# If $f:[0,1]\to \mathbb{C}$ be continuous with $f(0)=0$ and $f(1)=2$, then $|f(t_0)|=1$ for some $t_0 \in [0,1]$

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 $$g(0)< g(c) = 1 for some $$c \in [0,1]$$. That is, $$g(c)=|f(c)|=1 \in T$$. We are done.

Is this correct?

Thank You!

• Yes. ${}{}{}{}{}$ – copper.hat Apr 22 at 17:25
• Yes, this is correct. – 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.