0
$\begingroup$

I am trying to find the image of $\{z \in \mathbb{C} : |z| < 1\}$ under $f(z) = \frac{1}{z}$

Let $$w = \frac{1}{z} \Rightarrow z = \frac{1}{w} = \frac{1}{u+iv} = \frac{u-iv}{u^2 + v^2}$$

if $w = u +iv$.

Then considering the boundary first, $$|z| =1 \Rightarrow \Big|\frac{u}{u^2+v^2} -i\frac{v}{u^2+v^2}\Big| =1 $$ Using Pythagoras gives $$\left(\frac{u}{u^2 +v^2}\right)^2 + \left(\frac{v}{u^2 +v^2}\right)^2 = 1$$ but this is not the equation of a unit circle because I have this factor of $(u^2 + v^2)^2$ lying about, which is what the answer should be...

What's happening?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Of course, it makes no sense to talk about the image of $0$ under $\frac1z$, but the image of $\{z\in\mathbb{C}\,|\,\lvert z\rvert<1\}\setminus\{0\}$ is $\{z\in\mathbb{C}\,|\,\lvert z\rvert>1\}$.

$\endgroup$
2
  • $\begingroup$ Why does my method not yield the unit circle as the boundary? $\endgroup$
    – PhysicsMathsLove
    Jan 22, 2019 at 14:11
  • 1
    $\begingroup$ \begin{align}\left(\frac{u}{u^2 +v^2}\right)^2 + \left(\frac{v}{u^2 +v^2}\right)^2 = 1&\iff\frac{u^2+v^2}{(u^2+v^2)^2}=1\\&\iff\frac1{u^2+v^2}=1\\&\iff u^2+v^2=1.\end{align} $\endgroup$
    – José Carlos Santos
    Jan 22, 2019 at 14:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .