Solving $\frac{df}{dt}=\frac{i\cdot f}{|f|}$ where $f: \mathbb{R^+} \mapsto \mathbb{C}$ I've never seen a complex DE before, so this is uncharted territory for me. But it's separable so it's easy to turn it into an integral:
$$f(t) = \int_0^t\frac{i \cdot f}{|f|} dt$$ Can this be solved?
Oh and the boundary condition is $f(0) = \frac{1}{\sqrt{2}}$
 A: Write
$$f(t)=r(t) e^{i\phi(t)}$$
with unknown real-valued functions $t\mapsto r(t)>0$ and $t\mapsto \phi(t)$. Then you have
$$f'(t)=\bigl(r'(t)+ i r(t)\phi'(t)\bigr)e^{i\phi(t)}\ .$$
Plugging this into the given differential equation leads to the condition
$$r'(t)+ i r(t)\phi'(t)=i\ .$$
Now separate real and imaginary parts.
A: Hint.
Your differential equation (that I denote (E)), is in a vectorial space, namely in $\mathbb C$.
So if $f(t)= u(t) + i v(t)$ is a solution, with $u,v$ real functions, they must hold following equations:
$$u^\prime(t)= -\frac{v(t)}{\sqrt{u^2(t)+v^2(t)}} \tag 1$$ and $$v^\prime(t)= \frac{u(t)}{\sqrt{u^2(t)+v^2(t)}} \tag 2$$ you get those equations by going to the real and imaginary coordinates.
From there you get (providing the right hypothesis...): $$\frac{u^\prime(t)}{v^\prime(t)} = - \frac{v(t)}{u(t)} \tag{1'}$$ and $$(u^\prime(t))^2 + (v^\prime(t))^2 = 1 \tag{2'}$$
(1') implies that $(u(t))^2 + (v(t))^2$ is constant and equal to $\vert f(0) \vert^2$.
You can probably move forward from there.
