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The question I met is to show that if $z=\cos (\theta)+i\sin(\theta)$ with $i=\sqrt{-1}$,then $ Re(\frac{z-1}{z+1})=0$

In the normal way, we found that: $$\frac{z-1}{z+1}=\frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}\\ =\frac{\bigl(\cos(\theta)-1+i\sin(\theta)\bigl)\bigl(\cos(\theta)+1-i\sin(\theta)\bigl)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\ =\frac{\cos^2(\theta)+\cos(\theta)-\cos(\theta)-1+\sin^2(\theta)+2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\ =\frac{2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}$$

So $Re(\frac{z-1}{z+1})=0$

If we do it in another way: $$ \frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}=\frac{-2\sin^2(\frac{\theta}{2})+2i\cos(\frac{\theta}{2})\sin(\frac{\theta}{2})}{2cos^2(\frac{\theta}{2})+2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\ =-\tan(\frac{\theta}{2})\frac{\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ =-\tan(\frac{\theta}{2})\frac{\cos\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)+i\sin\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\ =-\tan(\frac{\theta}{2})\bigl(\cos(-\theta-\frac{\pi}{2})+i\sin(-\theta-\frac{\pi}{2})\bigl) $$

So the real part of it will be $-\tan(\frac{\theta}{2})\cos(-\theta-\frac{\pi}{2})$ which is not $0$.

Which step I made mistake or they are equivalent?

yuanming luo
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  • There must be a sign error, the $\theta$ should simplify instead, and the fraction turn to $e^{i\pi/2}$. –  Jan 08 '19 at 09:44
  • Demoivre's theory, the arguments of two complex number are added. – yuanming luo Jan 08 '19 at 09:47
  • How did you get from $$-\tan(\frac{\theta}{2})\frac{\cos\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)+i\sin\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}$$ to $$-\tan(\frac{\theta}{2})\bigl(\cos(-\theta-\frac{\pi}{2})+i\sin(-\theta-\frac{\pi}{2})\bigl)?$$ – 5xum Jan 08 '19 at 09:55
  • @5xum That's what I asked earlier, and the OP's comment above is about. It's simply subtraction of angles, from dividing unit length complex numbers by one another. – Arthur Jan 08 '19 at 09:56
  • @Arthur " That's what I asked earlier"... I don't see your question anywhere... – 5xum Jan 08 '19 at 09:58
  • @5xum For some reason I deleted that comment once I got the clarification. Could've saved you the trouble if I had kept it. Sorry. – Arthur Jan 08 '19 at 09:58

2 Answers2

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You should be more careful when applying transformations of sines into cosines.

It's much easier than that: $$ \sin\frac{\theta}{2}-i\cos\frac{\theta}{2}= i\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right) $$ You could write this in trigonometric form $$ =\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) \left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right) =\cos\left(\frac{\theta}{2}+\frac{\pi}{2}\right)+ i\sin\left(\frac{\theta}{2}+\frac{\pi}{2}\right) $$ and now you can go and chase for your mistake.

Another way to do the same: the conjugate of your number $w=\frac{z-1}{z+1}$ is $$ \bar{w}=\frac{\bar{z}-1}{\bar{z}+1} $$ but, since $|z|=1$, we have $\bar{z}=z^{-1}$; therefore $$ \bar{w}=\frac{z^{-1}-1}{z^{-1}+1}=\frac{1-z}{1+z}=-w $$ From $\bar{w}=-w$ it follows that $\operatorname{Re}(w)=0$.

If you want to find the imaginary part in term of $\theta$, you can consider $z=u^2$, where $u=\cos(\theta/2)+i\sin(\theta/2)$; then $$ w=\frac{z-1}{z+1}=\frac{u^2-1}{u^2+1}=\frac{u-u^{-1}}{u+u^{-1}} =\frac{u-\bar{u}}{u+\bar{u}}= \frac{2i\sin(\theta/2)}{2\cos(\theta/2)}=i\tan\frac{\theta}{2} $$

egreg
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We have $$ \sin\left(\frac\theta2\right) - i\cos\left(\frac\theta2\right) = -\sin\left(-\frac\theta2\right) - i\cos\left(\frac\theta2\right)\\ = -\cos\left(-\frac\theta2-\frac\pi2\right) - i\sin\left(\frac\theta2 + \frac\pi2\right)\\ = -\left(\cos\left(\frac\theta2 + \frac\pi2\right) + i\sin\left(\frac\theta2 + \frac\pi2\right)\right) $$ so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.

Arthur
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