Question: Let $T=\{z\in \mathbb{C}:|z|=1\}$ and $f:[0,1] \to \mathbb{C}$ be continuous with $f(0)=0$, $f(1)=2$. Show that there exists at least one $t_0$ in $[0,1]$ such that $f(t_0)$ is in $T$.

Attempt: We write $f(x)=u(x)+i v(x)$, $ \ i= \sqrt{-1}$.

Both of $u(x)$ and $v(x)$ must be continuous on $[0,1]$. Therefore, $g(x)=|f(x)|= \sqrt{(u(x))^2+(v(x))^2} $ must also be continuous on $[0,1]$. Moreover, $g(x)$ is a real valued function.

Now, $g(0)=0$ and $g(1)=2$. Then, the Intermediate Value Theorem (for continuous functions) guarantees that $g(0)< g(c) = 1<g(1)$ for some $c \in [0,1]$. That is, $g(c)=|f(c)|=1 \in T$. We are done.

Is this correct?

Thank You!

  • 1
    $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – copper.hat Apr 22 at 17:25
  • $\begingroup$ Yes, this is correct. $\endgroup$ – avs Apr 22 at 17:42

Very well done.

I would have proved it more or less the same way; I notice you went out of your way to prove that $\vert f(t) \vert$ is continuous. I would simply have observed that the norm function $\vert \cdot \vert: \Bbb C \to \Bbb R$ is continuous, and so $\vert f(t) \vert$, being the composition of the continuous functions $f(t)$ and $\vert \cdot \vert$, is continuous.


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