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Let z=x+i y be a complex number where xa...

Let `z=x+i y` be a complex number where `xa n dy` are integers. Then, the area of the rectangle whose vertices are the roots of the equation `z (\bar {z })^3+ \bar{z} z^3=350` is 48 (b) 32 (c) 40 (d) 80

A

48

B

32

C

40

D

80

Text Solution

Verified by Experts

The correct Answer is:
A
Since `zbarz(z^2+barz^2)=350`
`rArr 2(x^2+y^2)(x^2-y^2)=350`
`rArr (x^2+y^2)(x^2-y^2)=175` since `x,y ne I,` the only possible case which gives intergal solution is
`x^2+y^2=25`
`x^2-y^2=7`
From Eqs.(i) and (ii)
`x^2=16 y^2=9 rArr x=pm4,y= pm 3`
`therefore ` Area of rectangle `=8xx6=48`
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