Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$
How would I do this?
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$$\frac{z-1}{z+1}=\frac{z-1}{z+1}\frac{\overline z+1}{\overline z+1}=\frac{|z|^2-1+2i\,\text{Im}\,z}{|z+1|^2}\implies\text{Re}\,\left(\frac{z-1}{z+1}\right)=\frac{|z|^2-1}{|z+1|^2}$$
But we know that $\;z=\cos\theta+i\sin\theta\implies |z|=1\;$ , so...
DonAntonio
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What is that line above z in the second expression, I have never seen that before except for on recurring decimals? – Joe Slater Nov 24 '13 at 12:40
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Er...the complex conjugate of the complex number $;z;$ . You must have studied this, imo, if you're attempting to do this exercise as it is one of the most basic and elementary definitions in the subject of complex numbers! – DonAntonio Nov 24 '13 at 12:42
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1oh ok, yes i know what the conjugate is. Thanks for the answer. – Joe Slater Nov 24 '13 at 12:44
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In the denominator of last expression do you mean |z+1| squared instead of |z-1| squared? – Joe Slater Nov 24 '13 at 13:09
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Yes, a typo. Thanks, editing on its way. – DonAntonio Nov 24 '13 at 13:10
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If $w=a+ib,\bar w=a-ib\implies w+\bar w=2a$
So, $w$ will be purely imaginary iff $ w+\bar w=0$
Here $$\frac{z-1}{z+1}+\overline{\left(\frac{z-1}{z+1}\right)}$$ $$=\frac{z-1}{z+1}+\frac{\bar z-1}{\bar z+1}=\frac{z\bar z-\bar z+z-1+(z\bar z+\bar z-z-1)}{(z+1)(\bar z+1)}$$
$$=\frac{2(z\bar z-1)}{z\bar z+(z+\bar z)+1}$$
Now, $z\bar z=|z|^2=|\cos\theta+i\sin\theta|^2=1$
lab bhattacharjee
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$$ \begin{equation} \begin{split} \frac{z-1}{z+1} &= \frac{\cos\theta + i\sin\theta-1}{\cos\theta + i\sin\theta+1} \\ &= \frac{\cos\theta-1+i\sin\theta}{\cos\theta+1+i\sin\theta} \\ &=\frac{-2\sin^2\frac{\theta}{2} + i\sin\theta}{2\cos^2\frac{\theta}{2}+i\sin\theta} \\ &= \frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}\cdot\frac{-\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}} \\ &=\tan\frac{\theta}{2}\cdot i\cdot \frac{i\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}\\ &= i\tan\frac{\theta}{2} \end{split} \end{equation} \\ $$ Clearly the above expression is purely imaginary.
Priyatham
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I like this answer the best by the way, for it is the simplest and looks the cleanest :) – Mr Pie Feb 18 '18 at 12:21
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Using $\displaystyle\cos2A=2\cos^2A-1=1-2\sin^2A $ and $\displaystyle\sin2A=2\cos A\sin A,$
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{-2\sin^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}{2\cos^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}$$
$$=\frac{2i\sin\frac{\theta}2\left(\cos\frac{\theta}2+i\sin\frac{\theta}2\right)}{2\cos\frac{\theta}2\left(\cos\frac{\theta}2+i\sin\frac{\theta}2\right)}=i\tan\frac{\theta}2$$ as $\displaystyle\cos\frac{\theta}2+i\sin\frac{\theta}2\ne0$
lab bhattacharjee
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