Find the real value of x,y if (1/(x^2+y^2))+(1/(x+yi))=1

Question

asked Feb 11, 2019 153k views

1 Answer

Mlhazan · Answer 1 · 2019-02-15T13:46:11+0000

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

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