What is the polynomial function of lowest degree with a lead coefficient of 1 and roots 1 and 1i?

To find the polynomial function of lowest degree with a lead coefficient of 1 and roots 1 and 1i, we first need to note that polynomials with real coefficients require complex roots to be in conjugate pairs. Thus, if 1i is a root, its conjugate, -1i, must also be a root.

The roots of the polynomial are:

• 1 (a real root)
• 1i (a complex root)
• -1i (the conjugate of 1i)

Using the root-to-factor theorem, we can write the factors corresponding to these roots:

• (x – 1) for the root 1
• (x – 1i) for the root 1i
• (x + 1i) for the root -1i

The polynomial can be expressed as:

P(x) = (x - 1)(x - 1i)(x + 1i)

Now, we can simplify the expression. First, let’s multiply the factors for the complex roots:

(x - 1i)(x + 1i) = x^2 + 1

Now we substitute this result back into our polynomial:

P(x) = (x - 1)(x^2 + 1)

Next, we expand this:

P(x) = x(x^2 + 1) - 1(x^2 + 1)
      = x^3 + x - x^2 - 1
      = x^3 - x^2 + x - 1

So, the polynomial function of lowest degree, with a lead coefficient of 1 and the given roots, is:

P(x) = x^3 - x^2 + x - 1

This polynomial meets the criteria of having a lead coefficient of 1 and includes all specified roots.