$$(-i)^{\frac 5 2} = (e^{-\frac{\pi i}{2}})^{\frac 5 2} = e^{-\frac{5\pi i}{4}}$$
How do we justify the last step before going to $e^{\frac{3\pi i}{4}}$? I think the approach is supposed to be
$$(-i)^{\frac 5 2} = e^{\frac 5 2Ln(-i)} = e^{\frac 5 2(\ln(-i)+iArg(-i))} = e^{\frac 5 2(0+iArg(-i))} = e^{\frac 5 2(\frac{-\pi i}{2})} = e^{-\frac{5\pi i}{4}}$$
My brother is taking "complex methods" and is expected to evaluate $(-i)^{\frac 5 2}$ without knowing complex logarithm (So obviously, we can't define $a^b$ without knowing complex logarithm). There's probably something in his notes we have overlooked. I don't think complex exponential is enough. Is it? We're actually still waiting for confirmation from the professor that they can actually evaluate $e$ raised to complex numbers that are not purely real or purely imaginary. My brother says he doesn't remember any complex exponential. Still, even if you know complex exponential, without complex logarithm, I don't see how you can rigorously prove $e_\mathbb C^5=e_\mathbb R^5$.
By the way, $$(-i)^{\frac 5 2} \ne (-i)^{\frac 1 2}$$ right (Wolfram Alpha confirms)?
I am afraid that things like $$(e^{-\frac{\pi i}{2}})^{\frac 5 2} = e^{-\frac{5\pi i}{4}}$$
will lead students to think like
$$(-i)^{\frac 5 2} = ((-i)^5)^{\frac 1 2}$$
We might have a second look at his notes, but based on this question, $(-i)^{\frac 5 2}$ might be defined as multi-valued.
Best Answer
Use de Moivre's theorem and $-i=\cos\theta+i\sin\theta,\,\theta: to=-\pi /2$. Yes, there are multi-valued issues in extending the theorem to fractional powers, but it's all that can be done with pre-exponential methods.