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Could you help me fill in a gap in my understanding of maths?

As long as the exponent is rational, I can decompose it into something that always works with primitives I know.

Say, I see a power: $x^{-{a\over b}}$ for natural a,b.

Using the basic rules:

$$\begin {align} x^{-a} =& {1 \over x^a} \\ x^{1\over b} =& \sqrt[b]{x} \\ ({x^a})^b =& {x^{a b}} \end{align}$$

I can always decompose rational exponents into $ x^{{1\over b} \cdot -a} = \sqrt[b]{x} ^ {-a} = {1 \over \sqrt[b]{x}^a }$- and integer degree roots are something well within my grasp. Integer root is always just simply a number of multiplications, $\sqrt[a]{x}$ is just some $y \cdot y \cdot y \cdot ... \cdot y$, as many $y$'s as $a$ says.

If the exponent is irrational, it's trickier, but I can always take $e^{\pi} \approx e^{314159265358979323 \over 100000000000000000}$ and take more $\pi$ digits if needed, and even though the number of underlying multiplications becomes ridiculous, that's still something I can understand.

But I'm completely at loss how to understand stuff like $x^{i\pi \over 4}$. I just can't perform an imaginary number of multiplications. I know how to perform conversion of complex numbers between the exponential and $a+ib$ form, but I perform it like a mysterious voodoo magic recipe without ability to wrap my mind around how that kind of exponentiation is supposed to work. I can still apply the old conversion formulas, $x^a \cdot x^b = x^{a+b}$; $({x^a})^b = {x^{a b}};$ and obtain correct results but I just don't understand the underlying mechanism - as $x^n$ is just $n$ multiplications of $x$, how can understand how that works when $n$ is imaginary?

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  • $\begingroup$ Graphically, real axis is horizontal, and imaginary axis is vertical. Most good mathematicians are able to picture the mathematics visually through graphics. $\endgroup$ – 0tyranny 0poverty Nov 4 '17 at 16:44
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    $\begingroup$ Have you seen this youtube.com/watch?v=mvmuCPvRoWQ ? $\endgroup$ – onurcanbektas Nov 4 '17 at 17:11
  • $\begingroup$ Yeah, and if you really want to grok the absurdity of it all, imagine dividing $i$ into four equal parts. $\endgroup$ – theDoctor Nov 5 '17 at 0:58
  • $\begingroup$ If you want to calculate the exponential of a complex number, you can do either of the following: (1) use DeMoivre's identity, which passes the buck to the trig functions, or (2) use the power series for the exponential, which converges rapidly due to k! in the denominator of the terms. Otherwise, you can approximate your argument as close as you like with a rational number and do the ridiculous number of steps. $\endgroup$ – richard1941 Nov 8 '17 at 3:23
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On the real line, real numbers act additively by shifting, and multiplicatively by scaling away from zero.

On the complex plane, real numbers act additively by shifting along the real axis (horizontally), imaginary numbers act additively by shifting along the imaginary axis (vertically). Note that orbits of these two actions are orthogonal. Horizontal lines versus vertical.

Real numbers act multiplicatively by stretching away from the origin, while imaginary numbers by rotating $90º$. Note that the orbits of these two actions are also orthogonal. Orbits of scalings are radial lines; orbits of rotations are circles.

Now the most fundamental identity of exponentials is $a^{x+y} = a^xa^y$. Exponentiation turns adders into multipliers. It turns real adders (i.e. horizontal shifts) into real multipliers, i.e. scalings away from zero. Horizontal lines into radial lines.

And therefore what must exponentiation transform the orthogonal imaginary shifts, i.e. vertical shifts into? They must transform into those multipliers which are orthogonal to the radial expansions. Which are the rotations. Vertical lines into circles. So exponentiation with an imaginary exponent must be a rotation.

The base of the exponentiation sets the size scale of these stretchings and rotations, and exponentiation with base $e$ does natural rotations in radians, but this picture works with any base of exponentiation, so long as $a>1.$

This intuition is encoded in Euler's identity $e^{i\theta}=\cos\theta + i\sin\theta$. A special case is $e^{i\pi} = -1$, which just says that rotation by $180º$ is the same thing as reflection. This intuitive point of view for understanding Euler's identity is explained in a popular 3Blue1Brown video.

