MHT CET · Maths · Complex Number
If \(|z|=1\) and \(w=\frac{z-1}{z+1}\) (where \(z \neq-1\) ), then \(\operatorname{Re}(\mathrm{w})\) is
- A 0
- B \(-\frac{1}{|z+1|^2}\)
- C \(\left|\frac{z}{z+1}\right| \cdot \frac{1}{|z+1|^2}\)
- D \(\frac{\sqrt{2}}{|z+1|^2}\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(\mathrm{W} =\frac{\mathrm{z}-1}{\mathrm{z}+1} \)
\( =\frac{x+\mathrm{i} y-1}{x+\mathrm{i} y+1} \)
\( =\frac{(x-1+\mathrm{i} y)}{(x+1+\mathrm{i} y)} \times \frac{(x+1-\mathrm{i} y)}{(x+1-\mathrm{i} y)}\)
\(=\frac{\left(x^2+y^2-1\right)+(-x +1+x y+1) \mathrm{i}}{(x+1)^2+y^2} \)
\(=\frac{x^2+y^2+1}{(x+1)^2+y^2}+\frac{2 y \mathrm{i}}{(x+1)^2+y^2}\)
Given that \(|z|=1\)
\(\Rightarrow x^2+y^2=1\)
\(\Rightarrow x^2+y^2-1=0\)
\(\therefore \operatorname{Re}(\mathrm{w})=0\)
\( =\frac{x+\mathrm{i} y-1}{x+\mathrm{i} y+1} \)
\( =\frac{(x-1+\mathrm{i} y)}{(x+1+\mathrm{i} y)} \times \frac{(x+1-\mathrm{i} y)}{(x+1-\mathrm{i} y)}\)
\(=\frac{\left(x^2+y^2-1\right)+(-x +1+x y+1) \mathrm{i}}{(x+1)^2+y^2} \)
\(=\frac{x^2+y^2+1}{(x+1)^2+y^2}+\frac{2 y \mathrm{i}}{(x+1)^2+y^2}\)
Given that \(|z|=1\)
\(\Rightarrow x^2+y^2=1\)
\(\Rightarrow x^2+y^2-1=0\)
\(\therefore \operatorname{Re}(\mathrm{w})=0\)
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