Graph $\text{Im}\left(\frac{1}{z}\right)=1$

Question

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?

Answer 1

$$x^2+y^2=-y\iff x^2+\left(y+\dfrac12\right)^2=\left(\dfrac12\right)^2$$ which is a circle with centre $\left(0,-\dfrac12\right)$ and radius $=\dfrac12$ unit

Answer 2

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 $1/2$

Answer 3

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 $\cfrac{-i}{2}$.