How does a complex function plots a given circle line? If I have a complex function $w = \frac{1}{z}$ and I have to show how does it display a circle line: $x^2+y^2 +2x-4y+1 = 0$. I am not sure how to do it. Here is my try:
$z = x+yi \Rightarrow w = \frac{1}{x + yi}$ Then $u(x;y) = \frac{1}{x}$ and $v(x;y) = \frac{1}{y}$. Then I express x from the circle line: $x(x+2) = y^2 -4y +1 \Rightarrow x_1 = y^2 -4y +1 $ and $x_2 = y^2 -4y -1$. From there I get a system: $$\begin{matrix} u = y^2-4y +1 \\ v = \frac{1}{y}   \end{matrix} \Rightarrow y = \frac{1}{v} \Rightarrow u = \frac{v^2}{v-1}$$. I know that I am doing something completely wrong, because $u$ is the real part of the function and not a circle line, so I don't know how to show how would this function display the circle line indicated above. Any help would be appreciated!

I'll find the (complex) locus of $x^2+y^2+2x-4y+1=0$.
Notice we can complete the square:
$$(x+1)^2+(y-2)^2=4$$
Now this is a circle with radius 2 centered at $(-1,2)$. We can represent this as
$$|z-(-1+2i)|=2$$
$$|z+1-2i|=2$$
EDIT: To find the locus of $w=\frac{1}{z}$:
$$z=\frac{1}{w}$$
$$|\frac{1}{w}+1-2i|=2$$
$$|1+w-2wi|=2|w|$$
Let $w=u+vi$
$$|(1+u+2v)-(2u-v)i|=2|u+vi|$$
$$(1+u+2v)^2+(2u-v)^2=4(u^2+v^2)$$
$$1+u^2+4v^2+2u+4v+4uv+4u^2-4uv+v^2=4u^2+4v^2$$
$$1+u^2+2u+4v+v^2=0$$
$$(u+1)^2+(v+2)^2=4$$
Therefore, the locus of $w$ is a circle around $(-1,-2)$ with radius $2$. In other words, $|w+1+2i|=2$.
