Question

Solve the following equation for the real number

(1+i/1-i)^2+ 1/x+iy = 1+i

Answer 1

It looks like the equation says

`\left((1 + i)/(1 - i)\right)^2 + \frac1{x+iy} = 1+i`

Simplify the first term.

`(1 + i)/(1 - i) = (e^(i\pi/4))/(e^(-i\pi/4)) = e^(i\pi/2) = i`

Alternatively,

`(1+i)/(1-i) \cdot (1+i)/(1+i) = ((1+i)^2)/(1-i^2) = \frac{1+2i+i^2}2 = i`

Then

`i^2 + \frac1{x+iy} = 1+i`

`-1 + \frac1{x+iy} = 1+i`

`\frac1{x+iy} = 2+i`

`x+iy = \frac1{2+i}`

Simplify the right side.

`\frac1{2+i} \cdot (2-i)/(2-i) = (2-i)/(2^2-i^2) = \frac{2-i}5`

Then

`x+iy = \frac25 - i\frac15 \implies \boxed{x=\frac25 \text{ and } y = -\frac15}`