Question

Let z be non-real complex number lying on the circle I z I = 1 . Then show that

$\displaystyle{z=\dfrac{1+i\tan\left(\dfrac{arg \ z}{2}\right)}{1-i\tan\left(\dfrac{arg \ z}{2}\right)} }$

Answer 1

Given:

z be non-real complex number lying on the circle | z | = 1.

To Prove:

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

Proof :

Let,

$\huge{z = r \: {e}^{i \alpha} }$

Where,

r = |z| and,

α = arg(z)

But, |z| = 1

=> r = 1

Therefore, we get,

$\huge{= > z = {e}^{i \alpha }}$

According yo Euler's Formula,

we get,

${ = > z = \cos( \alpha ) + i \sin( \alpha ) }$

Now,

$\huge{z = {e}^{i \alpha } = \frac{ {e}^{i \alpha } }{ {e}^{ - i \alpha } }}$

So, from Euler's formula,

we get,

$\huge{ = > z = \frac{ \cos \frac{ \alpha }{2} + i \sin( \frac{ \alpha }{2} ) }{ \cos \frac{ \alpha }{2} - i \sin \frac{ \alpha }{2} } }$

Multiplying Numerator and Denominator by (cos α/2 )

we get,

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

Hence, Proved