Why is $\frac{e^{i\pi n/50} + 1}{e^{i\pi n/50}-1} = \frac{e^{i\pi n/100} + e^{-i\pi n/100}}{e^{i\pi n/100} - e^{i\pi n/100}}$

Question

Can someone explain why this is the case? I am trying to use this in one of my proofs.

Answer 1

Let $z=e^{i\theta} $ and $\theta =\frac{nπ}{100}$ $$\therefore LHS=\frac{Cos(2\theta)+1+iSin(2\theta)}{Cos(2\theta)+iSin(2\theta)-1}$$ $$=\frac{2Cos^2\theta +2iSin\theta Cos\theta}{-2Sin^2\theta +2iSin\theta Cos\theta}$$ $$=\frac{2Cos\theta (Cos\theta +iSin\theta )}{2iSin\theta (Cos\theta +iSin\theta)}$$ $$=\frac{2Cos\theta }{2iSin\theta }$$ $$=\frac {2Re(z)}{2iIm(z)}$$ $$=\frac{z+\bar z}{z- \bar z}$$

Answer 2

Hint:

Let $$\begin{align}A = \exp\left(\frac{i\pi n}{100}\right) + \exp\left(-\frac{i\pi n}{100}\right) &= \exp\left(\frac{i\pi n}{100}\right) + \frac{1}{\exp\left(\frac{i\pi n}{100}\right)} \\ &= \frac{\exp\left(\frac{i\pi n}{100}\right)\exp\left(\frac{i\pi n}{100}\right) + 1}{\exp\left(\frac{i\pi n}{100}\right)} \\ &= \frac{\exp\left(\frac{i\pi n}{100} + \frac{i\pi n}{100}\right) + 1}{\exp\left(\frac{i\pi n}{100}\right)} = \,?\end{align}$$

Similarly, let $$B = \exp\left(\frac{i\pi n}{100}\right) - \exp\left(-\frac{i\pi n}{100}\right) = \,?$$ What do you get for $\dfrac AB$?

Answer 3

Start from $$ \begin{align} \frac{ \frac{1}{e^{i\pi n/100}} (e^{i\pi n/50}+1) }{\frac{1}{e^{i\pi n/100}}(e^{i\pi n/50}-1)} \end{align} $$

Answer 4

For all $x$, $$\frac{x+1}{x-1}=\frac{x+1}{x-1}\cdot 1 =\frac{x+1}{x-1}\cdot\frac{x^{-1/2}}{x^{-1/2}}=\frac {x^{1/2}+x^{-1/2}}{x^{1/2}-x^{-1/2}}$$ So now let $x=e^{i\pi n/50}$.

Incidentally, $e^{i\pi n/50}=(-1)^{n/50}$ $\stackrel {\small\rm or}{=} i^{n/25}$ by Euler's identity.