# What are x and y? y/1+i = x^2 + 4/x+2i

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### 1 Answer

`y/(1+i)=x^2+4/(x+2i)`

`(y(1-i))/((1+i)(1-i))=x^2+(4(x-2i))/((x+2i)(x-2i))`

`(y(1-i))/2=x^2+(4(x-2i))/(x^2+4)`

`y/2-(y/2)i=x^2+(4x)/(x^2+4)-(8/(x^2+4))i`

Comparing real and imaginary parts

`y/2=x^2+(4x)/(x^2+4)`

`y/2=8/(x^2+4)`

Thus we have

`x^2+(4x)/(x^2+4)=8/(x^2+4)`

`x^4+4x^2+4x-8=0`

Solving above biquadratic equation, we have two real roots and other two roots are complex. Since x can not be complex so ignore them. Real roots are

`x approx.93`

`xapprox-1.50`

So `yapprox 3.29` when `x approx .93` and

`yapprox2.56` when `xapprox-1.50`