How do we evaluate $(-i)^{\frac 5 2}$ without complex logarithms? $$(-i)^{\frac 5 2} = (e^{-\frac{\pi i}{2}})^{\frac 5 2} = e^{-\frac{5\pi i}{4}}$$
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.
 A: 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.
A: We can evaluate complex square roots purely by algebraic means.  Herein I describe the procedure for a general case.  The reader should be able to Express a quantity such as $(-i)^5$ with no radicals in the form $a+bi$ readily.
We are to solve
$(x+yi)^2=a+bi$
for real numbers $x,y$ given real numbers $a,b$.  First expand the square on the left side:
$(x^2-y^2)+(2xy)i=a+bi$
We now take realbparts and imaginary parts, and also the absolute value of the equation remembering that $(x^2-y^2)+2+(2xy)^2=(x^2+y^2)^2$:
$x^2-y^2=a\text{.....Eq. 1}$
$2xy=b\text{.....Eq. 2}$
$x^2+y^2=\sqrt{a^2+b^2}\text{.....Eq. 3}$
Then by taking half the sum of Eqs. 1 and 3 we get a solution for $x^2$:
$x^2=\dfrac{\sqrt{a^2+b^2}+a}{2}\text{.....Eq. 4}$
Half the difference between Eqs. 1 and 3 gives $y ^2$:
$y^2=\dfrac{\sqrt{a^2+b^2}-a}{2}\text{.....Eq. 5}$
This is almost the answer.  Because we may potentially select signs independently for $x$ and $y$ we are left with four candidates, this being too many, when $b$ is nonzero.  To handle that use Eq. 2 for the relevant sign information:
$\text{x and y have the same sign when b is positive}$
$\text{x and y have opposite signs when b is negative}$
$\text{There is no sign issue when b is zero because then Eqs. 4 and 5 give no extraneous roots}$
A: This matter is of definitions. Likely, the class defines roots to be multi-valued. Therefore, asking to evaluate a complex square root root is equivalent to solving a complex quadratic.
