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  • If (z-1)/(z+1) is a purely imaginary number (z ≠ –1), then find the value of |z|.

 If \frac{z-1}{z+1}  is a purely imaginary number(z \neq -1), then find the value of |z|.

Answers (1)

Let z=x+iy

\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}

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

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

\frac{z-1}{z-1} is purely imaginary

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

x^2-1+y^2=0

x^2+y^2=1 Hence,    |z|=1

Posted by

infoexpert21

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