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  • If |z1| = 1 (z1 ≠ –1) and z_2= (z_1-1)/(z_1-1) then show that the real part of z2 is zero

If |z1| = 1 (z1 ≠ –1) and z_2= \frac{z_1-1}{z_1+1}  then show that the real part of z2 is zero

Answers (1)

Let z_1=x+iy  |z_1 |= \sqrt{x^2+y^2} =1

z_2= \frac{z_1-1}{z_1+1}=\frac{x+iy-1}{x+iy+1}

=\frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy] }

= \frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2}

\frac{x^2+y^2-1+2iy}{(x+1)^2+y^2 }=0

Since,    x^2+y^2=1

\frac{1-1+2iy}{(x+1)^2+y^2}=\frac{0+2iy}{(x+1)^2+y^2}

Therefore, the real part of z_2 is zero

 

Posted by

infoexpert21

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