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let z=x+iy be the complex number where x and y are integers .then what is the area of rectangle whose vertices are the roots of the equation Z(z)^3+z(Z)^3=350 (where Z represents the conjugate of z)

let z=x+iy be the complex number where x and y are integers .then what is the area of rectangle whose vertices are the roots of the equation Z(z)^3+z(Z)^3=350 (where Z represents the conjugate of z)

Grade:12

5 Answers

Ashwin Sinha
520 Points
13 years ago

Dear Pratish,

Let Z= Complex no.=x + iy.

Let Z= Z bar=x-iy.

 Now,    ZZ3 + ZZ3  = 352

         » ZZ.Z2   +  ZZ.Z2  = 350

         » ZZ (Z2   +  Z2 )  = 350

         » (x+iy).(x-iy) [ (x-iy)2 + (x+iy)2 ] =350

         » (x2 + y2).(2x2 - 2y2) =350

         » (x2 + y2).(x2 - y2) =175

         » 25.7=175

         » x2 + y2 = 25  &  x2 - y2 = 7 

         »x2=16 & y2=9

         »x=+4 or -4     &      y=+3 or-3

         »So 4 pts of rectangle are  (4,3) , (4,-3) , (-4,3) , (-4,-3).

         »Hence, length of rectangle=8 units       &      breadth of rectangle=6 units.

         »So, area of rectangle=length . breadth =8.6=48 sq. units.

 

                       Hope you understood it well!!!!!!!

                       Plz click Yes below!!!!!!

                                                Good Luck!!!!!!!

Yash Sinha
32 Points
13 years ago

DEAR PRATISH!

Your ANSWER IS


let z = x+iy

and

    z = x-iy

therefore,

→ zz3 + zz3 = 350

or,zz(z2 + z2) = 30

or,|z|2 [(x+iy)2+(x-iy)2] = 350                (if z=x+iy)

or,(x2+y2)[x2-y2] = 350/2 = 175 = 25*7   (by analogy)

or,(x2+y2)=25 & (x2-y2)=7

        or, x=+4 or -4     &      y=+3 or-3

or,So 4 pts of rectangle are  (4,3) , (4,-3) , (-4,3) , (-4,-3).

         or,Hence, length of rectangle=8 units       &      breadth of rectangle=6 units.

         or,So, area of rectangle=length . breadth =8.6=48 sq. units. ANSWER

Hope you understood it fully!

rohan sharma
15 Points
13 years ago

coordinates are (4,3)      (-4,3)     (-4,-3)        

(4,-3)        area is    48 units

aku -- kumar
38 Points
13 years ago

dear student,

 

Let Z= Complex no.=x + iy.
Let Z= Z bar=x-iy.
 Now,    ZZ3 + ZZ3  = 352
         » ZZ.Z2   +  ZZ.Z2  = 350
         » ZZ (Z2   +  Z2 )  = 350
         » (x+iy).(x-iy) [ (x-iy)2 + (x+iy)2 ] =350
         » (x2 + y2).(2x2 - 2y2) =350
         » (x2 + y2).(x2 - y2) =175          » 25.7=175
         » x2 + y2 = 25  &  x2 - y2 = 7 
         »x2=16 & y2=9
         »x=+4 or -4     &      y=+3 or-3
         »So 4 pts of rectangle are  (4,3) , (4,-3) , (-4,3) , (-4,-3).
         »Hence, length of rectangle=8 units       &      breadth of rectangle=6 units.
         »So, area of rectangle=length . breadth =8.6=48 sq. units.

hope you're now clear with the problem now, and don't ever try to challenge the faculty, you should know to respect them

Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
Let Z= Complex no.=x + iy.
Let Z= Z bar=x-iy.
Now,    ZZ3 + ZZ3  = 352
=> ZZ.Z2   +  ZZ.Z2  = 350
=> ZZ (Z2   +  Z2 )  = 350
=> (x+iy).(x-iy) [ (x-iy)2 + (x+iy)2 ] =350
=> (x2 + y2).(2x2 - 2y2) =350
=> (x2 + y2).(x2 - y2) =175
=> x2 + y2 = 25  &  x2 - y2 = 7     [since, 175 = 25 . 7]
=> x2=16 & y2=9
=> x = +4 or –4     &      y = +3 or – 3So 4 pts of rectangle are  (4,3) , (4,-3) , (-4,3) , (-4,-3).
Hence, length of rectangle = 8 units       &      breadth of rectangle = 6 units.
So, area of rectangle = length . breadth = 8.6 = 48 sq. units.
 
Thanks and regards,
Kushagra

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