Complex Fourier Series Question

**Paymemoney** · February 27th 2011, 01:12 AM

Hi
Can someone explain to be how they got the line the arrow is pointing to?

P.S

**NOX Andrew** · February 27th 2011, 01:27 AM

Starting from the preceding line:

$\dfrac{1}{2\pi}\left(\dfrac{\left(-1\right)^n\left(e^{\pi} - e^{-\pi}\right)}{1-nj}\right)$

Move $\left(e^{\pi} - e^{-\pi}\right)$ to the numerator of $\dfrac{1}{2\pi}$ (commutative property of multiplication).

$\dfrac{\left(e^{\pi} - e^{-\pi}\right)}{2\pi}\left(\dfrac{\left(-1\right)^n}{1-nj}\right) = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{\left(-1\right)^n}{1-nj}$

Move $(-1)^n$ to the far right of the expression (commutative property of multiplication).

$\dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{\left(-1\right)^n}{1-nj} = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\left(-1\right)^n$

Multiply by $\dfrac{1+nj}{1+nj}$ .

$\dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\dfrac{1+nj}{1+nj}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2j^2}\left(-1\right)^n$

I assume $j = i = \sqrt{-1}$ , in which case $j^2 = i^2 = -1$ .

$\dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2j^2}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2\left(-1\right)}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 + n^2}\left(-1\right)^n$

**Paymemoney** · February 28th 2011, 06:33 PM

Can you explain why would you multiply by $\frac{1+nj}{1+nj}$

**tonio** · February 28th 2011, 06:39 PM

Originally Posted by Paymemoney

Can you explain why would you multiply by $\frac{1+nj}{1+nj}$

That's the way you do to find an expression of the form $a+bi$ for a fraction of the

form $\frac{1}{\alpha+i\beta}\,,\,\,a,b,\alpha,\beta\in\ mathbb{R}$ .

The only weird thing is why you people use the letter "j" to denote what the huge majority

of mathematicians in the world denote by "i" = the imaginary unit, though I found that

in a very old book...

Tonio

**CaptainBlack** · February 28th 2011, 06:56 PM

Originally Posted by Paymemoney

Hi
Can someone explain to be how they got the line the arrow is pointing to?

P.S

All that is happening is that the top and bottom of the right hand side of the previous line are multiplied by $(1+nj)$ which is the complex conjugate of the denominator of the previous line. This renders the denominator real and moves all the complex terms into the numerator. Other than that there is some minor rearrangement of the other terms in the previous expression.

CB

**CaptainBlack** · February 28th 2011, 06:58 PM

Originally Posted by tonio

That's the way you do to find an expression of the form $a+bi$ for a fraction of the

form $\frac{1}{\alpha+i\beta}\,,\,\,a,b,\alpha,\beta\in\ mathbb{R}$ .

The only weird thing is why you people use the letter "j" to denote what the huge majority

of mathematicians in the world denote by "i" = the imaginary unit, though I found that

in a very old book...

Tonio

Who are the "you people" to whom you refer? In engineering the complex unit is universally referred to as "j"

CB

**tonio** · March 1st 2011, 01:37 AM

Originally Posted by CaptainBlack

Who are the "you people" to whom you refer? In engineering the complex unit is universally referred to as "j"

CB

Oh, there you go! Engineers...****sigh****

....and I bet they have an "excellent"

explanation why they use that notation, don't they?

Oh, well...

Tonio

**CaptainBlack** · March 1st 2011, 02:36 AM

Originally Posted by tonio

Oh, there you go! Engineers...****sigh****

....and I bet they have an "excellent"

explanation why they use that notation, don't they?

Oh, well...

Tonio

To avoid confusion with the symbol for current, and pending further research I will bet that its use can be traced to Oliver Heaviside.

CB

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