AP EAMCET · Maths · Complex Number
\(P\) is a point denoting \(z\) in the argand diagram and if \(\frac{z-i}{z-1}\) is always purely imaginary, then locus of \(P\) is
- A the circle with centre \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\)
- B the circle with centre \(\left(-\frac{1}{2},-\frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\)
- C the points on the circle with centre \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\), excluding the points \((1,0)\) and \((0,1)\)
- D the points on the circle with centre \(\left(-\frac{1}{2},-\frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\), excluding the origin
Answer & Solution
Correct Answer
(C) the points on the circle with centre \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\), excluding the points \((1,0)\) and \((0,1)\)
Step-by-step Solution
Detailed explanation
Let \(z=x+i y\), then \(\frac{z-i}{z-1}=\frac{x+i(y-1)}{(x-1)+i y}\) So, \(\quad \operatorname{Re}\left(\frac{z-i}{z+1}\right)=\frac{x(x-1)+y(y-1)}{(x-1)^2+y^2}\) \(\because \frac{z-i}{z+1}\) is purely imaginary, So,…
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