Question

z=x+iy
w= 1-iz/z-i

Prove that (mod w)|w|=1,
and hence z is real. ​

Answer 1

hope it's helpful :-)))

Answer 2

Answer:

ıllıllı ʜᴇʏᴀ ıllıllı

Step-by-step explanation:

Given

$W=(\frac{1-iz}{z-i} )^{2}$

Squaring on both Sides,

$W^{2} =(\frac{1-iz}{z-i} )^{2}$

$=> 1= \frac{1+i^{2}z^{2}-2iz  }{z^{2}i^{2}-2iz } </strong></p><p><strong>(∵║w║ 1 ⇒ [tex]w^{2} =1$

⇒$z^{2} -1-2iz=1-z^{2} -2iz$

$2z^{2}=2$

$z^{2} =1$

$(x+iy^{2} )=1$

$x^{2}i^{2} y^{2}  +2ixy=1$

$x^{2}-y^{2}  +2ixy=1$

Imaginary Part is Zero.so Z is Purely Real.