Let $z \in C$ be such that $\frac{z^2 +3i}{z-2+i} = 2+3i$. Then the sum of all possible values of $z^2$ is
$-19-2i$
We are given a complex equation $\frac{z^2 + 3i}{z - 2 + i} = 2 + 3i$ and we need to find the sum of all possible values of $z^2$. This involves solving a quadratic equation in $z$ and then calculating the sum of the squares of its roots.
Step 1: Simplify the Equation
First, we clear the denominator by multiplying both sides by $(z - 2 + i)$:
$z^2 + 3i = (2 + 3i)(z - 2 + i)$
$z^2 + 3i = (2 + 3i)z - (2 + 3i)(2 - i)$
Now, let's compute the product $(2 + 3i)(2 - i)$:
$(2 + 3i)(2 - i) = 4 - 2i + 6i - 3i^2$
Since $i^2 = -1$, we have:
$4 + 4i + 3 = 7 + 4i$
Substitute this back into the equation:
$z^2 + 3i = (2 + 3i)z - (7 + 4i)$
Step 2: Form the Quadratic Equation
Rearrange all terms to one side to form a standard quadratic equation $az^2 + bz + c = 0$:
$z^2 - (2 + 3i)z + (7 + 4i + 3i) = 0$
$z^2 - (2 + 3i)z + (7 + 7i) = 0$
Step 3: Calculate the Sum of the Squares of the Roots
Let the roots of the quadratic equation be $z_1$ and $z_2$. These are the possible values of $z$. We want to find the sum of all possible values of $z^2$, which is $z_1^2 + z_2^2$.
From Vieta's formulas, we know:
- Sum of roots: $z_1 + z_2 = \frac{-b}{a} = 2 + 3i$
- Product of roots: $z_1 z_2 = \frac{c}{a} = 7 + 7i$
Using the algebraic identity $z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1 z_2$:
$z_1^2 + z_2^2 = (2 + 3i)^2 - 2(7 + 7i)$
Expand $(2 + 3i)^2$:
$(2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i$
Expand $2(7 + 7i)$:
$2(7 + 7i) = 14 + 14i$
Subtract the two results:
$z_1^2 + z_2^2 = (-5 + 12i) - (14 + 14i)$
$z_1^2 + z_2^2 = -5 - 14 + 12i - 14i = -19 - 2i$
Final Answer
The sum of all possible values of $z^2$ is $-19 - 2i$.
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