Why are there complex numbers in the exact solution of $\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}$? Knowing that
\begin{align} 
\cot(z)=\frac{1}{z}-2z\cdot\sum_{n=1}^{\infty} \dfrac{1}{\pi^2n^2-z^2} 
\end{align}
we can easily calculate the value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}
\end{align}
by just plugging in $z=i\pi$. Therefore:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}=\frac{1}{2}\cdot\left(\pi \coth(\pi)-1\right)
\end{align}
In this case, the "$i's$" cancel eachother out nicely. 
I also wanted to calculate the exact value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
with an equal approach. After partial fraction decomposition:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}=\frac{1}{2i}\left(\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\dfrac{1}{n^2+i}\right)=\frac{1}{2i}\left(\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\sum_{n=1}^{\infty}\dfrac{1}{n^2+i}\right)
\end{align}
By plugging in $z=\sqrt{i}\pi$ and $z=i\sqrt{i}\pi$, I arrive at:
\begin{align}
\frac{1}{2i}\left[\sum_{n=1}^{\infty}\dfrac{1}{n^2-i}-\sum_{n=1}^{\infty}\dfrac{1}{n^2+i}\right]&=\frac{1}{2i}\left[\frac{1}{2i}-\frac{\pi}{2\sqrt{i}} \cdot \cot(\sqrt{i}\pi)-\left(-\frac{1}{2i}-\frac{1}{2i\sqrt{i}}\cdot \cot(i\sqrt{i}\pi)\right)\right]=\\&=\frac{1}{4}\left[-2+\pi\sqrt{i}\cot(\sqrt{i}\pi)-\frac{\pi}{\sqrt{i}}\cot(i\sqrt{i}\pi)\right]
\end{align}
In this case I can't get completely rid of the "$i's$" and the exact value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
is a complex number. But the answer must obviously be a real number.
Wolfram Alpha gets:
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}\approx0.57848+0.\times10^{-22}\,i
\end{align}
You now can argument that the imaginary part is negligible and so the value is a real number, but that doesn't satisfy my question, because there always will be a imaginary part in the answer.
If anyone can explain to me, why it is the case, that the exact value for
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}
\end{align}
is a complex number, I would be very glad.

Addendum to Somos' answer: by expanding $\cot(x+iy)$ via trig identities, one can determine the real and imaginary parts of $\cot(z)$, allowing the answer to be expressed purely in terms of real-valued terms:
$$-\frac{1}{2}-\frac{\pi  \sin \left(\sqrt{2} \pi \right)}{2 \sqrt{2} \left(\cos
   \left(\sqrt{2} \pi \right)-\cosh \left(\sqrt{2} \pi \right)\right)}-\frac{\pi  \sinh
   \left(\sqrt{2} \pi \right)}{2 \sqrt{2} \left(\cos \left(\sqrt{2} \pi \right)-\cosh
   \left(\sqrt{2} \pi \right)\right)}$$
Cleaning up the result a bit, with $u = \sqrt{2}\pi$ this can be written as:
$$-\frac{1}{2} + \frac{\pi}{2\sqrt{2}}\frac{\sin(u)+\sinh(u)}{\cosh(u)-\cos(u)}$$
