Let z be a non-real complex number lying on the circle |z| = 1. Then prove that : z = \(\frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}\)

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Let z be a non-real complex number lying on the circle |z| = 1. Then prove that : z = \(\frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}\)

Mathematics
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\[\large{z = \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}}\]
non-real complex? do you mean just imaginary?

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In which grade do you learn this?
@ParthKohli in 11th grade
I was able to prove by taking RHS and simplifying and simplifying it into such a form which was exactly equivalent to the LHS i.e. 'z' But , the book says this is not a method of proving infact it is a method of "verifying" ... :(
:P Well lemme post what I think,,,, it will take time...
(nothing to do here)
First thing to do is rationalize the denominator. Multiply top and bottom by 1 + i . tan(Arg(z)/2)
you can also do this by taking modulus of given equation
|dw:1360507411970:dw|
you will get magnitude as 1
@JamesJ sir but that will be verifying --- my book as well as the teacher says
@AravindG yes I tried that and got : \[\large{\frac{z}{\cos \theta} = 1 + i \tan \theta }\]
how u gt that ??
I have 1 more method in stock ..bt I am keeping it as a novelty for end of discussion Let me see your working here
@AravindG : Say : z = \(\cos \theta + i \sin \theta \) Divide both sides by cos theta \(\large{\frac{z}{\cos \theta} = \frac{\cos \theta + i\sin \theta}{\cos \theta}}\)
^ This will give us : \[\large{\frac{z}{\cos\theta} = 1 + i\tan\theta}\]
which I can also write as : \[\large{\frac{z}{\cos\theta} = 1 + i \tan (arg(z))}\]
ok so I divide both sides by 2 now... lemme do it now .. I am curious that I can do it just in the guidance of you all
of course we are watching..do ur best
\[\Large \frac{1}{\cos\phi}z*=1-i\tan\phi \]
Oh sorry, I was slow in my typing, you got that far already.
complex number is one of my fav topics ...do tag me next time u hav a qn on it
Ok so I got : \[\large{\frac{ z + \cos \theta}{z \cos \theta} = 1 + i \tan \frac{\theta}{2}}\]
why did you do that??
you are looking for z , just equate for z rest will come up :)
To get the numerator (see the RHS of required to prove equation) ... now I will go for denominator
What I am suggesting is this. Let the left hand side be called \( w \). Then \[ w = \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})} . \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 + i \tan ( \frac{arg(z)}{2})} \] \[ = \frac{ (1 + i \tan ( \frac{ arg(z)}{2} ))^2}{1 + \tan^2(\arg(z)/2)} \] \[ = \frac{1 - \tan^2(\arg(z)/2) + 2i \tan(\arg(z)/2)}{\sec^2(\arg(z)/2)} \] Then the tan of the argument of \( w \) is \[ \tan(\arg(w)) = \frac{2 \tan(\arg(z)/2)}{1 - \tan^2(\arg(z)/2)} = \tan(\arg(z)) \] and hence \( w \) and \(z\) have the same same arguments Now you just need to show that \( |w| = |z| \) and then it will follow that \( w = z \).
OK mistake there it is : \[\large{\frac{z + \cos\theta}{2\cos\theta} = 1 + i \tan \frac{\theta}{2}}\]
still y did you go for this ? you could have got it from 1st eqn itself using \(\tan 2 \theta \) formula
and then I will subtract \(i\tan\frac{\theta}{2}\) both sides so that I get : \[\large{\frac{z+ \cos\theta - 2i\tan\theta}{2\cos\theta} = 1 -i\tan\frac{\theta}{2}}\]
ok nvm u can continue
Oh yes @AravindG let me try that ... great thanks but dont' leave now , I am yet to finish :)
Oh! Sorry everyone got to go now.. but I promise I will be back and write my solution what I get and @AravindG thanks for that trick^^^ and @JamesJ sir I will get back to you too what I not understand from your soln ... Sorry friends but it is urgent Thanks everyone
ok all the best !
i will post a better method if I get one
ok got it :)
Let the complex number be \[\bf \large z=\cos \theta +i \sin \theta\] \[\bf \large z=\cos^2 \theta-\sin^2 \theta +i(2\sin \dfrac{\theta}{2}\cos \dfrac{\theta}{2})\] \[\bf \large z=(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})^2\] \[\bf \large =(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})\] =\[\bf \large =\dfrac{\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}-i \sin \dfrac{\theta}{2}}\] \[\bf \large =\dfrac{1+ i \tan \dfrac{\theta}{2}}{1- i \tan \dfrac{\theta}{2}}\] \[\bf \large z=\dfrac{1+i \tan \dfrac{\arg(z)}{2}}{1+i \tan \dfrac{\arg(z)}{2}}\]
oh in second step its \[\large \cos ^2\dfrac{\theta}{2}-\sin^2 \dfrac{\theta}{2}\]
Yes, that's good.
Hence proved :)

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