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Question

Let |z|2+(34i)z+(3+4i)¯z92=0 and (1i)z+(1+i)¯z16=0 intersect at z1 and z2. Then the sum of the areas of two parallelograms having origin as one common vertex, and (z1+z2) as diagonal of one parallelogram and (z1z2) as diagonal of the other parallelogram is

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Solution

|z|2+(34i)z+(3+4i)¯z92=0 (1)
(1i)z+(1+i)¯z16=0 (2)
Let z=x+iy¯¯¯z=xiy
z+¯¯¯z=2x,z¯¯¯z=2iy (3)

From (1) and (3),
|z|2+3(z+¯¯¯z)4i(z¯¯¯z)92=0
x2+y2+6x+8y92=0 (4)
From (2),
(z+¯¯¯z)i(z¯¯¯z)16=0
2x+2y16=0
x+y=8
y=8x (5)

From (4) and (5),
x2+(8x)2+6x+8(8x)92=0
x2+6416x+x2+6x+648x92=0
2x218x+36=0
x29x+18=0
(x6)(x3)=0
x=3,6
From (5), y=5,2

z2=3+5i(3,5) and z1=6+2i(6,2)


Clearly, quadrilaterals OABC and ODEA are the parallelograms of same area.
Area of OAC is 12|6532|=12
Area of parallelogram OABC is 2×12=24

Total area =2× area(OABC)=48 sq. units

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