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Question

Let C1 and C2 be two curves on the complex plane defined as C1:z+¯¯¯z=2|z1| and C2:arg(z+1+i)=α, where α(0,π2) such that curves C1 and C2 touch each other exactly at one point at P(z0). A particle starts from the point P(z0). It moves horizontally away from the origin by 2 units and then vertically away from the origin by 3 units to reach at a point P(z1). Let A denote the area bounded by C1 and the line joining P(z0) and Q(z0), where Q(z0) is the new position of P(z0) when P(z0) is rotated about the origin through an angle of 2α in clockwise direction. Then

A
|z0|=2
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B
|z0|=22
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C
|z1|=5
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D
A=23 sq. units
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Solution

The correct option is D A=23 sq. units
C1:z+¯¯¯z=2|z1|
2x=2|x1+iy|
x2=(x1)2+y2
y2=2x1y2=2(x12)

C2:arg(z(1i))=α
Curve C2 is a ray emanating from (1,1) and making an angle α from the positive real axis.
Since C1 and C2 have exactly one common point,
C2 must be a tangent to C1.
C2:y+1=m(x+1)

Solving C1 and C2, we get
y2=2(y+1m112)
my22y+3m2=0
Put D=0
44m(3m2)=0
3m22m1=0(3m+1)(m1)=0(3m+1)(m1)=0m=13,1
As α(0,π2),
m=1

C2:y+1=x+1y=x
Putting y=x in the curve C1,
x2=2x1(x1)2=0x=1
P(z0)(1,1)
|z0|=2


From above graph, P(z1)(3,4)
|z1|=5



Area of the shaded region,
A=211/22x1 dx
=2⎢ ⎢ ⎢(2x1)3/232×2⎥ ⎥ ⎥11/2
=23[10]=23 sq. units

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