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

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

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

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

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

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

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

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

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

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

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

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

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

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

AravindG
 one year ago
Best ResponseYou've already chosen the best response.1I have 1 more method in stock ..bt I am keeping it as a novelty for end of discussion Let me see your working here

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

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

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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.0ok 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mathslover
 one year ago
Best ResponseYou've already chosen the best response.0Oh! 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

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

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

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