Find the image and preimage of some functions I have functions
\begin{align}
f: \mathbb{R} \rightarrow \mathbb{R};&\ \  f(x) = x^2-4x-1, \\
g:\mathbb{C} \rightarrow \mathbb{C};&\ \  g(x) = x^5
\end{align}
and the set $D=\{z \in \mathbb{C} | |z|<1\}$.
I need to find $f((1,\infty))$, $f^{-1}((1,\infty))$ and $g(D)$. My instructor taught me a method in which I need to intuitively find the image/preimage and then to show formally that my intuition is right by double-inclusion. For example I say that $f((1,\infty))=[-4,\infty)$, and then I would show this by double-inclusion. I find this method very confusing. Is there any other method? If not, could you please give me some indications? Thank you.
 A: I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:

$A=(1,0)$ give us the point $B=(1,f(1)=-4)$ and then is easy to see that $f((1,\infty))=[-5,\infty[$. For $f^{-1}((1,\infty))$ we have to see what are the values of $x$ such that $f(x) \in (1,\infty)$. The point $F$ represent the initial point. It is the point such that 
$f(x)=1 \Rightarrow x^2-4x-1=1 \Rightarrow x=2+\sqrt{6}$ and $x=2-\sqrt{6}$ 
That means if we take $x<2-\sqrt{6}$ or $x>2+\sqrt{6}$  we will get $f(x) \in (1,\infty)$.
$2)$ If $|z|<1$ write $z=r(\cos\alpha+i\sin\alpha)$, with $r<1$ and $\alpha \in [0,2\pi]$ and then $g(z)=z^5=r^5(\cos5\alpha+i\sin5\alpha))$ with, $r^5<1$ and $5\alpha \in [0,5\pi]$. 
A: I agree with your instructor. His approach  is reasonable here. Now a few hints.
1) Draw a graph of the parabola $f(x) = x^2-4x-1=(x-2)^2-5$. 
What are your candidates for
$f((1,\infty))$ and $f^{-1}((1,\infty))$?
2) As regards $g(z)=z^5$ try to show that $g(D)=D$. 
$g(D)\subseteq D$ because if $|z|<1$ then $|g(z)|=|z^5|=|z|^5<1$.
For the other inclusion, we have that
if $re^{it}\in D$ then $0\leq r<1$ and $t\in [0,2\pi)$.
Can you find $z\in D$ such that $z^5=re^{it}$?
