Question

If $z=x+iy$ and $w=\frac{(1-iz)}{(z-i)}$, then $|w|=1$ implies that, in the complex plane

• A. $z$ lies on the imaginary axis
• B. $z$ lies on the real axis
• C. $z$ lies on the unit circle
• D. None of these

Answer 1

Answer: (B)

$z$ lies on the real axis

Given $|w|$$=1$

$1=|\frac{1-iz}{z-i}|$

$\Rightarrow |z-i|=|1-iz|$.....(i)

Now let $z=x+iy$

$\Rightarrow iz=-y+ix$

Put in (i)

$\Rightarrow x^2+(y-1{)}^2=(y+1{)}^2+x^2$

$\Rightarrow (y+1{)}^2-(y-1{)}^2=0$

$\Rightarrow \left(2y\right)\left(2\right)=0$

$\Rightarrow y=0$.

This is the equation of the x-axis.