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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.

Git Gud
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snowman
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5 Answers5

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For the RHS you have: $$ \dfrac{1+i}{-1+(1+i)t}=\dfrac{1+i}{t-1+it}=\dfrac{(1+i)(t-1-it)}{(t-1)^2+t^2}= \dfrac{2t-i-1}{2t^2-2t+1} $$

Emilio Novati
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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*}

Ángel Mario Gallegos
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Below put $\ a = 1\!+\!i\ \Rightarrow\ \bar a = 1\!-\!i,\,\ a\bar a = 2 = a+\bar a$

$\ \dfrac{a\bar a t - a}{a\bar a\, t^2\! -(a\!+\!\bar a)\,t +1 } = \dfrac{\quad\!\ \ \ a\,\ \ \ \ (\bar a t-1)}{(at-1){(\bar at-1)}}= \dfrac{a}{at-1}$

Bill Dubuque
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    What do you mean by "now cross multiply" This is an identity to be proven. You don't use cross multiplication for such a proof. – imranfat Feb 23 '15 at 20:58
  • @imranfat Sure you can. But I gave a more general proof before I saw your comment. – Bill Dubuque Feb 23 '15 at 21:09
  • No you can't. When you have to prove an identity, you work on one side and use the algebra to arrive at the other side. That's the proper way of proving identities. Cross multiplication is nothing more than an indicator that the identity is correct, but doesn't serve as a valid proof. The edit you presented now is much better though... – imranfat Feb 23 '15 at 21:09
  • @imranfat $\ a/b = c/d \iff ad = bc,$ and $,b,d \ne 0\ $ – Bill Dubuque Feb 23 '15 at 21:15
  • Yes, that is very true. We use that to solve rational equations. For identities however, we have to start on one side of the equal sign and work our way to the other side through algebraic manipulations. I mean no offense (you got more stars than I do), but that how it is done here, don't know about where you work. Perhaps over there where you work/live it is an acceptable form of proof... – imranfat Feb 23 '15 at 21:25
  • @imranfat The OP did not ask how to derive the RHS from the LHS. Rather, they asked how to verify the equality, for which cross multiplication works fine. But it's a moot point by now. – Bill Dubuque Feb 23 '15 at 21:31
  • Yes, from that standpoint certainly agreed. – imranfat Feb 24 '15 at 03:56
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$$2t^2-2t+1=((1+i)t-1)((1-i)t-1)\\2t-1-i=(1+i)((1-i)t-1)$$

Ross Millikan
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Factor the denominator we have $2t^2-2t+1=(2t-1-i)(t-\frac{1-i}{2})$ so our fraction is equal to $$\frac{1}{t-\frac{1-i}{2}}=\frac{1+i}{-1+(1+i)t}$$

marwalix
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