Calculating $i^{3/2}$ in two ways gives different results [closed]

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I have very little familiarity with complex numbers, but I now have to work with $$i^{3/2}$$ where $i$ is the imaginary unit.

The on-line WolframAlpha machine gives $$i^{3/2} = (-1)^{3/4} = -\frac 1 {\sqrt{2}} + \frac i {\sqrt{2}}.$$

From my part, thinking naively I do $$i^{3/2} = \sqrt{i^3} = \sqrt{i\cdot i^2} = \sqrt {-i}.$$

But the machine gives

$$\sqrt {-i} = -(-1)^{3/4} = \frac 1 {\sqrt{2}} - \frac i {\sqrt{2}} = -i^{3/2}.$$

So either the machine is wrong (I doubt it), or I am doing something wrong in how I manipulated $i^{3/2}$ into $\sqrt{-i}$.

Any hint of how should I go about understanding the mistake?

• Xander Henderson: What is your definition of $z^x,ドル where $z$ is a complex number? Do you understand that this is not a uniquely defined number? And that the theorems about exponentiation (e.g. $(ab)^x = a^x b^2$) don't necessarily translate to the complex case?
• J. W. Tanner: non-zero complex numbers have two square roots. and cf. this question
• Lee Mosher: Fractional powers of complex numbers that are not on the positive real line (such as negative numbers, or complex numbers with a nonzero imaginary part), are not well-defined. This site has many posts on this topic which you might wish to look over. See for example here and here and here.
• A rural reader: It sounds like the best place for you to begin is here. It seems odd you have little experience working with complex numbers, yet somehow run into $i^{3/2}$ ...
• Alecos Papadopoulos: @Aruralreader It is because up to now my involvement with statistical distributions did not really push me to work with their characteristic functions.

Answer 1

There is no such thing as the square root of a number. Every (nonzero) number has exactly two square roots, which differ by a factor of $-1$. When the input is a real number, we have a nice convention that the square root of a positive number is positive real, and of a negative number is a positive number times $i$. However, this convention could just as easily be reversed, particularly in the second case.

The square root, like all non-integer powers, is a multi-valued function when taken over the complex numbers. This means that the normal power identities like $a^{bc} = (a^b)^c$ (an identity you're implicitly using in your computations here) don't necessarily work as you expect.

$i^{3/2} = e^{i\pi\frac{3}{4}} = -\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ is certainly one square root of $-i$. $\sqrt{-i}=\frac{1}{\sqrt{2}} -i\frac{1}{\sqrt{2}}$ is the other.

If you want to learn more about this topic, read into principal values and branch cuts of complex multi-valued functions.

Answer 2

Recall DeMoivre's Formula (which is actually a special case of Euler's).

$$(\cos(\theta) + i\sin(\theta))^n = \cos(n\theta) + i\sin(n\theta)$$

Plug in $n = \frac{3}{2}$.

$$(\cos(\theta) + i\sin(\theta))^{3/2} = \cos\left(\frac{3\theta}{2}\right) + i\sin\left(\frac{3\theta}{2}\right)$$

OK, so now we just need a $\theta$ so that $\cos(\theta) = 0$ and $\sin(\theta) = 1$. Fortunately, this is one of those "special" angles: $\theta = \frac{\pi}{2}$.

$$\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right)^{3/2} = \cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right)$$ $$i^{3/2} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} = \frac{-1+i}{\sqrt{2}}$$

Easy peasy. This answer agrees with WolframAlpha and with Python (enter 1j ** (3/2) into the REPL).

Except one little thing: $\cos$ and $\sin$ are periodic. So can add or subtract 2ドル\pi$ to $\theta$, and it should also work. So let's try $\theta = \frac{-3\pi}{2}$.

$$\left(\cos\left(\frac{-3\pi}{2}\right) + i\sin\left(\frac{-3\pi}{2}\right)\right)^{3/2} = \cos\left(\frac{-9\pi}{4}\right) + i\sin\left(\frac{-9\pi}{4}\right)$$ $$i^{3/2} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} = \frac{1 - i}{\sqrt{2}}$$

What gives? The Fundamental Theorem of Algebra. If you define $i^{3/2}$ as the complex number $z$ such that $z^2 = i^3$, or $z^2 + i = 0$, then there are two roots. One is the principal square root, and the other is the additive inverse of it.

Answer 3

The symbol $i^{\frac{3}{2}}$ represents two complex numbers due to the multi-valued nature of the complex logarithm. $$\begin {align} i^{\frac{3}{2}}&\equiv e^{\frac{3}{2},円\log i} \\ &=e^{\frac{3}{2}\left(\frac{π}{2}+2kπ\right)i} \\ &=e^{\left(\frac{3π}{4}+3kπ\right)i}\;\;\;\;\;\;\;k\in \mathbb{Z}\;, \end{align}$$ so, for k=0, we get $$\displaystyle -\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}$$ and, for k=1, $$\displaystyle \dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}}\;.$$

Answer 4

While collecting material to study, here is why the Wolfram Alpha machine gives \begin{align} i^{3/2} &= -\frac 1 {\sqrt{2}} + \frac i {\sqrt{2}},\\ \sqrt {-i} &= \frac 1 {\sqrt{2}} - \frac i {\sqrt{2}}. \end{align}

In https://mathworld.wolfram.com/ComplexExponentiation.html the machine gives us a formula to exponentiate a complex number by another complex number,

$$ (a+bi)^{(c+di)}=(a^2+b^2)^{(c+id)/2}\cdot \exp\{i\cdot (c+id)\cdot{\rm arg}(a+ib)\}. $$

In both of our cases, $$a = 0, d = 0,$$

so the machine's expression reduces to

$$ (bi)^{c}=(b^2)^{c/2}\cdot \exp\{(i\cdot c \cdot {\rm sgn}(b)\cdot \pi/2\}. $$

A. For $i^{3/2}$ we have $$b=1 \implies {\rm sgn}(b)=1, \;c= 3/2.$$

Plugging in we get $$i^{3/2} = \exp\{i\cdot 3\pi/4\} = {\rm cos}(3\pi/4) +i\cdot {\rm sin}(3\pi/4) = -\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}.$$

B. For $\sqrt{-i}$ we have $$b=-1 \implies {\rm sgn}(b)=-1, \;c= 1/2.$$

Plugging in, $$\sqrt{-i} = \exp\{-i\cdot \pi/4\} = {\rm cos}(\pi/4) -i\cdot {\rm sin}(\pi/4) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}.$$

Answer 5

Think of multiplying by $i$ as rotating 90ドル^\circ$. So you rotate from $(1,0)$, 90ドル^\circ$ counter clock then 45ドル^\circ$ more which is the angle on unit circle representing 135ドル^\circ$.

Answer 6

As was pointed out, a square root (half power) has two values. I would do this: write -i as $e^{-i\pi/2}$. Then $(-i)^{3/2}= (e^{i\pi/2})^{3/2}= e^{-3i\pi/4}= cos(3\pi/4)- i sin(3\pi/4)= -\sqrt{2}/2+ i\sqrt{2}/2$.

If we add 2ドル\pi$ to the exponent we go all the way around the unit circle, going back to -i. -i also equals $e^{(-\pi/2+ 2\pi)i}= e^{3\pi i/2}$. Now $(-i)^{3/2}= (e^{3\pi i/2})^{3/2}= e^{9\+ pi i/4}= cos(9\pi/4)+ i sin(9\pi/4)= \sqrt{2}/2+ i\sqrt{2}/2$.