Algebraically show that a circle that contains a point at the origin is a line when mapped with $f(z)=1/z$ Is there a way to algebraically show that a circle that contains a point at the origin is a line when mapped with $f(z)=1/z$? I tried to work through an example, such as $|z+2i|=2$, I still got a circle. Is this more of an inspection thing or...?
$$|z+2i|=2$$
Since $w=f(z)=\frac{1}{z}$ we know that $\frac{1}{w}$.
$$|\frac{1}{w}+2i|=2$$
$$\frac{|1+2wi|}{|w|}=4$$
$$|1+2wi|=4|w|$$ 
Then, let $w=u+vi$
$$(1-2v)^2+(2u)^2=4u^2+4v^2$$
Then we get
$$4v^2-4v+1+4u^2=4u^2+4v^2$$
$$-4v+1=0$$
which gives a line. Thank you all :)
 A: Let's begin by studying $|z+2i|=2$.  This describes a circle centered at $-2i$ of radius $2$.  If we were to put this into $(x,y)$ coordinates, we get $x^2+(y+2)^2=4$.
Observe that $\frac{1}{z}$ is its own inverse, so the image of the circle under the map $\frac{1}{z}$ is the same as the solutions to
$$
\left|\frac{1}{z}+2i\right|=2.
$$
Squaring both sides, we get
$$
\left(\frac{1}{z}+2i\right)\left(\frac{1}{\overline{z}}-2i\right)=4.
$$
Multiplying this out, one gets
$$
\frac{1}{z\overline{z}}+\frac{2i}{\overline{z}}-\frac{2i}{z}+4=4
$$
or
$$
\frac{1}{z\overline{z}}+\frac{2i}{\overline{z}}-\frac{2i}{z}=0
$$
Multiplying through by $|z|$, we get
$$
1+2zi-2\overline{z}i=0.
$$
Making the substitution $z=x+iy$, we get
$$
1-4y=0
$$
which is a line.
The error in the original question is that when the equation was squared, the $2|w|$ did not become $4|w|^2$, the $2$ was not squared.  When you make that change to the original question, you get $1-4v=0$ after simplifying, which agrees with this answer.
A: A circle with center at $a$ containing the origin is described by
$$
 \lvert z - a \rvert = \lvert  a \rvert \, .
$$
The substitution $z = 1/w$ gives
$$
 \lvert \frac 1w - a \rvert = \lvert  a \rvert \, .
$$
or
$$
  \lvert  a w \rvert^2 =  \lvert 1 - aw \rvert^2 = 
1 + \lvert  a w \rvert^2 - 2 \operatorname{Re}( a w) \\
\Longleftrightarrow \operatorname{Re}( a w) = \frac 12 \\
\Longleftrightarrow w = \frac{1}{ a}(\frac 12 + iy) \text{ for some } y \in \Bbb R
$$
which is a line.
A: You may want to try to look at polar coordinates. A circle that contains the Origin can be described as $r=acos\theta+bsin\theta$, where $a$ and $b$ are numbers (not both zero). I'll leave it up to you to verify that such circle has some $\theta$ for which it passes through the Origin. Now reciprocating gives a new polar form $r=\frac{1}{asin\theta+bcos\theta}$ which can be written as $a(rsin\theta)+b(rcos\theta)=1$ from which we get $ay+bx=1$, a line (not through the Origin). For more information, you may want to google Mobius Transformations, that also highlight other cases.
