<s>[INST] how does this expression cancel out How does $$\frac{2t-1-i}{2t^2-2t+1}=\frac{1+i}{-1+(1+i)t}$$ I just can't see how this works...
I typed the LHS in WFA and it gave the RHS but I don't know how anything can cancel.
 [/INST] By using quadratic formula we get the roots of $2t^2-2t+1$, they are $\frac{1+i}{2}$ and $\frac{1-i}{2}$, then, we can factor out $2t^2-2t+1$ as 
\begin{align*}
2t^2-2t+1&=2\left(t-\frac{1+i}{2}\right)\left(t-\frac{1-i}{2}\right)\\
&=(2t-1-i)\left(t-\frac{1-i}{2}\right)
\end{align*}
Hence
\begin{align*}
\frac{2t-1-i}{2t^2-2t+1}&=\frac{2t-1-i}{(2t-1-i)\left(t-\frac{1-i}{2}\right)}\\
&=\frac{1}{t-\frac{1-i}{2}}\cdot\frac{1+i}{1+i}\\
&=\frac{1+i}{(1+i)t-\frac{(1-i)(1+i)}{2}}\\
&=\frac{1+i}{(1+i)t-1}
\end{align*}
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