2

$$A=\frac{e^{i \pi k}-1}{e^{i \pi k h}-1}$$

Where

$k=p+q$, $p$ and $q$ natural numbers

$h\in]0,1[$ real number

We can consider that the denominator is never $0$.

The result may be someway like this:

if $k$ is even then $Re(A)= 0$ (or $1$, I dont know but it should be $0$ or $1$ or $-1$)

if $k$ is odd then...

Maths Student
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  • Personally I'd multiply by $\frac zz$ where $z$ is the complex conjugate of the denominator. Then identify the real part of the numerator. – Mark Bennet Mar 10 '14 at 19:51
  • I used this http://math.stackexchange.com/questions/707053/real-part-of-a-quotient post , and the result is if $k=2n$ then $Re(A)= 0$ and if $k=2n+1$ then $Re(A)=1$. I think it is correct, If someone can confirm this answering this post will be great, thank you! – Maths Student Mar 10 '14 at 19:55
  • Usage of these rules should help:

    (I): $Re\left(a\cdot b\right)=Re\left(a\right)\cdot Re\left(b\right)-Im\left(a\right)\cdot Im\left(b\right), Im\left(a\cdot b\right)=Re\left(a\right)\cdot Im\left(b\right)+Im\left(a\right)\cdot Re\left(b\right)$ (II) $Re\left(z\right)=\frac{z+\bar{z}}{2}, Im\left(z\right)=\frac{z-\bar{z}}{2i}$ (III)for $z=Ae^{ib}$ holds $Re\left(z\right)=A cos\left(b\right)$ and $Im\left(z\right)=A sin\left(b\right)$

    then start with $A=Re\left(A\right)+iIm\left(A\right)$ and get the denominator on the other side.

     – Max Mar 10 '14 at 19:58
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    Do I miss something here? If we multiply both numerator and denominator by $e^{i \pi k}+1$ the numerator becomes $e^{i 2 \pi k}-1=\cos 2 \pi k + i \sin 2 \pi k -1 = 1-1=0$? – Alex Mar 10 '14 at 20:00
  • @Alex since $e^{i\pi}=-1$ you could be dividing by zero. On the other hand the numerator is $0$ or $-2$ – Mark Bennet Mar 10 '14 at 20:06
  • @MarkBennet: I'm talking about the numerator, the denominator is not $0$ – Alex Mar 10 '14 at 21:16
  • @Alex - if you multiply by $\frac 00$ you are dividing by $0$ – Mark Bennet Mar 10 '14 at 21:22
  • @MarkBennet: oops sorry, I understand now: I didn't realize first $e^{i \pi k} +1$ is also $0$ – Alex Mar 10 '14 at 21:30

1 Answers1

2

$$A = { (e^{i\pi k} -1) (e^{-i\pi kh} -1) \over (e^{i\pi kh} -1) (e^{-i\pi kh} -1) } = { e^{-i\pi k(h-1)} - e^{-i\pi kh} - e^{i\pi k} + 1 \over 2(1-\cos (kh \pi)) }$$

From which we obtain:

$$ \operatorname{re}A = { \cos({\pi k(h-1)}) - \cos({\pi kh}) - \cos({\pi k}) + 1 \over 2(1-\cos (kh \pi)) } $$

As Robjohn mentioned, the basic idea here is that if $z \neq 0$, then ${ w \over z} = { w \bar{z} \over z \bar{z}} = { w \bar{z} \over |z|^2}$.

copper.hat
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