Question

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

Answer 1

Hint: Put $\left| w \right|=1$ i.e. $\left| \dfrac{1-iz}{z-i} \right|=1$ . Now, use the property $\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}$ and cross multiply the equation and further square both sides. Apply the property of complex number, given as $\left| z \right|=z\overline{z}$ .

Complete step-by-step answer:
Now, simplify the expression and hence, put $z=x+iy$ and $\overline{z}=x-iy$ to get the relation. And use the fact that any complex number $z=x+iy$ will be real if $y=0$ .

As, it is given in the problem that $w=\dfrac{1-iz}{z-i}$ ………………………………(i)
Where, $z=x+iy$
And hence, we have to prove that if $\left| w \right|=1$ then z is purely real i.e.
$\left| w \right|=1\Rightarrow z$ is purely real.
So, let us suppose $\left| w \right|=1$ …………………………………….(ii)
Hence, taking modulus of equation (i) to both sides, we get
$\left| w \right|=\left| \dfrac{1-iz}{z-i} \right|$
As $\left| w \right|=1$ from the equation (ii). So, we can re-write the above equation as
$\Rightarrow \left| \dfrac{1-iz}{z-i} \right|=1$ ……………………………….(iii)
Now, as we know the property related to modulus of complex number is
$\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}$ ………………………………..(iv)
Hence, equation (iii) with the help of equation (iv) is given as
$\Rightarrow \dfrac{\left| 1-iz \right|}{\left| z-i \right|}=1$
On cross multiplying the above equation, we get
$\left| 1-iz \right|=\left| z-i \right|$
Squaring both sides of the above equation, we get
${{\left| 1-iz \right|}^{2}}={{\left| z-i \right|}^{2}}$ ……………………………………….(v)
Now, as we know an identity related to modulus of complex number is given as,
$\Rightarrow {{\left| z \right|}^{2}}=z\overline{z}$ ………………………………..(vi)
Where $\overline{z}$ is conjugate of z.
Hence, we get the equation (v) with the help of equation (vi) as
$\Rightarrow \left( 1-zi \right)\left( \overline{1-iz} \right)=\left( z-i \right)\left( \overline{z-i} \right)$ ………………………………(vii)
Now, we know $\overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+.....{{z}_{n}}}=\overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}}+.....\overline{{{z}_{n}}}$ …………………………(viii)
Hence, we get the equation (viii) as
$\Rightarrow \left( 1-zi \right)\left( \overline{1}-\overline{iz} \right)=\left( z-i \right)\left( \overline{z}-\overline{i} \right)$ ………………………………..(ix)
Now, we know $\overline{{{z}_{1}}{{z}_{2}}}=\overline{{{z}_{1}}}\cdot \overline{{{z}_{2}}}$ …………………………(x)
And if $z=x+iy$, then $\overline{z}=x-iy$
So, we get equation (ix) as
$\Rightarrow \left( 1-iz \right)\left( 1-\overline{i}\overline{z} \right)=\left( z-i \right)\left( \overline{z}+i \right)$
$\Rightarrow \left( 1-iz \right)\left( 1+i\overline{z} \right)=\left( z-i \right)\left( \overline{z}+i \right)$
Now, on simplifying the above equation, we get the above equation as
$\Rightarrow 1+i\overline{z}-iz-{{i}^{2}}z\overline{z}=z\overline{z}+zi-i\overline{z}-{{i}^{2}}$
As, ${{i}^{2}}=-1$ ,so, we get
$1+i\overline{z}-iz+z\overline{z}=z\overline{z}+zi-i\overline{z}+1$
Now, cancelling out the same terms from both the sides, we get the above equation as
$\Rightarrow i\overline{z}-iz=+zi-i\overline{z}$
$\Rightarrow i\overline{z}+i\overline{z}=zi+zi$
$\Rightarrow 2\overline{z}i=2zi$
$\Rightarrow \overline{z}=z$
Now, putting $z=x+iy$ to the above equation, we get
$\Rightarrow x-iy=x+iy$
$\Rightarrow 2iy=0$
$\Rightarrow y=0$
Hence, we get $z=x+iy=x+i0=x$
So, $z=x$
Hence, complex number z is real as the imaginary part of z is 0. So, if $\left| w \right|=1\Rightarrow z$ is real.

Note: One may directly put $z=x+iy$ to the given expression $w=\dfrac{1-iz}{z-i}$ and hence, simplify the relation and get the equation in $a+ib$ form, now put the modulus of that expression as 1 because $\left| w \right|=1$. Hence, it can be another approach but will take time and expression may become complex by involvement of x and y. So, the property of complex numbers will always make the solution of these kinds of problems much more flexible and less time taking.
${{\left| z \right|}^{2}}=z\overline{z}$ is the most important property for this kind of problem and remember this property for future reference. There are a lot of good problems based on this single property. And using it in this problem is the key point as well.