Finding the residue of a function with a surd variable $$f(z) = \frac{1}{z-2\sqrt{z}+2}$$
Is this the correct way of doing this, please advice - Thanks, what i did was try to rationalise the expression first as follows:
$$f(z) = \frac{1}{z-2\sqrt{z}+2} = \frac{1}{(z+2)-2\sqrt{z}}*\frac{(z+2)+2\sqrt{z}}{(z+2)+2\sqrt{z}}$$
$$\therefore$$
$$f(z) = \frac{(z+2)+2\sqrt{z}}{z^2+4}$$
My poles are at $z= ±4i$
$$\operatorname{Res}(f(z)e^{zt}, 4i) = \lim_{z\to 4i} (z-4i)\frac{e^{zt}((z+2)+2\sqrt{z})}{(z-4i)(z+4i)}=\lim_{z\to 4i}  \frac{e^{zt}(z+2+2\sqrt{z})}{(z+4i)}$$
$$\operatorname{Res}(f(z)e^{zt}, 4i) =\frac{e^{4it}(4i+2+2\sqrt{4i})}{8i}$$
and
$${Res}(f(z)e^{zt}, -4i) = \lim_{z\to -4i} (z+4i)\frac{e^{zt}((z+2)+2\sqrt{z})}{(z-4i)(z+4i)}=\lim_{z\to-4i}  \frac{e^{zt}(z+2+2\sqrt{z})}{(z-4i)}$$
$$\operatorname{Res}(f(z)e^{zt}, -4i) =\frac{e^{-4it}(-4i+2+2i\sqrt{4i})}{-8i}$$
$$\sum \operatorname{Res}(f(z)e^{zt}, 4i)=\frac{e^{4it}(4i+2+2\sqrt{4i})}{8i}+\frac{e^{-4it}(-4i+2+2i\sqrt{4i})}{-8i}$$

Why are calculated  The function {f\left( x \right)e^x }
\begin{array}{l}
 {\mathop{\rm Re}\nolimits} s\left( {f\left( x \right); - 2i} \right) = \mathop {\lim }\limits_{z \to  - 2i} \left( {z + 2i} \right)\left( {f(z)} \right) \\ 
  = \mathop {\lim }\limits_{z \to  - 2i} \frac{{z + 2 + 2\sqrt z }}{{z - 2i}} = \frac{{ - 2i + 2 + 2\sqrt { - 2i} }}{{ - 4i}} \\ 
 \end{array}
\begin{array}{l}
 {\mathop{\rm Re}\nolimits} s\left( {f\left( x \right);2i} \right) = \mathop {\lim }\limits_{z \to  - 2i} \left( {z - 2i} \right)\left( {f(z)} \right) \\ 
  = \mathop {\lim }\limits_{z \to  + 2i} \frac{{z + 2 + 2\sqrt z }}{{z + 2i}} = \frac{{ + 2i + 2 + 2\sqrt { + 2i} }}{{ + 4i}} \\ 
 \end{array}
