1z+b_0)$ and according with the problem the third factor has as only condition $b_5\ne0$ and also of real coefficients. So you can choose it as you want and ,clearly, without forgetting that the degree is odd.