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.
 A: For the RHS you have:
$$
\dfrac{1+i}{-1+(1+i)t}=\dfrac{1+i}{t-1+it}=\dfrac{(1+i)(t-1-it)}{(t-1)^2+t^2}=
\dfrac{2t-i-1}{2t^2-2t+1}
$$
A: 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*}
A: Below put $\ a = 1\!+\!i\ \Rightarrow\ \bar a = 1\!-\!i,\,\ a\bar a = 2 = a+\bar a$
$\ \dfrac{a\bar a t - a}{a\bar a\, t^2\! -(a\!+\!\bar a)\,t +1 } = \dfrac{\quad\!\ \ \ a\,\ \ \ \ (\bar a t-1)}{(at-1){(\bar at-1)}}= \dfrac{a}{at-1}$
A: $$2t^2-2t+1=((1+i)t-1)((1-i)t-1)\\2t-1-i=(1+i)((1-i)t-1)$$
A: Factor the denominator we have $2t^2-2t+1=(2t-1-i)(t-\frac{1-i}{2})$ so our fraction is equal to $$\frac{1}{t-\frac{1-i}{2}}=\frac{1+i}{-1+(1+i)t}$$
