# 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 $t_0=x$?

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Hint: consider $|f|$. – Antonio Vargas May 4 '12 at 6:29
I dont get, sorry? – Un Chien Andalou May 4 '12 at 6:30
$|f|$ is a continuous real-valued function from $[0,1]$ to an interval containing $[0,2]$ - invoke IVT on it. (You can't simply say $f(x)=1$ in the complex plane because the image of $f$ could trace out a path around the point $1$ in $\Bbb C$; considering drawing examples to help your intuition.) – anon May 4 '12 at 6:32
You need to find a point $t_0$ such that $|f(t_0)| = 1$. – copper.hat May 4 '12 at 6:38
Shouldn't this be tagged as complex analysis instead of real analysis ? – Belgi May 4 '12 at 13:14

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 $g(t)=1$...
You can't have Intermediate value theorem in Complex analysis because inequalities like $a+bi < c+di$ does not apply. – user9413 May 4 '12 at 13:56