Results of dot product for complex functions Suppose we are given a $C^1$ function $f(t):\mathbb{R} \rightarrow \mathbb{C}$ with $f(0) = 1$, $\|f(t)\| = 1$ and $\|f'(t)\| = 1$. I have already proven that $\langle f(t), f'(t)\rangle = 0$ for all $t$. Now I have to show that either $f'(t) = if(t)$ or $f'(t) = -i f(t)$. How do I go about showing this? 
(I am terribly sorry for the horrible title, I could not think of a good one).
 A: Presumably by $\langle f(t) , f'(t) \rangle = 0$, you mean that $\text{Re} f(t) \overline{f'(t)} = 0$ (if $z_1,z_2 \in \mathbb{C}$ and $z_1 \overline{z_2} = 0$, then you must have either $z_1 = 0$ or $z_2 = 0$).
If $\text{Re} f(t) \overline{f'(t)} = 0$, then $f(t) \overline{f'(t)} = i \zeta(t)$. where $\zeta$ is real valued. $\zeta$ is continuous, and furthermore, $|f(t) \overline{f'(t)}| = 1 \ =|\zeta(t)|$. Consequently, $\zeta$ is either the constant $1$ or $-1$. Multiplying $f(t) \overline{f'(t)} = i \zeta(t)$ on both sides by $f'(t)$ gives $f(t) = i \zeta(t) f'(t)$, from which the result follows.
A: For every $t$ we have
$$
0=\frac{d}{dt}|f(t)|^2=f(t)\overline{f'(t)}+\overline{f(t)}f'(t)=2\Re(f(t)\overline{f'(t)})=2\langle f(t),f'(t)\rangle,
$$
i.e. $f(t) \perp f'(t)$ for every $t$. Notice that the map 
$$
\mathbb{C}\mapsto \mathbb{C},\ z \mapsto iz
$$
corresponds to the rotation around the origin with angle $\pi/2$.
Therefore for every $t$ there is a real number $\lambda(t)$ such that 
$$
f'(t)=i\lambda(t) f(t).
$$
Since $|f(t)|=1=|f'(t)|$ for every $t$, we have $|\lambda(t)|=1$, i.e. $\lambda(t)=\pm 1$. The function 
$$
\mathbb{R} \to \mathbb{R},\ t \mapsto \frac{f'(t)}{if(t)}=\lambda(t)
$$
is continuous and so we have either $\lambda(t)=-1$ for every $t$, or $\lambda(t)=+1$ for every $t$.
