Question

Find the real value of x,y if

(1/(x^2+y^2))+(1/(x+yi))=1

Answer 1

It sounds like
`x,y` are supposed to be real numbers. If so, then we can do the following.

`\frac1{x^2+y^2}+\frac1{x+yi}=1`

Multiply the second term's numerator and denominator by the conjugate of the denominator:

`\frac1{x^2+y^2}+(x-yi)/((x+yi)(x-yi))=1`

`\frac1{x^2+y^2}+(x-yi)/(x^2+y^2)=1`

`(x+1-yi)/(x^2+y^2)=1`

Since the left hand side is equal to 1, this means it has no imaginary part, so that
`y=0`. Then the real parts of both sides of the equation give us

`(x+1)/(x^2)=1\implies x+1=x^2\implies x^2-x-1=0\implies x=\frac{1\pm\sqrt5}2`