Proving contour integral equal to zero Let $G$ be the path traversed once as shown:

Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i}  \right)$
The singular points $v = \pm 1, \pm i$ all lie inside the contour $G$. Thus, from this theorem (*), we have
\begin{align*}
\int_{G}{\dfrac{1}{v^4-1} \text{d}v} &= \dfrac{1}{4} \left( \int_{G}{\dfrac{1}{v-1}\text{d}v} - \int_{G}{\dfrac{1}{v+1}\text{d}v} + \int_{G}{\dfrac{i}{v-i}\text{d}v} - \int_{G}{\dfrac{i}{v+i}\text{d}v}  \right) \\
&= \dfrac{1}{4}\left( 2\pi i - 2\pi i + i\left( 2\pi i \right) - i \left( 2\pi i \right) \right) \\
&= \dfrac{1}{4} \left( 0 \right) \\
&= 0
\end{align*}
(*) Theorem: Let $C$ be a simple closed contour with a positive orientation such that $v_0$ lies interior to $C$, then $\displaystyle{\int_{C} {\dfrac{dv}{(v-v_0)^n}} = 2\pi i}$ for $n =1$ and $0$ when $n \neq 1$ is an integer.
Is that proof correct? If so, could you also point out if there are still theorems I have to mention to make it more accurate?
I'm trying to solve (perhaps overthink) this with the other approach:
We see that it is analytic except at $\pm 1$ and $ \pm i$.
Also, we can apply deformation of the contour $G$ by forming a leaf-like contour and forming the respective circles $C_1, C_2, C_3,$ and $C_4$. As shown here:

The integration can then be evaluated as
$$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} $$
And,
$$\int_{C_n}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4} \left( \int_{C_n}{\dfrac{1}{v-1}\text{d}v} - \int_{C_n}{\dfrac{1}{v+1}\text{d}v} + \int_{C_n}{\dfrac{i}{v-i}\text{d}v} - \int_{C_n}{\dfrac{i}{v+i}\text{d}v}  \right) $$
Note that when $v_n$ lies exterior to $C_n$, then by Cauchy-Goursat theorem, $\displaystyle{\int_{C_n}{\dfrac{dv}{v-v_n}} = 0}$.
Thus, for $n = 1,$,
$$\int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0-0 + i(2\pi i)- 0) = \dfrac{- \pi }{2} $$
for $ n = 2,$
$$\int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}( 2\pi i  - 0 + 0-0) = \dfrac{ \pi i}{2}$$
for $ n = 3,$
$$\int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 - 0 + 0 - i (2\pi i) ) = \dfrac{\pi }{2}$$
for $ n = 4,$
$$\int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 -(2\pi i) + 0-0 ) = \dfrac{ - \pi i }{2}$$
Therefore, $$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{- \pi }{2} + \dfrac{ \pi i}{2} + \dfrac{\pi }{2} + \dfrac{ - \pi i }{2} = 0$$
Did I just overcomplicate it? Is my first proof already enough? If any of these proofs are correct, could you also point out if there are still theorems I have to mention for them to make it more accurate?
 A: The shortest proof is as follows:
Let $\displaystyle f(z)=\frac{1}{z^4-1}$. Let $\displaystyle I=\int \limits _Cf(z)\,dz$.
Note that both the contour $C$ and the function $f$ are invariant under the rotation operation $z\mapsto iz$. Therefore, $I=iI$. Then $I=0$.
A: Since OP's first solution works just fine, I will provide yet another solution:
We "inflate" the contour $G$ so it becomes a CCW-oriented circle of radius $r > 1$.

Then
$$ \int_{G} \frac{\mathrm{d}z}{z^4 - 1}
= \int_{|z| = r} \frac{\mathrm{d}z}{z^4 - 1}
\stackrel{(w=1/z)}{=} \int_{|w|=\frac{1}{r}} \frac{-\mathrm{d}w/w^2}{w^{-4} - 1}
= - \int_{|w|=\frac{1}{r}} \frac{w^2}{1 - w^4} \, \mathrm{d}w $$
In the last integral, $\frac{w^2}{1-w^4}$ has no poles inside the circle $|w| = \frac{1}{r}$ since $\frac{1}{r} < 1$. Therefore the integral evaluates as $0$ by the Cauchy's integral theorem.
Remark. This is an example of the  "residue at infinity".
A: You could do it easier by using the special symmetry your function and your path have: They both behave nicely under rotations of $e^{i\frac{\pi}{2}}$. For the path it is evident that $C_k = e^{k\cdot i\frac{\pi}{2}}C_1$, and for the function it is the same:
\begin{align}f(e^{i\frac{\pi}{2}}z) &= \frac{1}{(e^{i\frac{\pi}{2}}z-1)(e^{i\frac{\pi}{2}}z-i)(e^{i\frac{\pi}{2}}+i)(e^{i\frac{\pi}{2}}+1)(e^{i\frac{\pi}{2}}+i)} \\ &= e^{-i\frac{\pi}{2}}\frac{1}{(z-e^{i\frac{\pi}{2}})(z-ie^{i\frac{\pi}{2}})(z+e^{i\frac{\pi}{2}})(z+ie^{i\frac{\pi}{2}})}\\ &= e^{-i\frac{\pi}{2}}\frac{1}{(z-1)(z-i)(z+1)(z+i)}\\ &= e^{-i\frac{\pi}{2}}f(z)
\end{align}
So with the same notation as in your own post, you can write
\begin{align}\int\limits_{C_k} f(z) dz = \int\limits_{C_1}f(e^{k\cdot i\frac{\pi}{2}}z) dz = e^{-k\cdot i\frac{\pi}{2}} \int\limits_{C_1}f(z) dz,\end{align}
so by adding up the different terms you end up with
$$\int\limits_G f(z) dz = \int\limits_{C_1}f(z) dz\cdot\sum\limits_{k=1}^4 e^{-k\cdot i\frac{\pi}{2}} = \int\limits_{C_1}f(z) dz\cdot(1+i-1-i) = 0$$
