Some Equations

Given:

${ x }^{ 2 }+{ y }^{ 2 }=1$ and $x+y=i$

Start with the expansion for the square of a binomial:

${ (x+y) }^{ 2 }={ x }^{ 2 }+2xy+{ y }^{ 2 }$

which can be rewritten as

${ (x+y) }^{ 2 }={ (x }^{ 2 }+{ y }^{ 2 })+2xy$

substitute the given:

$i^{ 2 }=1+2xy$

Solve:

$-1=1+2xy\\ -2=2xy\\ xy=-1$

Since we have $x^{2} + y^{2} = 1$ , then we can add 2xy in the both sides: $x^{2} +2xy + y^{2} = 1 +2xy$ ; $(x + y)^{2} = 1 + 2xy$ ; $2xy = (x + y)^{2} -1$ ; $xy = ((x + y)^{2} - 1)/2$ ;

Now since we have x + y = i, then we can do $(x + y)^{2} = i^{2}$ So: $xy = (i^{2} - 1)/2$ ; We know that $i^{2} = -1$ . Then: xy = -1

1. x=i-y
2. x²+y²=1
3. Putting value of x
4. (i-y)²+y²=1
5. i²-2iy+y²+y²=1
6. -1-2iy+2y²=1
7. 2y²-2iy=2
8. y²-iy=1
9. y(y-i)=1
10. y(i-y)=-1
11. But i-y=x
12. Therefore xy=-1

$\color{#20A900}{(x+y)^2=i^2}$ We know that $i^2=-1$ .So the above equation becomes: $\color{#3D99F6}{x^2+2xy+y^2=(x^2+y^2)+2xy=-1}$ We know that $x^2+y^2=1$ .Substituting this into the above equation becomes: $\color{#BA33D6}{1+2xy=-1\rightarrow2xy=-1-1=-2\rightarrow xy=\frac{-2}{2}=\boxed{-1}}$

nice solution:

\left\{ { x }^{ 2 }+{ y }^{ 2 }=1\quad .........(A)\\ x+y=i\quad ............(B) \right \\ From\quad (A)\\ { x }^{ 2 }+{ y }^{ 2 }=1\quad or\\ { x }^{ 2 }+{ y }^{ 2 }+2xy-2xy=1\quad or\\ { (x+y) }^{ 2 }-2xy=1........\quad (C)\quad \quad \quad \quad As\quad (x+y)=i\\ So,\quad (C)\quad Become\\ { i }^{ 2 }-2xy=1\quad or\quad \quad As\quad { i }^{ 2 }=-1\\ -1-2xy=1\quad or\\ -2=2xy\quad or\\ \boxed { xy=-1 } \\

From $x+y=i$ $\Longrightarrow$ ${ (x+y) }^{ 2 }={ i }^{ 2 }$ $\Longrightarrow$ ${ x }^{ 2 }{ + }{ y }^{ 2 }+2xy={ -1 }$ $\longrightarrow$ (1)

$\because$ ${ x }^{ 2 }+{ y }^{ 2 }=1$

$\therefore$ From (1), $1+2xy=-1$ $\Longrightarrow$ $2xy=-2$ $\Longrightarrow$ $xy=-1$

Answer: $xy$ = $\boxed { -1 }$