$f(z)=z+{1\over z},z\in\mathbb C,z\neq 0$,then which of the following is true?
$1.f$ is analytic on $\mathbb C\backslash \{0\}.$
$2.f$ is conformal on $\mathbb C\backslash \{0\}.$.
$3.f$ maps the unit circle to a subset of the real axis.
$4.$ The image of any circle in $\mathbb C\backslash \{0\}$ is again a circle.
$1.$ is easily seen to be true. I think $2.$ is also true because derivative of $f$ over $\mathbb C\backslash \{0\}.$ is non-zero and $f$ is analytic in that region.
$3.$ is true. Because the points of the unit circle can be written as $z=(\cos \theta,\sin \theta)$ Then ${1\over z}=(\cos\theta,-i\sin\theta).$ Hence $$f(z)=z+{1\over z}\\=(\cos \theta,\sin \theta)+(\cos\theta,-i\sin\theta)\\=(\cos\theta,0).$$ So this is a subsetof the real line.I believe it is the interval $[-1,1]$ of $\mathbb R.$
$4.$ is false.Because, in similar method as $3.$,we can show that any circle under this map goes to a subset of the real line and not to any cirle.
Are my answers and corresponding methods correct$?$ Thank you.
Best Answer
Yes all the arguments are correct except $2$.
Consider two unit circles intersecting at some point in $\Bbb C$ with an angle different from $0,\pi$.
Apply $f$ on the both the circles .Since $f$ maps each circle to $\Bbb R$ so angle between them will be either $0$ or $\pi$.Hence the mapping is not conformal
For $3$ .
Hence $f(z)=z+\dfrac{1}{z}=z+\dfrac{\overline z}{|z|^2}=z+\overline z=2\text {Re} =2x$ if $z=x+iy$
Since $|z|=1$ so $x^2+y^2=1\implies x\in [-1,1]$
So $f(z)\in [-2,2]$