11TH•Mathematics•Complex Numbers•Question 16

Solution

If $z=x+iy$ and $w=\dfrac{1-iz}{z-i}$, show that $|w|=1 \Rightarrow z$ is real.

Given:

w=\frac{1-iz}{z-i},\quad z=x+iy

If $|w|=1$ then:

\left|\frac{1-iz}{z-i}\right|=1\Rightarrow |1-iz|=|z-i|

Now compute each modulus.

First:

1-iz=1-i(x+iy)=1-ix+y=(1+y)-ix

So:

|1-iz|^2=(1+y)^2+x^2

z-i=x+i(y-1)

So:

|z-i|^2=x^2+(y-1)^2

Given $|1-iz|=|z-i|$ implies squares are equal:

(1+y)^2+x^2=x^2+(y-1)^2

Cancel $x^2$ :

(1+y)^2=(y-1)^2

Expand:

y^2+2y+1=y^2-2y+1\Rightarrow 4y=0\Rightarrow y=0

Hence $z=x+0i$ is real.

\boxed{|w|=1\Rightarrow z\in\mathbb{R}}

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