So how do we understand an expression like $x^{\frac{i\pi}{4}}$? Well assuming $x$ is real with $x>0$, since the exponent is imaginary, it's a rotation. How big a rotation? well it depends on the base $x$, and the exponent. Computing this magnitude could be thought of as an exercise in operations derived from repeated multiplication, as you set out in your question, but doing so is of limited utility.

The notion of exponentiation as repeated multiplication is, for natural numbers $n,m$ equivalent to the identity $a^{m+n}=a^ma^n$. It immediately gives $$a^{n} = a^{\underbrace{1+\dotsb+1}_{n\text{ times}}}=\underbrace{a\cdot\dotsb\cdot a}_{n\text{ times}}.$$ But the identity also makes sense for naturals, integers, rationals, reals, complexes, even matrices and more, whereas the repeated multiplication notion only makes sense for $n$ natural, and is only extended to zero, negatives, and rationals via the above identity (or other similar). Therefore we should view the identity $a^{m+n}=a^ma^n$ not just as a consequence of exponentiation as repeated multiplication, but as a complete and fundamental conceptual replacement. Exponentiation is, by definition and fundamental conception, the operation that turns addition into multiplication.

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    $\begingroup$ re "rotation by 180º 180º is the same thing as reflection": NO! It's not the same. reflection reverses polarity ("handedness") and rotation doesn't. So-called "reflection through the origin" is a bastard concept, best left ignored. $\endgroup$ – Pieter Geerkens Nov 3 '17 at 19:11
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    $\begingroup$ @PieterGeerkens: reflection through the origin is orientation reversing in odd dimensions, but orientation preserving in even dimensions. In the case of the complex plane, you can simply observe Euler's identity. $\endgroup$ – ziggurism Nov 3 '17 at 19:27
  • $\begingroup$ I think I understand. I can first temporarily ignore the fact the exponent is imaginary (just pocket the $i$ for later use), and use all the old tricks to coerce rational into integer and irrational into rational, but once I come up with the proper amount of multiplications, I restore the $i$ which changes the meaning of the multiplication operator - apply them as a rotation operators, not as stretching operators. Will that work? $\endgroup$ – SF. Nov 4 '17 at 10:30
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    $\begingroup$ @SF.: No that will not work. Firstly, complex exponentiation is totally different from real exponentiation, and commonly one takes a branch cut, which means that you cannot use identities for exponentiation with positive real base. After all, why should they hold to begin with? Secondly, $x^i = \exp(i·\ln(x))$ and hence depends on $x$ in a way that you cannot think of the $i$-th power as a separate thing. For example $(1+i)^i = \exp(i·(\ln(2)/2+iπ/4))$ and clearly the real and imaginary parts of $\ln(1+i)$ affect the result differently! $\endgroup$ – user21820 Nov 4 '17 at 13:12
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    $\begingroup$ @ziggurism: It is very weird and arguably useless to care only about complex exponentiation with real base... But I honestly think your saying "will work" to the asker was misleading, because exponentiation by $i$ has nothing to do with rotation operators contrary to what the asker apparently thinks (see his/her last comment). I have no problem with your answer itself (especially the "limited utility" part). $\endgroup$ – user21820 Nov 4 '17 at 13:33
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I think it would be helpful for you to let go of the idea (at least in complex analysis) that $a^b$ means multiplying $a$ $b$ times by itself. Rather, exponentiation starts at the function $\exp : \mathbb C \to \mathbb C$, which you probably usually denote as $\exp(a) = e^a$. Then (once you have chosen a branch of the complex logarithm) you can define $a^b = \exp(\log(a) \cdot b)$. With that definition, all the repeated multiplication rules that you know come down to the identities $\exp(a + b) = \exp(a)\exp(b)$ and $\exp(0) = 1$.

Rather than trying to press complex exponentiation into the mold of repeated multiplication, see complex exponentiation -- or more fundamentally, the function $\exp$ -- as its own thing, that in special cases can be interpreted as repeated multiplication, thanks to the functional relation $\exp(a)\exp(b) = \exp(a + b)$.

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I guess we agree on $ e^{a + i \phi} = e^{a} \cdot e^{i \phi} $. So let's look at the latter term with imaginary exponents.

