Let z=x+iy be a complex number where x and y are intergers. The area of the rectangle whose vertices are the roots of the equation ¯¯¯zz3+z¯¯¯z3=350 is
A
48 sq. units
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B
32 sq. units
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C
40 sq. units
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D
80 sq. units
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Solution
The correct option is A48 sq. units Given equation is ¯¯¯zz3+z¯¯¯z3=350 ⇒z¯z(¯z2+z2)=350
Putting z=x+iy, we get ⇒(x+iy)(x−iy)[2x2−2y2]=350⇒(x2+y2)(x2−y2)=175
By observation, we write ⇒(x2+y2)(x2−y2)=25×7
Or, ⇒(x2+y2)(x2−y2)=175×1
Or, ⇒(x2+y2)(x2−y2)=35×5
We only get integral values of x and y, when x2+y2=25,x2−y2=7
Therefore, x=±4,y=±3
The length of rectangle =8
The width of rectangle =6