$$ | z - 3 + i |

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$$ | z - 3 + i | < | z + 2 i | $$

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To solve the inequality $$ | z - 3 + i | < | z + 2 i | $$, we start by letting $$ z = x + yi $$, where $$ x $$ and $$ y $$ are real numbers. We can rewrite the inequality in terms of $$ x $$ and $$ y $$: 1. The left-hand side becomes: $$ | z - 3 + i | = | (x - 3) + (y + 1)i | = \sqrt{(x - 3)^2 + (y + 1)^2} $$ 2. The right-hand side becomes: $$ | z + 2i | = | x + (y + 2)i | = \sqrt{x^2 + (y + 2)^2} $$ Now, we can rewrite the inequality: $$ \sqrt{(x - 3)^2 + (y + 1)^2} < \sqrt{x^2 + (y + 2)^2} $$ Next, we square both sides to eliminate the square roots (noting that both sides are non-negative): $$ (x - 3)^2 + (y + 1)^2 < x^2 + (y + 2)^2 $$ Expanding both sides: - Left-hand side: $$ (x - 3)^2 + (y + 1)^2 = (x^2 - 6x + 9) + (y^2 + 2y + 1) = x^2 + y^2 - 6x + 2y + 10 $$ - Right-hand side: $$ x^2 + (y + 2)^2 = x^2 + (y^2 + 4y + 4) = x^2 + y^2 + 4y + 4 $$ Now we can set up the inequality: $$ x^2 + y^2 - 6x + 2y + 10 < x^2 + y^2 + 4y + 4 $$ Subtracting $$ x^2 + y^2 $$ from both sides: $$ -6x + 2y + 10 < 4y + 4 $$ Rearranging gives: $$ -6x + 10 < 2y + 4 $$ $$ -6x + 6 < 2y $$ $$ -3x + 3 < y $$ $$ y > -3x + 3 $$ Thus, the solution to the inequality $$ | z - 3 + i | < | z + 2 i | $$ is given by the region above the line $$ y = -3x + 3 $$ in the complex plane. In summary, the solution is: $$ y > -3x + 3 $$ The solution is $$ y > -3x + 3 $$.

\( | z - 3 + i | < | z + 2 i | \)

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