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Let a belong to R and n belong to N, n≥2. Then the number of complex roots of the equation ….

Options:
a. n−1
b. n
c. n+1
d. 2n

1 Answer

3 votes

Final answer:

The number of complex roots of the equation is n.

Step-by-step explanation:

In this problem, we are asked to determine the number of complex roots of the equation. The equation is not specified, so we will generalize the result. Let's consider a polynomial equation of the form:

anzn + an-1zn-1 + ... + a1z + a0 = 0

where an, an-1, ..., a0 are real coefficients and z is a complex number.

The Fundamental Theorem of Algebra states that a polynomial of degree n has n complex roots, counting multiplicities. Therefore, the number of complex roots of the given equation is n.

The question asks about the number of complex roots of a polynomial equation when a belongs to the real numbers (R) and n belongs to the natural numbers (N), with n ≥2. According to the Fundamental Theorem of Algebra, any non-constant single-variable polynomial with real or complex coefficients has as many complex roots as its degree, counted with multiplicity.

Hence, if the degree of the polynomial is n, it will have n complex roots. This includes real roots (since real numbers are a subset of complex numbers) and non-real complex roots.

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