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

- mathslover

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- mathslover

wait

- mathslover

\[\large{z = \frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}}\]

- anonymous

non-real complex? do you mean just imaginary?

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## More answers

- ParthKohli

In which grade do you learn this?

- mathslover

@ParthKohli in 11th grade

- mathslover

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" ... :(

- mathslover

:P Well lemme post what I think,,,, it will take time...

- ParthKohli

(nothing to do here)

- JamesJ

First thing to do is rationalize the denominator. Multiply top and bottom by
1 + i . tan(Arg(z)/2)

- AravindG

you can also do this by taking modulus of given equation

- AravindG

|dw:1360507411970:dw|

- AravindG

you will get magnitude as 1

- mathslover

@JamesJ sir but that will be verifying --- my book as well as the teacher says

- mathslover

@AravindG yes I tried that and got :
\[\large{\frac{z}{\cos \theta} = 1 + i \tan \theta }\]

- AravindG

how u gt that ??

- AravindG

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

- mathslover

@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

^ This will give us : \[\large{\frac{z}{\cos\theta} = 1 + i\tan\theta}\]

- mathslover

which I can also write as :
\[\large{\frac{z}{\cos\theta} = 1 + i \tan (arg(z))}\]

- mathslover

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

- AravindG

of course we are watching..do ur best

- anonymous

\[\Large \frac{1}{\cos\phi}z*=1-i\tan\phi \]

- anonymous

Oh sorry, I was slow in my typing, you got that far already.

- AravindG

complex number is one of my fav topics ...do tag me next time u hav a qn on it

- mathslover

Ok so I got :
\[\large{\frac{ z + \cos \theta}{z \cos \theta} = 1 + i \tan \frac{\theta}{2}}\]

- AravindG

why did you do that??

- AravindG

you are looking for z , just equate for z rest will come up :)

- mathslover

To get the numerator (see the RHS of required to prove equation) ... now I will go for denominator

- JamesJ

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 \( w = z \).

- mathslover

OK mistake there it is :
\[\large{\frac{z + \cos\theta}{2\cos\theta} = 1 + i \tan \frac{\theta}{2}}\]

- AravindG

still y did you go for this ? you could have got it from 1st eqn itself using \(\tan 2 \theta \) formula

- mathslover

and then I will subtract \(i\tan\frac{\theta}{2}\) both sides so that I get :
\[\large{\frac{z+ \cos\theta - 2i\tan\theta}{2\cos\theta} = 1 -i\tan\frac{\theta}{2}}\]

- AravindG

ok nvm u can continue

- mathslover

Oh yes @AravindG let me try that ... great thanks but dont' leave now , I am yet to finish :)

- mathslover

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

- AravindG

ok all the best !

- AravindG

i will post a better method if I get one

- AravindG

ok got it :)

- AravindG

Let the complex number be
\[\bf \large z=\cos \theta +i \sin \theta\]
\[\bf \large z=\cos^2 \theta-\sin^2 \theta +i(2\sin \dfrac{\theta}{2}\cos \dfrac{\theta}{2})\]
\[\bf \large z=(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})^2\]
\[\bf \large =(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})(\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2})\]
=\[\bf \large =\dfrac{\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}-i \sin \dfrac{\theta}{2}}\]
\[\bf \large =\dfrac{1+ i \tan \dfrac{\theta}{2}}{1- i \tan \dfrac{\theta}{2}}\]
\[\bf \large z=\dfrac{1+i \tan \dfrac{\arg(z)}{2}}{1+i \tan \dfrac{\arg(z)}{2}}\]

- AravindG

oh in second step its \[\large \cos ^2\dfrac{\theta}{2}-\sin^2 \dfrac{\theta}{2}\]

- JamesJ

Yes, that's good.

- AravindG

Hence proved :)

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