Question

(x - i )( y- xi ) + (1 + i )( x - iy ) = 3- 9i => xy=?

Answer 1

(x - i )( y- xi ) + (1 + i )( x - iy ) = 3- 9i
x*y-(x^2)*i-y*i+x*(i^2)+x-y*i+x*i-y*(i^2)=3-9*i
(i^2)=-1
x*y+y+(x-(x^2)-2y)*i=3-9*i
(1)  x*y+y=3
(2)  x-(x^2)-2y=-9
обозначим от 1 ур. : y=3/(x+1)
подставляем на 2 ур. и получаем это: -(x^3)+10*x+3=0
решаем 
ответы :x1=-3 , x2=1.5+sqrt(13)/2 , x3=1.5-sqrt(13)/2
нам нужно x*y=3*x/(x+1)
для 1 x :  (x*y)1=4.5
для 2 x :  (x*y)2=(9+3*sqrt(13))/(5+sqrt(13))
для 3 x :  (x*y)3=(9-3*sqrt(13))/(5-sqrt(13))
sqrt это корень
 

Answer 2

To solve this equation, we'll first expand the left-hand side and then equate the real and imaginary parts to the real and imaginary parts of the right-hand side.

Given equation: $(x - i)(y - xi) + (1 + i)(x - iy) = 3 - 9i$

Let's start by expanding the left-hand side:

$(x - i)(y - xi) + (1 + i)(x - iy) = xy - xyi - x^2i + xi^2 + x - iy + xi - i^2y$

Now, recall that $i^2 = -1$:

$(x - i)(y - xi) + (1 + i)(x - iy) = xy - xyi - x^2i + xi^2 + x - iy + xi + y$

Simplify further by combining like terms:

$xy + (x - y)i + (1 - x^2 + y) = 3 - 9i$

Now, we can equate the real and imaginary parts separately:

1. Equate the real parts: $xy + 1 - x^2 + y = 3$

2. Equate the imaginary parts: $x - y = -9$

Now, we have a system of equations. Let's solve it:

From equation (2): $x = y - 9$

Substitute this value of $x$ into equation (1):

$(y - 9)y + 1 - (y - 9)^2 + y = 3$

Expand and simplify:

$y^2 - 9y + 1 - (y^2 - 18y + 81) + y = 3$

$y^2 - 9y + 1 - y^2 + 18y - 81 + y = 3$

$10y - 80 = 3$

Now, isolate $y$ by moving the constant term to the other side:

$10y = 3 + 80$

$10y = 83$

$y = \frac{83}{10}$

Now that we have the value of $y$, we can find $x$ using the second equation:

$x = y - 9 = \frac{83}{10} - 9 = \frac{83}{10} - \frac{90}{10} = -\frac{7}{10}$

So, $x = -\frac{7}{10}$ and $y = \frac{83}{10}$.

Finally, to find $xy$:

$xy = \left(-\frac{7}{10}\right)\left(\frac{83}{10}\right) = -\frac{581}{100}$