Question 5 - In Class Fun

We are asked to compute the integral
$$
\int_{C_1(i)}\frac{1}{z^4+1}dz.
$$
I am confident that
$$
\begin{align}
z^4+1&=\bigg(z-\bigg(\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\bigg)\bigg)\bigg(z-\bigg(-\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\bigg)\bigg)\\ &\times\bigg(z-\bigg(-\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}\bigg)\bigg)\bigg(z-\bigg(\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}\bigg)\bigg).
\end{align}
$$
The point $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ is inside the circle of radius $1$ centered at $i$. By the Cauchy integral formula, this integral is equal to
$$
\begin{align}
&2\pi i\cdot\Bigg[\bigg(z-\bigg(-\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\bigg)\bigg)\bigg(z-\bigg(-\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}\bigg)\bigg)\\&\bigg(z-\bigg(\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}\bigg)\bigg)\Bigg]^{-1}=\frac{\pi(1-i)}{2\sqrt{2}},
\end{align}
$$
where $z=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$.
However, when I try to use mathematica to solve this integral using the contour $\gamma(t)=i+e^{it},0\leq t<2\pi$ I get the answer $\frac{\pi}{\sqrt{2}}$. Which is the right answer and why?

Edited out the typos. Thanks, @complexcharacter.

Your factorization has a typo or small mistake - should the factors not be $(z-[\text{stuff}])$ rather than $(z^4-[\text{stuff}])$?

I neglected to mention that both $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ and $-\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ are points inside the circle of radius $1$ centered at $i$.
Let $z_1=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$, $z_2=-\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$, $z_3=-\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$, and $z_4=\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$.
So the integral evaluates to
$$
\begin{align}
&2\pi i\cdot\Big[(z_1-z_2)(z_1-z_3)(z_1-z_4)\Big]^{-1}+2\pi i\cdot\Big[(z_2-z_1)(z_2-z_3)(z_3-z_4)\Big]^{-1}\\&=\frac{\pi(1-i)}{2\sqrt{2}}+\frac{\pi(1+i)}{2\sqrt{2}}=\frac{\pi}{\sqrt{2}},
\end{align}
$$
which is precisely what we wanted!

So, when I factor the bottom I first get $$z^4 + 1 = (z^2 + 1)(z^2 -1) = (z+i)(z-i)(z+1)(z-1)$$
And because our path is $C_1(i)$, we only enclose the pole at $z_0=i$. This gives me
$$\begin{array}{rcl}
f(z)&=&\frac{1}{(z+1)(z-1)(z+i)}\\
f(z_0)&=&\frac{1}{(1+i)(i-1)(2i)}\\
&=&\frac{1}{(-1+i-i+i^2)(2i)}\\
&=&\frac{1}{-4i}
\end{array}$$

Here, I apply the Cauchy integral formula to get:

$$\int_{C_1(i)}\frac{1}{z^4+1}=\int_{C_1(i)}\frac{f(z)}{z-i}=2\pi i\cdot f(z_0)=\frac{2\pi i}{-4i}$$

My problem is, my current understanding would lead me to believe this is in fact $-\frac{\pi}{2}$.

I don't think $z^4+1=(z^2+1)(z^2-1)$ is true.

So just to make sure I completely understand this process, when we factor a function and have two points which lie inside our curve, we treat each of the points as a new "$z_o$" and evaluate the integral at both those points. In other words say we have an integral $$\int_\gamma\frac{1}{(z-p_1)(z-p_2)f(z)}$$

where $p_1,p_2$ are inside $\gamma$ and $f(z)$ is some function. The integral becomes
$$\int_\gamma\frac{1}{(z-p_1)(z-p_2)f(z)}=2\pi i\ \frac{1}{(p_1-p_2)f(p_1)+({p_2-p_1})f(p_2)}?$$

We would then use the same process for how many ever points are inside $\gamma$ right?

$\int_{\gamma} \frac{1}{(z-p_1)(z-p_2)f(z)}dz=\int_{C_1} \frac{g(z)}{(z-p_1)}dz +\int_{C_2} \frac{h(z)}{(z-p_2)}dz$ (Rational expressing to do here) where $C_1$ and $C_2$ are two circles with the centers $p_1$ and $p_2$ respectively and are inside $\gamma$, then you can apply Cauchy theorem on each term.

That makes sense. Thanks