AP EAMCET · Maths · Complex Number
If \(z=x+i y\) and \(x^2+y^2=1\), then \(\frac{1+x+i y}{1+x-i y}=\)
- A \(\bar{z}\)
- B z
- C \(z+1\)
- D \(z-1\)
Answer & Solution
Correct Answer
(B) z
Step-by-step Solution
Detailed explanation
\(\frac{1+x+i y}{1+x-i y} = \frac{1+z}{1+\bar{z}}\) Since \(x^2+y^2=1\), \(|z|^2=1 \Rightarrow z\bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z}\) \(\frac{1+z}{1+\frac{1}{z}} = \frac{1+z}{\frac{z+1}{z}}\) \((1+z) \cdot \frac{z}{z+1} = z\)
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