AP EAMCET · Maths · Complex Number
If \(Z \neq \pm 1\) is a complex number and \(\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right)=\frac{\pi}{4}\), then the locus of \(Z\) in the Arg and plane is
- A \(x^2+y^2-2 y-1=0\)
- B \(x^2+y^2+2 y-1=0\)
- C \(x^2+y^2-2 x+1=0\)
- D \(x^2+y^2+2 x+1=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-2 y-1=0\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } z=x+i y \\ & \begin{aligned} \frac{z-1}{z+1} & =\frac{x+i y-1}{x+i y+1} \\ & =\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y} \\ & =\frac{x^2+y^2-1+2 i y}{(x+1)^2+y^2} \\ \arg \left(\frac{z-1}{z+1}\right) & =\tan ^{-1}\left(\frac{2…
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