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Question

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

The correct answer 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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