The given equation is:

$z^5 + bz^4 + cz^3 + dz^2 + ez + f = 0$

We are told that three of the roots are $z = -2$, $z = 2i$, and $z = 1 + i$. Since the coefficients of the equation are real, the complex roots must occur in conjugate pairs. Therefore, the additional roots are $z = -2i$ (the conjugate of $2i$) and $z = 1 - i$ (the conjugate of $1 + i$).

Thus, the five roots are: $z = -2, \quad z = 2i, \quad z = -2i, \quad z = 1 + i, \quad z = 1 - i$

Step 1: Forming the Polynomial from Roots

The polynomial with these roots can be written as: $(z + 2)(z - 2i)(z + 2i)(z - (1+i))(z - (1-i)) = 0$

Step 2: Simplifying Each Factor

1. $(z - 2i)(z + 2i) = z^2 + 4$ (Difference of squares)
2. $(z - (1+i))(z - (1-i)) = (z - 1 - i)(z - 1 + i) = (z - 1)^2 + 1 = z^2 - 2z + 2$

Thus, the equation becomes: $(z + 2)(z^2 + 4)(z^2 - 2z + 2) = 0$

Step 3: Expanding the Polynomial

First, expand $(z^2 + 4)(z^2 - 2z + 2)$: $(z^2 + 4)(z^2 - 2z + 2) = z^4 - 2z^3 + 2z^2 + 4z^2 - 8z + 8$ $= z^4 - 2z^3 + 6z^2 - 8z + 8$

Now multiply this by $(z + 2)$: $(z + 2)(z^4 - 2z^3 + 6z^2 - 8z + 8) = z^5 - 2z^4 + 6z^3 - 8z^2 + 8z + 2z^4 - 4z^3 + 12z^2 - 16z + 16$ $= z^5 + 0z^4 + 2z^3 + 4z^2 - 8z + 16$

Step 4: Comparing with the Given Polynomial

The expanded polynomial is: $z^5 + 0z^4 + 2z^3 + 4z^2 - 8z + 16 = 0$

By comparing this with the original equation $z^5 + bz^4 + cz^3 + dz^2 + ez + f = 0$, we can find the values of $b$, $c$, $d$, $e$, and $f$:

• $b = 0$
• $c = 2$
• $d = 4$
• $e = -8$
• $f = 16$

Thus, the values are: $b = 0, \quad c = 2, \quad d = 4, \quad e = -8, \quad f = 16$

Would you like a detailed explanation of any part of this solution?

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Here are 5 related questions for further practice:

1. What happens if one of the roots is a real multiple of another?
2. How would the equation change if all coefficients were complex?
3. How do you find the polynomial for any set of real and complex roots?
4. What would the solution look like if the roots included irrational numbers?
5. How can we generalize this method to higher-degree polynomials?

Tip: When roots are complex, remember to always account for their conjugates if the polynomial has real coefficients.