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Post by njbellord on Feb 13, 2017 19:21:37 GMT
I would be grateful if you could explain how this part of the calculation is performed:
$2i\cdot\text{Im}(S_n) = (A − \bar{B})(1 + i)^n + (B − \bar{A})(1 − i)^n$
I can follow all the rest both before and after but that bit has had me puzzled. Sorry but I have not been able to create the sign for conjugates i.e. line over A and B.
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Post by Admin on Feb 13, 2017 19:45:00 GMT
Hi Nicholas
Just subtract the expression for $\overline{S}_n$ from the expression for $S_n$ given on the last line of part (ii) and then re-arrange the terms.
If that's not enough detail, post again.
Vasco
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Feb 13, 2017 20:07:19 GMT
I would be grateful if you could explain how this part of the calculation is performed: $2i\cdot\text{Im}(S_n) = (A − \bar{B})(1 + i)^n + (B − \bar{A})(1 − i)^n$ I can follow all the rest both before and after but that bit has had me puzzled. Sorry but I have not been able to create the sign for conjugates i.e. line over A and B. Nicolas, Note that $\overline{(1+i)^n} = (1-i)^n$ and $\overline{(1-i)^n} = (1+i)^n$ because $\overline{z^n} = \overline{z} \overline{z} ...$. The LHS of course is just a version of $(x+yi) - (x-yi) = 2yi$. I see you found \bar{}. \overline{} also works for the conjugate line. Gary |
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Post by njbellord on Feb 15, 2017 18:53:55 GMT
Many thanks. I find my problem was that I had copied an earlier part incorrectly.
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