ExamBro
ExamBro
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 

  1. A \(\frac{3}{2}\)
  2. B \(4\)
  3. C \(2\)
  4. D \(\frac{1}{2}\)
Verified Solution

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 \(\operatorname{Re}( u )+\operatorname{Im}( u )=1\)…
Same subject
Explore more questions on app