Intuition why the function $f(z) = \frac{1}{z}$ maps lines to circles For the function $f(z) := \frac{1}{z}$, is there an intution why it maps complex numbers $z$ to points on a circle?
I did this question where $z$ varied on a line $x=1$, giving $f(z) = w$ with $|w-\frac{1}{2}| = \frac{1}{2}$.
The proof for the above came from just considering the LHS... but surely if I were to just find $w$ algebraically, I would not have realised that $w$ would be a pointon a circle. This is why if a question asked me: "What circle would be the image of $f$ is $z$ varied on $x=30$?" I would have a bit of trouble.
 A: Let $z=re^{i\theta} $ so that $\frac 1z = \frac1r e^{-i\theta}$
The equation of any line parallel to the y-axis is given by $r=x_0 \sec\theta$
So 
$$ \begin{eqnarray*}  
\frac 1z &=& \frac 1{x_0}\cos\theta  e^{-i\theta}
\\   &=& \frac 1{2x_0}( e^{i\theta}+e^{-i\theta}  ) e^{-i\theta}
\\   &=& \frac 1{2x_0}( 1+e^{-2i\theta}  )
\\   &=& \frac 1{2x_0} +\frac 1{2x_0}e^{-2i\theta}  
\end{eqnarray*} $$
Which is clearly a circle (traversed twice) with centre at $\frac 1{2x_0}$ and radius $\frac 1{2x_0}$
This can be generalized to yield circles for any line of the form $y=mx+b$ unless $b=0$ in which case the transformation maps lines onto lines and is equivalent to a reflection in the $x$ axis  
A: Let $L=\{z \in  \mathbb C: Re(z)=1\}$. If $z \in L$, then $z=1+it$ for some $t \in \mathbb R$. Hence
$f(z)=\frac{1}{1+it}$. Therefore
$|f(z)- \frac{1}{2}|=|\frac{1}{1+it}- \frac{1}{2}|=|\frac{-1-it}{2(1+it)}|=\frac{1}{2}.$
A: The line $x=30$ in the complex plane (that can be thought of as a circle passing by $30$ and infinity and symmetrical wrt to the real axis) is mapped by $1/z$ into a circle passing by $1/30$ and $0$ with the symmetry preserved.
A: First of all mirroring doesn't alter this so you can instead consider the map $\overline f(z) = 1/\overline z$ which is inversion in the unit circle. This means that we can interpret it (the image of a point) geometrically (as a point on the line through origin that is at the distance $1/|z|$ from the point).
Now it's a two steps, the first is to establish the image and then to see that it has the required property. I assume that you're familliar with how to construct the image(*) so I skip right to showing how it has the required property. 
You have a circle around $M$ that passes through $O$ (the origin in complex plane), you also have a line $L$. We also have that the (extension of) the diameter crosses $L$ under a right angle (that's part of the skipped step), call this point $X$ 
Now pick a point $A$ on the line and consider the line through $OA$ and let $B$ be the other intersection with the circle. Let $C$ be the midpoint of the line segment $OB$.

Now consider the right triangles $OCM$ (right because it's congruet to $BCM$) and $OXA$, these are similar because they are right and share the angle at $O$. This means that the $OC$ is to $OM$ as $OX$ is to $OA$. 
Algebraically this means that
$${OB \over OM} = {2\cdot OC \over OM} = {2\cdot OX\over OA}$$
or 
$$OB = 2\cdot OM\cdot OX{1\over OA}$$
But from the skipped step we know that $2\cdot OM\cdot OX$ is the square of the radius of the unit cirle.
(*) You actually have that construct on wikipedia.
A: Lines and circles are objects given by equations of the form $az\bar z + b z + \bar b \bar z + c = 0$ where $b$ is a complex number and $a$ and $d$ are real numbers, and where $a,b,c$ are not all simultaneously zero.
If $z$ is nonzero, then letting $w= 1/z$, you get $(az\bar z + b z + \bar b \bar z + c)w\bar w = a + b\bar w + \bar b w +dw\bar w$, which is an equation on $w$ of the same form.
So inversion turns lines and circles into lines and circles.
In particular, the lines are the objects you obtain when $a=0$, so they are transformed into the objects you obtain with $d=0$, so lines or circles that go through $0$. So in a sense, the lines are the objects that go through $1/0$, or "the point at infinity".
