continuous function from $[0,1]$ to $\mathbb{C}$ Let $T=\{z\in \mathbb{C} :|z|=1\}$ and $f:[0,1]\rightarrow \mathbb{C}$ be continuous with $f(0)=0$ and $f(1)=2$.  I need to show that there exists at least one $t_0\in [0,1]$ such that $f(t_0)\in T$.  Do I need to apply the Intermediate Value Theorem?  I guess there will be some point $x\in [0,1]$ for which $f(x)=(1,0)\in \text{ i.e } 1\in T$ choose $t_0=x$?
 A: The intermediate value theorem is applicable in real calculus but does not hold for complex numbers. We no longer have only intervals in the complex plane - we have paths - and unlike in one dimension there are many different paths between two complex numbers.
Indeed, in our case a function $f:[0,1]\to\Bbb C$ could trace out a path from $0$ to $2$ that goes around the point $w=1$, so it is not necessarily the case that there is a $t$ for which $f(t)=1$. Geometrically, we can describe the set $\{|f|=1:f\in\Bbb C\}$ as the unit circle. The point $0$ is inside the circle and the point $2$ is outside, so what we're looking to prove, intuitively, is that any path from the inside to the outside of the unit circle must intersect the unit circle somewhere.
The trick to doing this is to reduce the situation down to real variables where we can invoke IVT. To do this, consider the function $g(t)=|f(t)|$ that maps $[0,1]\to[0,\infty)$. It is a real function such that we have the endpoints $g(0)=0$ and $g(1)=2$, the absolute value function is continuous and the composition of continuous maps is continuous, so $g$ is continuous, and finally we want to say there's a $t$ for which $g(t)=1$...
