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I'm evaluating $frac{1}{2pi i}oint_C frac{e^{zt}}{z(z^2+1)}dz$, $t > 0$ where $C$ is defined by the vertices $2+2i$, $-2+2i$, $-2-2i$, and $2-2i$. I'm pretty sure this is done by finding the poles in $C$ and the residues at the poles and then adding them together, but the answer is supposed to be $1-cos t$. Here is what I've done:

$$frac{1}{2pi i}oint_Cf(z)dz=a_{-1}$$
$$a_{-1}=lim_{z to a}frac{1}{(m-1)!}frac{d^{m-1}}{dz^{m-1}}{(z-a)^mf(z)}$$

The poles are found where $z(z^2+1)$ would be zero, which are $z=0,i,-i$. I need to take the above limit as $z$ goes to each pole.
$$lim_{z to 0}frac{1}{0!}frac{d^0}{dz^0}{(z-0)^1frac{e^{zt}}{z(z^2+1)}}=lim_{z to 0}frac{e^{zt}}{z^2+1}=1$$

$$lim_{z to i}frac{1}{0!}frac{d^0}{dz^0}{(z-i)^1frac{e^z}{z(z^2+1)}}=lim_{z to i}frac{e^{zt}+(z-i)e^{zt}t}{2z}=frac{e^{it}}{2i}$$

$$lim_{z to -i}frac{1}{0!}frac{d^0}{dz^0}{(z+i)^1frac{e^z}{z(z^2+1)}}=lim_{z to -i}frac{e^{zt}+(z+i)e^{zt}t}{2z}=frac{e^{-it}}{-2i}$$

If I were correct, the answer would be $1+frac{e^{it}}{2i}-frac{e^{-it}}{2i}$.

I think that you made a simple mistake in evaluating the second limit, which can be rewritten as:

$lim_{zto i}frac{(z-i)e^{zt}}{z(z+i)(z-i)} = -frac{1}{2}e^{it}$ so you missed a minus sign