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).

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.
