Graph $\text{Im}\left(\frac{1}{z}\right)=1$ I used the identity $z=x+iy$ which resulted in $\text{Im}\left(\frac{1}{x+iy}\right)$. Multiplying by the conjugate, I found that this was equal to $\text{Im}\left(\frac{x-iy}{x^2+y^2}\right)$, which by splitting this fraction into two terms is $\frac{-y}{x^2+y^2}=1$. Multiplying I found $-y=x^2+y^2$ which I think is the equation of a circle. Can somebody tell how to graph this equation in the complex plane?

Since $\operatorname{Im}z = \frac{1}{2i}(z-\bar z)$ the relation can be rewritten with equivalences at each step:
$$
\frac{1}{z}-\frac{1}{\bar z} = 2i \\
\bar z - z = 2i z \bar z \\
z \bar z-\frac{i}{2}z + \frac{i}{2}\bar z = 0\\
\left(z+\frac{i}{2}\right)\left(\bar z - \frac{i}{2}\right) = \frac{1}{4} \\
\left|z+\frac{i}{2}\right|^2 = \frac{1}{4}
$$
The latter is obviously the equation of a circle of radius $\cfrac{1}{2}$ centered at $\cfrac{-i}{2}$.
