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$$(-i)^{\frac 5 2} = (e^{-\frac{\pi i}{2}})^{\frac 5 2} = e^{-\frac{5\pi i}{4}}$$

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.

  • In complex variables, $z^a$ is multi-valued if $a$ is not an integer. – Eclipse Sun Nov 13 '18 at 18:04
  • @EclipseSun Does this depend on the textbook? My textbook says $z^a$ is defined as the principal value given by $e^{a Ln|z|}$, although there are of course the non-principal values of $e^{a \ln|z|}$ So in some textbooks $z^a$ is actually a set, and it's the set of all $e^{a \ln|z|}$? –  Nov 13 '18 at 18:08
  • Yes and no. If you define using a principal value, you need to remove some points in the complex plane (for example non-positive real numbers). So it will not be the function we want. (We want to define it on $\mathbb{C}\setminus{0}$.) Moreover, each author has his/her favorite principal value. It is more reasonable to take it as a multi-valued function instead. – Eclipse Sun Nov 13 '18 at 18:15
  • Why is it that you are involved in your brother's homework assignment, and why is it that you want to discuss at this forum what the professor might have said? Clearly this is an elementary exercise and everything will be clarified soon enough. – M. Wind Nov 13 '18 at 18:27
  • @EclipseSun Hang on. Why aren't square roots of real numbers multi-valued? Would you say $4^{1/2}={2,-2}$? Why not say (something1)^(something2) is single-valued and then if you want to get all the (1/something2)-roots of something1 you say "solve z^(1/something2)=something1"? I think this question should instead ask for the square roots of $i^5$ or the solutions to $z^2=i^5$ –  Nov 13 '18 at 18:56
  • @M.Wind I want to see if there might be something we might have overlooked or if the professor is wrong. By asking online, I (re)discover the multi-valuedness of complex exponentiation to a non-integer real number. It is about asking the right question, not yet about finding the right answer. –  Nov 13 '18 at 18:59
  • @JackBauer For positive real numbers you have preference to choose the positive one, called principal square root. This is not the case for complex numbers. – Eclipse Sun Nov 13 '18 at 21:35

3 Answers3

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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.

J.G.
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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:

$\text{x and y have the same sign when b is positive}$

$\text{x and y have opposite signs when b is negative}$

$\text{There is no sign issue when b is zero because then Eqs. 4 and 5 give no extraneous roots}$

Oscar Lanzi
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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.