Let z be a non-real complex number lying on the circle |z| = 1. Then prove that :

z = $\frac{ 1 + i \tan ( \frac{ arg(z)}{2} )}{1 - i \tan ( \frac{arg(z)}{2})}$

Question

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 · Answer

wait

mathslover · Answer

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

anonymous · Answer

non-real complex? do you mean just imaginary?

parthkohli · Answer

In which grade do you learn this?

mathslover · Answer

@ParthKohli in 11th grade

mathslover · Answer

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 · Answer

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

parthkohli · Answer

(nothing to do here)

jamesj · Answer

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

aravindg · Answer

you can also do this by taking modulus of given equation

aravindg · Answer

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