The Highlander
Senior Member
- Joined
- Feb 18, 2022
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"We" do not "do the conjugate of \(\displaystyle i + 1\) is \(\displaystyle 1 - i\)"! The conjugate of (i + 1) is (i - 1)! We (your helpers) used the conjugate of (1 + i ) not (i + 1). Why?
Well, because, if you look back over the first 8 posts, you will notice that, in his final piece of advice to you, @topsquark rearranged the denominator in your expressions from (i + 1) to (1 + i)!
I expect he did this (just as I would have|) for two reasons: firstly, as someone with a wealth of experience in these matters (and thorough familiarity with the Difference of Squares rule), he would have the foresight to recognize that this rearrangement would produce a positive result in the denominator (we like to avoid negatives whenever that choice is an option) and, secondly, you had already stated (in Post #3) that your desired result would have (1 - i) in the numerator and so multiplying (top & bottom) by (1 - i) was the way to achieve that (which meant you had to have (1 + i) rather than (i + 1) as the denominator before rationalizing it.
I trust you understand that while Addition is Commutative, Subtraction is not. Therefore...
While (i + 1) = (1 + i), (i - 1) ≠ (1 - i)!
So it was perfectly OK for the denominator to be rearranged from (i + 1) to (1 + i) in order to get the desired result.
You could, just as legitimately, have retained (i + 1) as the denominator and multiplied the top and bottom by its conjugate: (i - 1), however that would have given you -2 as the resulting denominator BUT, multiplying the numerator by (i - 1) would also have changed its sign too!
Therefore: \(\displaystyle \frac{1+e^{-\pi}}{-2}(i-1)=\frac{1+e^{-\pi}}{2}(1-i)\)
So, if you had pressed ahead with (i + 1) as the denominator and multiplied top & bottom by
its conjugate, (i - 1), then you would (or should) have arrived at: \(\displaystyle \frac{1+e^{-\pi}}{-2}(i-1)\) as your answer!
Now that is a perfectly correct answer to the problem.... it just doesn't look like the one you were hoping to arrive at.
Proof (before you ask,
):-\(\displaystyle \frac{1+e^{-\pi}}{-2}(i-1)=\frac{1+e^{-\pi}(i-1)}{-2}=\frac{1+e^{-\pi}(i-1)}{-2}\times\frac{-1}{-1}=\frac{1+e^{-\pi}\times(i-1)\times (-1)}{-2\times (-1)}\\\;\\=\frac{1+e^{-\pi}\times \overbrace{(-1)\times(i-1)}^{\text{Focusing on this bit}}}{2}=\frac{1+e^{-\pi}\times\overbrace{(-i+1)}^{\&\text{ swap 'em}}}{2}= \frac{1+e^{-\pi}\times(1-i)}{2}=\frac{1+e^{-\pi}(1-i)}{2}=\frac{1+e^{-\pi}}{2}(1-i)\)
Hope that helps.
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i can write the answer in different form