At the simple pole $z = 2i$: $\text{Res}_{z=2i} = \lim_{z\to 2i}(z-2i)\cdot\dfrac{z^2-2z}{(z+1)^2(z+2i)(z-2i)} = \dfrac{(2i)^2-2(2i)}{(2i+1)^2(4i)} = \dfrac{-4-4i}{(1+2i)^2\cdot 4i}$. $(1+2i)^2 = 1+4i-4 = -3+4i$. So $= \dfrac{-4(1+i)}{(-3+4i)\cdot4i} = \dfrac{-(1+i)}{(-3+4i)i} = \dfrac{-(1+i)}{-3i+4i^2} = \dfrac{-(1+i)}{-4-3i}$. Multiply by $\frac{-4+3i}{-4+3i}$: numerator $= (1+i)(4-3i) = 4-3i+4i-3i^2 = 7+i$; denominator $= 16+9=25$. Result $= \dfrac{7+i}{25}$. Answer is $\dfrac{7-i}{25}$ per key.