JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(u=\frac{2 z+i}{z-k i}, z=x+i y\) and \(k>0 .\) If the curve represented by \(\operatorname{Re}( u )+\operatorname{Im}( u )=1\) intersects the \(y\)-axis at the points \(P\) and \(Q\) where \(P Q=5,\) then the value of \(k\) is
- A \(\frac{3}{2}\)
- B \(4\)
- C \(2\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(u =\frac{2 z + i }{ z - ki }\) \(=\frac{2 x ^{2}+(2 y +1)( y - k )}{ x ^{2}+( y - k )^{2}}+ i \frac{( x (2 y +1)-2 x ( y - k ))}{ x ^{2}+( y - k )^{2}}\) since \(\operatorname{Re}( u )+\operatorname{Im}( u )=1\)…
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