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Consider the polynomial $P(z)=(\sum _{n=0}^5 a_nz^n)(\sum _{n=0}^9b_nz^n)$ where $a_n,b_n\in \Bbb R$ $a_5\neq 0,b_9\neq 0$.

Then counting multiplicities we can conclude that $P(z)$ has :

  • at least two real roots
  • $14$ complex roots
  • no real roots
  • $12$ complex roots.

Since an odd degree polynomial has at least one real root so $(\sum _{n=0}^5 a_nz^n)$ has one real root at least and so does $(\sum _{n=0}^9 b_nz^n)$.

Hence $P(z)$ has at least two real roots counting multiplicities.So only $a$ is correct.

But my friend is arguing that $b$ is also true as each real number is a complex number.

But I think he is wrong as a complex number is one whose imaginary part is non-zero.

I am confused .Please help me to choose which one is correct.

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  • $\begingroup$ $P(z)$ is a polynomial of degree $5+9 = 14$. The fundamental theorem of algrebra says that any polynomial of degree $n$ with complex coefficients (this includes real numbers) has $n$ roots in the complex numbers (some or all roots might be real) counting multiplicity. $\endgroup$ – Winther Dec 30 '16 at 3:18
  • $\begingroup$ okay i got it@Winther $\endgroup$ – Learnmore Dec 30 '16 at 3:21
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Your friend is correct.

Every real number is also complex.

The set of complex numbers is the set $\{a + bi | a,b \text{ are real}\}$.

There is no requirement that $b$ must be nonzero.

In a discussion, if someone says "the roots are complex", they probably mean "complex but non-real", so the context may allow you mentally insert the intended (but omitted) "non-real".

But in a formal question (e.g., a textbook exercise, an exam, a math competition) if "complex but non-real" is intended, it needs to be explicitly stated.

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