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I have shown the following identity: $$ \frac{1+e^{i \theta}}{1-e^{i \theta}} = \frac{1}{2}\cot\left(\frac{\theta}{2}\right) $$ And I now need to use this to show that the the equation: $$ \left(\frac{z+i}{z-i}\right)^n = -1$$ has $\Im(z) = 0$.

I'm not sure where to get started on this, I have found a way to do it by conjugating the 2nd equation on both sides and showing that z is equal to its own conjugate, however this does not use the original identity, and I'd really like to see how it follows from that.

Any help would be much appreciated.

Lord_Farin
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Wooster
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  • If you conjugate the first fraction you get $(1+u^{-1})/(1-u^{-1})=-(1+u)/(1-u)$ (where $u=e^{i\theta}$), which means the number is purely imaginary, so your first identity is wrong. – egreg May 31 '13 at 12:14

2 Answers2

1

Something is wrong with your identity as $$\frac{1+e^{i2t}}{1-e^{2it}}=\frac{1+\cos2t+i\sin2t}{1-\cos2t-i\sin2t}=\frac{2\cos^2t+i2\sin t\cos t}{2\sin^2t-i2\sin t\cos t}$$

$$=\frac{2\cos t(\cos t+i\sin t)}{-2i\sin t(i\sin t+\cos t)}=i\cot t$$

Now using De Moivre's formula , $$\frac{z+i}{z-i}=(-1)^{\frac1n}=e^{\frac{(2r+1)i\pi}n}$$ where $0\le r<n$

Applying componendo and dividendo, $$\frac z{-i}=\frac{e^{\frac{(2r+1)i\pi}n}+1}{e^{\frac{(2r+1)i\pi}n}-1}=i \cot \frac{(2r+1)\pi}{2n}$$

$$\implies z= \cot \frac{(2r+1)\pi}{2n}$$

lab bhattacharjee
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$$ \frac {1+e^{i \theta}}{1-e^{i \theta}} = -\frac {e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} = -\frac {2\cos \frac \theta 2}{2i \sin \frac \theta 2} = -\frac 1i \cot \frac \theta 2 = i \cot \frac \theta 2 $$ You can recheck it here.

As for the second part $$ \left ( \frac {z+i}{z-i}\right )^n = -1 \\ \frac {z+i}{z-i} = e^{\frac {(2k+1)\pi i}n} \\ z + i = (z-i) e^{\frac {(2k+1)\pi i}n} \\ z \left ( 1-e^{\frac {(2k+1)\pi i}n}\right) = -i \left ( 1+e^{\frac {(2k+1)\pi i}n}\right) \\ z = -i\ \left ( \frac {1+e^{\frac {(2k+1)\pi i}n}}{1-e^{\frac {(2k+1)\pi i}n}} \right ) $$ Based on first result $$ \frac {1+e^{\frac {(2k+1)\pi i}n}}{1-e^{\frac {(2k+1)\pi i}n}} = i \cot \frac {(2k+1)\pi}{2n} $$ so $$ z = -i^2 \cot \frac {(2k+1)\pi}{2n} = \cot \frac {(2k+1)\pi}{2n} $$ which is real, i.e. $\Im(z) = 0$.

Kaster
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