how to solve this small question?

[The Highlander]

the link i give in my post 17 say the conjugate of $\displaystyle 7 + \sqrt{5}$ is $\displaystyle 7 - \sqrt{5}$ but we do the conjugate of $\displaystyle i + 1$ is $\displaystyle 1 - i$ not $\displaystyle i - 1$ why?

"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.

 

[logistic_guy]

thank very much The Highlander

this is too much information

give me time to read it

i appreciate it

 

[logistic_guy]

"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'm back

this idea is new i think i'm understanding

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.

yes i'm understand very well

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.

this i didn't seei can write the answer in different form

Proof (before you ask, ):-

$\displaystyle \frac{1+e^{-\pi}}{-2}(i-1)=\frac{1+e^{-\pi}(i-1)}{-2}=\frac{1+e^{-\pi}(i-1)}{-2}\times\frac{-1}{-1}=\frac{1+e^{-\pi}\times(i-1)\times (-1)}{-2\times (-1)}\\\;\\=\frac{1+e^{-\pi}\times \overbrace{(-1)\times(i-1)}^{\text{Focusing on this bit}}}{2}=\frac{1+e^{-\pi}\times\overbrace{(-i+1)}^{\&\text{ swap 'em}}}{2}= \frac{1+e^{-\pi}\times(1-i)}{2}=\frac{1+e^{-\pi}(1-i)}{2}=\frac{1+e^{-\pi}}{2}(1-i)$​

i'm understand the proof

Hope that helps.

yes very much. i read carefully what you write and i think you explained the solution nicely. thank very much The Highlander

two more question
what is reman integral?
what is the difference between line integral and complex integral?