JEE Advanced · Mathematics · 3. Complex Numbers
If \(w=\alpha+i \beta\), where \(\beta \neq 0\) and \(z \neq 1\) satisfies the condition that \(\left(\frac{w-\bar{w} z}{1-z}\right)\) is purely real, then the set of values of \(z\) is
- A \(|z|=1\) and \(z \neq 2\)
- B \(|z|=1\) and \(z \neq 1\)
- C \(z=\bar{Z}\)
- D None of these
Answer & Solution
Correct Answer
(B) \(|z|=1\) and \(z \neq 1\)
Step-by-step Solution
Detailed explanation
Let \(z_1=\frac{w-\bar{w}_z}{1-z}\), be purely real.
\(\Rightarrow z_1 =\bar{z}_1 \)
\( \therefore \frac{w-\bar{w} z}{1-z} =\frac{\bar{w}-w \bar{z}}{1-\bar{z}} \)
\( \Rightarrow w-w \bar{z}-\bar{w} z+\bar{w} z \cdot \bar{z} =\bar{w}-z \bar{w}~-\) \(w \bar{z}+w z \cdot \bar{z} \)
\( \Rightarrow (w-\bar{w})+(\bar{w}-w)|z|^2 =0 \)
\( \Rightarrow (w-\bar{w})\left(1-|z|^2\right) =0 \)
\( \Rightarrow |z|^2 =1 \)
\( \Rightarrow |z| =1 \text { and } z \neq 1 .\)
\(\text { [as, } w-\bar{w} \neq 0 \text {, since } \beta \neq 0 \text { ] }\)
\(\Rightarrow z_1 =\bar{z}_1 \)
\( \therefore \frac{w-\bar{w} z}{1-z} =\frac{\bar{w}-w \bar{z}}{1-\bar{z}} \)
\( \Rightarrow w-w \bar{z}-\bar{w} z+\bar{w} z \cdot \bar{z} =\bar{w}-z \bar{w}~-\) \(w \bar{z}+w z \cdot \bar{z} \)
\( \Rightarrow (w-\bar{w})+(\bar{w}-w)|z|^2 =0 \)
\( \Rightarrow (w-\bar{w})\left(1-|z|^2\right) =0 \)
\( \Rightarrow |z|^2 =1 \)
\( \Rightarrow |z| =1 \text { and } z \neq 1 .\)
\(\text { [as, } w-\bar{w} \neq 0 \text {, since } \beta \neq 0 \text { ] }\)
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