I have to draw the image of the complex map $z \mapsto 1/z$ with domain $\{z : \text{Re } z - \text{Im } z = 1\}$
So, for $z = (y+1)+iy$, the function takes $z$ to $\frac{1}{(y+1)+iy}=\frac{(y+1)-iy}{|z|^2}$
So, clearly the image set is $\{z:\text{Re } z + \text{Im } z = 1/|z|^2\}$; and therefore I have to sketch $x+y = 1/(x^2 + y^2)$. While I was able to obtain the curve on Wolfram Alpha, I'm trying to understand how I can sketch this myself. Can I get any help?
Edit (maybe a solution): Let $f$ be $f:z \mapsto1/z$, and let $w = u + iv = f(z) \implies z = 1/w$ Letting $z = x+iy$, I get $$x + iy = \frac{u}{u^2 + v^2}+i\frac{-v}{u^2 + v^2}$$ Since we know $x-y=1$ $$\implies \frac{u}{u^2 + v^2}-\frac{-v}{u^2 + v^2}=1 \implies u+v = u^2 + v^2 \\ \implies (u-1/2)^2 + (v-1/2)^2 = 1/2$$
Therefore, the image is $$\{w:|w-\frac{1}{2}(1+i)|=1/2\}$$
Is this correct? My first attempt was clearly incorrect
If $z=x+yi$ and $x-y=1$, then $z=y+1+yi$ and$$\frac1z=\frac x{x^2+y^2}-\frac y{x^2+y^2}i=\frac{y+1}{(y+1)^2+y^2}-\frac y{(y+1)^2+y^2}i.$$The set of all these numbers is the circle centered at $\frac12+\frac12i$ with radius $\frac1{\sqrt2}$ (except for the point $0$).