# Thread: Complex Fourier Series Question

1. ## Complex Fourier Series Question

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

P.S

2. 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 $j^2 = i^2 = -1$.

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

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

4. 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

5. 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

6. 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

7. 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

8. 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