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.