Complex Rectangle
Let
z
=
x
+
i
y
be a complex number where
x
and
y
are integers. Find the area of the rectangle whose vertices are the roots of the equation
z
z
ˉ
3
+
z
ˉ
z
3
=
3
5
0
.
Details: z ˉ = x − i y and i = − 1 .
The answer is 48.
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2 solutions
I think you have missed the point that the vertices of the rectangle are the roots of the equation.
Actually with ( x^2 + y^2 ) = 25 and ( x^2 - y^2 ) = 7
x^2 = 16 and y^2 = 9
then we get the solutions as x = + /- 4 and y = +/- 3
So the coordinates of the vertices are
( 4 , 3 ) , ( 4 , -3 ) , ( -4 , 3 ) and ( - 4 , - 3 )
so with that vertices we get the length and the width of rectangle
We have to draw the the rectangle in the coordinate plane by marking the 4 coordinate pairs . Then we can identify the length and width.
Then after calculation , we find that length = 8 and width = 6
Isn't there any way of doing it using algebra without substituting z = x + i y ?
I thought of starting with z z ˉ = ∣ z ∣ 2
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There must be various ways of doing this problem. I have posted my approach. I have solved it in another way too. I will post that solution after sometime. You can also try to solve this problem in different ways and post the solution here.
how is it possible to get two unknowns through a single equation? we cannot just guess . there must be another relation between x and y which can be most probably obtained by knowing that the vertices form a rectangle.
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The key is that x and y are integers. We know x 2 + y 2 ≥ x 2 − y 2 , which reduces the possibilities. By factoring 175 we reduce our options too. The only way that works is listed in the solution.
In the question z = x+iy, not x+iy(x^2+y^2)(x^2-y^2).
from where did you get this? z=x+iy(x^2+y^2)(x^2-y^2)= 175 and = to 5 x5x7
Is there an algebraic method to solve it?I mean without substituting z = x + i y ? @Sandeep Bhardwaj
Why to multiplicate area by 2
I think that who puts this problem forgets to coordinate the place of rectangle in the plane
Simplifying the equation, we get x 4 − y 4 = 1 7 5 . The only integer pair satisfying this equation is (x,y) = (4,3).
Area of the rectangle = 2 x ∗ 2 y = 4 x y = 4 8 .
Actually, you forgot to include the permutations of the signed integer pairs. there were no restrictions on the signs of x and y.
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I skipped that step... That was how I could deduce that the sides of the rectangle are 2x and 2y
Put z = x + i y ( x 2 + y 2 ) ( x 2 − y 2 ) = 1 7 5 = 5 × 5 × 7 Therefore ( x 2 + y 2 ) = 2 5 and ( x 2 − y 2 ) = 7 therefore x = 4 , y = 3 , A r e a = 8 × 6 = 4 8