Key Concepts and Formulas

• Algebra of Complex Numbers: Operations involving complex numbers, including addition, subtraction, multiplication, and division. Recall that $i^2 = -1$ and $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$.
• Quadratic Equations and Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}$.
• Algebraic Identity: $(a+b)^2 = a^2 + 2ab + b^2$ and $a^2 + b^2 = (a+b)^2 - 2ab$.

Step-by-Step Solution

Step 1: Transform the equation into a quadratic form

We are given the equation $\frac{z^2+3 i}{z-2+i}=2+3 i$. Our goal is to rearrange this equation into the standard quadratic form $az^2 + bz + c = 0$. We begin by multiplying both sides by the denominator:

$z^2 + 3i = (2+3i)(z-2+i)$

Expanding the right side, we get:

$z^2 + 3i = 2(z-2+i) + 3i(z-2+i)$ $z^2 + 3i = 2z - 4 + 2i + 3iz - 6i + 3i^2$

Since $i^2 = -1$, we have:

$z^2 + 3i = 2z - 4 + 2i + 3iz - 6i - 3$ $z^2 + 3i = 2z - 7 + 3iz - 4i$

Now, we rearrange the equation to get all terms on one side:

$z^2 - 2z - 3iz + 7 + 7i = 0$ $z^2 - (2+3i)z + (7+7i) = 0$

Step 2: Identify the coefficients and apply Vieta's formulas

Now we have a quadratic equation in the form $az^2 + bz + c = 0$, where $a=1$, $b = -(2+3i)$, and $c = 7+7i$. Let $z_1$ and $z_2$ be the roots of this equation. Using Vieta's formulas, we can find the sum and product of the roots:

Sum of roots: $z_1 + z_2 = -\frac{b}{a} = -\frac{-(2+3i)}{1} = 2 + 3i$

Product of roots: $z_1 z_2 = \frac{c}{a} = \frac{7+7i}{1} = 7 + 7i$

Step 3: Calculate the sum of the squares of the roots

We want to find the sum of the squares of the roots, which is $z_1^2 + z_2^2$. We can use the algebraic identity:

$z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2$

Substituting the values we found using Vieta's formulas:

$z_1^2 + z_2^2 = (2+3i)^2 - 2(7+7i)$

Step 4: Simplify the expression

First, we calculate $(2+3i)^2$:

$(2+3i)^2 = (2+3i)(2+3i) = 4 + 6i + 6i + 9i^2 = 4 + 12i - 9 = -5 + 12i$

Next, we calculate $2(7+7i)$:

$2(7+7i) = 14 + 14i$

Now, we substitute these values back into the expression for $z_1^2 + z_2^2$:

$z_1^2 + z_2^2 = (-5 + 12i) - (14 + 14i)$ $z_1^2 + z_2^2 = -5 + 12i - 14 - 14i$ $z_1^2 + z_2^2 = -19 - 2i$

Common Mistakes & Tips

• Be careful when expanding the product of complex numbers. Make sure to distribute correctly and remember that $i^2 = -1$.
• Vieta's formulas provide a powerful shortcut for finding the sum and product of roots without explicitly solving the quadratic equation.
• Double-check your calculations, especially when dealing with complex numbers and negative signs.

Summary

We transformed the given equation into a quadratic equation and used Vieta's formulas to find the sum and product of the roots. Then, using an algebraic identity, we calculated the sum of the squares of the roots, which is the desired result. The sum of all possible values of $z^2$ is $-19 - 2i$.

Final Answer

The final answer is $\boxed{-19-2 i}$, which corresponds to option (B).