Show that the function \[ f(z)=\left\{\begin{array}{c}\frac{z^{2}+2 i z}{z^{2}-i z+6,}, \text { if } z \neq-2 i \\ 4 i, z=-2 i\end{array}\right. \] is not contnions at \( z=-2 i \) (5mk

To show that the function \( f(z) \) is not continuous at \( z = -2i \), we need to examine the limit of \( f(z) \) as \( z \) approaches \( -2i \) and compare it to the value of the function at that point. First, we compute the value of \( f(z) \) at \( z = -2i \): \[ f(-2i) = 4i. \] Next, we need to determine the limit of \( f(z) \) as \( z \) approaches \( -2i \) from the other part of the function: \[ f(z) = \frac{z^2 + 2iz}{z^2 - iz + 6} \quad \text{for } z \neq -2i. \] We substitute \( z = -2i \): 1. Calculate \( z^2 \) for \( z = -2i \): \[ z^2 = (-2i)^2 = -4. \] 2. Substitute into the function: \[ f(z) = \frac{-4 + 2(-2i)i}{-4 - 2i + 6} = \frac{-4 + 4}{-4 - 2i + 6} = \frac{0}{2 - 2i} = 0. \] Now, we find that \( \lim_{z \to -2i} f(z) = 0 \). Since we have: \[ \lim_{z \to -2i} f(z) = 0 \quad \text{and} \quad f(-2i) = 4i, \] we conclude: \[ \lim_{z \to -2i} f(z) \neq f(-2i). \] Therefore, the function \( f(z) \) is not continuous at \( z = -2i \).