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mathslover
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Let z be a nonreal complex number lying on the circle z = 1. Then prove that :
z = \(\frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1  i \tan ( \frac{arg(z)}{2})}\)
 one year ago
 one year ago
mathslover Group Title
Let z be a nonreal complex number lying on the circle z = 1. Then prove that : z = \(\frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1  i \tan ( \frac{arg(z)}{2})}\)
 one year ago
 one year ago

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mathslover Group TitleBest ResponseYou've already chosen the best response.0
\[\large{z = \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1  i \tan ( \frac{arg(z)}{2})}}\]
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
nonreal complex? do you mean just imaginary?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
In which grade do you learn this?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli in 11th grade
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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" ... :(
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
:P Well lemme post what I think,,,, it will take time...
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
(nothing to do here)
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
First thing to do is rationalize the denominator. Multiply top and bottom by 1 + i . tan(Arg(z)/2)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
you can also do this by taking modulus of given equation
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
dw:1360507411970:dw
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
you will get magnitude as 1
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@JamesJ sir but that will be verifying  my book as well as the teacher says
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@AravindG yes I tried that and got : \[\large{\frac{z}{\cos \theta} = 1 + i \tan \theta }\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
how u gt that ??
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@AravindG : Say : z = \(\cos \theta + i \sin \theta \) Divide both sides by cos theta \(\large{\frac{z}{\cos \theta} = \frac{\cos \theta + i\sin \theta}{\cos \theta}}\)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
^ This will give us : \[\large{\frac{z}{\cos\theta} = 1 + i\tan\theta}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
which I can also write as : \[\large{\frac{z}{\cos\theta} = 1 + i \tan (arg(z))}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
of course we are watching..do ur best
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
\[\Large \frac{1}{\cos\phi}z*=1i\tan\phi \]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.0
Oh sorry, I was slow in my typing, you got that far already.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
complex number is one of my fav topics ...do tag me next time u hav a qn on it
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Ok so I got : \[\large{\frac{ z + \cos \theta}{z \cos \theta} = 1 + i \tan \frac{\theta}{2}}\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
why did you do that??
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
you are looking for z , just equate for z rest will come up :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
To get the numerator (see the RHS of required to prove equation) ... now I will go for denominator
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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 \( w = z \).
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
OK mistake there it is : \[\large{\frac{z + \cos\theta}{2\cos\theta} = 1 + i \tan \frac{\theta}{2}}\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
still y did you go for this ? you could have got it from 1st eqn itself using \(\tan 2 \theta \) formula
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
and then I will subtract \(i\tan\frac{\theta}{2}\) both sides so that I get : \[\large{\frac{z+ \cos\theta  2i\tan\theta}{2\cos\theta} = 1 i\tan\frac{\theta}{2}}\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
ok nvm u can continue
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Oh yes @AravindG let me try that ... great thanks but dont' leave now , I am yet to finish :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
ok all the best !
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
i will post a better method if I get one
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
ok got it :)
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
Let the complex number be \[\bf \large z=\cos \theta +i \sin \theta\] \[\bf \large z=\cos^2 \theta\sin^2 \theta +i(2\sin \dfrac{\theta}{2}\cos \dfrac{\theta}{2})\] \[\bf \large z=(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})^2\] \[\bf \large =(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})\] =\[\bf \large =\dfrac{\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}i \sin \dfrac{\theta}{2}}\] \[\bf \large =\dfrac{1+ i \tan \dfrac{\theta}{2}}{1 i \tan \dfrac{\theta}{2}}\] \[\bf \large z=\dfrac{1+i \tan \dfrac{\arg(z)}{2}}{1+i \tan \dfrac{\arg(z)}{2}}\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
oh in second step its \[\large \cos ^2\dfrac{\theta}{2}\sin^2 \dfrac{\theta}{2}\]
 one year ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
Yes, that's good.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
Hence proved :)
 one year ago
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