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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let the curve \(z(1+i)+\bar{z}(1-i)=4, z \in \mathrm{C}\), divide the region \(|z-3| \leq 1\) into two parts of areas \(\alpha\) and \(\beta\). Then \(|\alpha-\beta|\) equals :

  1. A \(1+\frac{\pi}{2}\)
  2. B \(1+\frac{\pi}{3}\)
  3. C \(1+\frac{\pi}{6}\)
  4. D \(1+\frac{\pi}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1+\frac{\pi}{2}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \text { Let } z=x+i y \\ & (x+i y)(1+i)+(x-i y)(1-i)=4 \\ & x+i x+i y-y+x-i x-i y-y=4 \\ & 2 x-2 y=4 \\ & x-y=2 \\ & |z-3| \leq 1 \\ & (x-3)^2+y^2 \leq 1 \end{aligned}\) Area of shaded region…
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