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?

You're almost there.  We have
$$x^2+y^2=-y\implies x^2+(y+1/2)^2=\frac14$$
In the complex plane, $x=\text{Re}(z)$ and $y=\text{Im}(z)$.  Therefore, this is a circle with center $(0,-1/2)$ and radius $1/2$