Imaginary exponents are indeed of a different quality then real ones. Maybe your best approach is to realize that the justification for the "magic" Euler equation

$$ e^{i \phi} = \cos(\phi) + i \sin(\phi) $$ only comes from the formal continuation of the exponential function from real to complex domain. Indeed, by just taking the series expansion of the exponential function, $e^x = \sum_{n=0}^\infty x^n/n!$, apply for $x = i \phi$, and grouping real and imaginary parts you see that this holds true. This gives you (for odd and even powers of $\phi$), with real $\phi$, of course:

$$ e^{i \phi} = \sum_{n=0}^\infty (i\phi)^n/n! = \sum_{n=0}^\infty (-1)^n \phi^{2n}/(2n)! + i \sum_{n=0}^\infty (-1)^n \phi^{2n +1}/(2n+1)! $$ and the two sums are just the expansions of $\cos \phi$ and $\sin \phi$, for real $\phi$.

So yes, it is unintuitive that you now stay on the unit circle in the complex domain, that it's cyclic in $2 \pi$, etc.


EDIT: (see also Paul Sinclair's comment)

If you know about the derivatives of $\cos \phi$ and $\sin \phi$, here's an approach without series expansions.

We know that $\cos' \phi = - \sin \phi$, and $\sin' \phi =\cos \phi$. Now define a function $ f(\phi) = \cos(\phi) + i \sin(\phi) $. Then you have $ f'(\phi) = - \sin(\phi) + i \cos(\phi) = i f(\phi) $.

On the other hand, we know for real constants $a$ that the equation $ f'(\phi) = a f(\phi) $ has a unique solution, which is $f(\phi) = c \cdot e^{a \phi}$ with some constant $c$ which is determined by some initial condition. Now comes the trick again of the formal continuation of the exponential function from real to complex domain. We allow now that $a$ may as well be complex. Then we have that $ f'(\phi) = i f(\phi) $ (as taken from the trig function above) has the unique solution $f(\phi) = c \cdot e^{i \phi}$. Determine the constant $c=1$ by observing $f(0) = c \cdot e^{i 0} = c$ and also $ f(0) = \cos(0) + i \sin(0) = 1 $.

So this explains Euler's equation as well.

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  • $\begingroup$ I'll need to get a better feel at what $x^{n}/n!$ does (doesn't even look all that convergent at a glance), and how these two sums manage to expand cos x and sin x. Looks like it could work, but still a bit too voodoo. $\endgroup$ – SF. Nov 3 '17 at 15:37
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    $\begingroup$ A more natural way to extend exponentiation to complex numbers is to note that $e^x$ is the unique solution to the differential equation $$y' = y, \quad y(0) = 1$$. Replace $y, x$ with $w, z$, and you have a differential equation of complex functions on the complex plane, which still has a unique solution. Further, this solution restricts to $e^x$ when $z = x +i0$. So we define $e^z$ to be that unique solution when $z$ is complex. The Taylor series (which is easily seen to be convergent, since $n!$ grows much faster than $x^n$ once $n > x$) follows directly from the differential equation. $\endgroup$ – Paul Sinclair Nov 3 '17 at 16:44
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    $\begingroup$ @SF. It sounds like that would be a useful thing for you to study, then: convince yourself that $\sum x^n/n!$ converges, and that it's equal to $e^x$, and that the series for $\cos x$ and $\sin x$ as given in the answer are correct - at least for real values of $x$. $\endgroup$ – David Z Nov 3 '17 at 21:30
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    $\begingroup$ @SF. : How do you define sine and cosine for arbitrary real arguments? $\endgroup$ – Eric Towers Nov 3 '17 at 22:46
  • $\begingroup$ Did you mean to write "intuitive"? $\endgroup$ – Eric Duminil Nov 5 '17 at 9:02
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$x^n$ is just $n$ multiplications of $x$

This is not understanding exponentiation; it is merely the reduction of exponentiation into other terms.

Reducing new concepts (in special cases) to familiar terms is a common first step to understanding something, but one eventually needs to proceed from there to synthesize an understanding about the concept itself, rather than use the other terms as a crutch.

I don't think you can intuitively understand complex exponentiation as a generalization of real exponentiation without first synthesizing an understanding of exponentiation.

You can, however, reduce complex exponentiation to other terms; the usual definition is

$$ z^w = \exp(w \log z)$$

so if you want to work with complex exponentiation in other terms, you can reduce the problem to understanding the complex exponential and logarithm.

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    $\begingroup$ sometimes the best answer is to reject the premise. $\endgroup$ – James S. Cook Nov 4 '17 at 3:01
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My intuition about this works best when I consider it as manipulating the vector $(1,0)$ on the complex plane by performing two separate actions on it:

  1. Scaling it based on the real part (with $\phi=0$ being the identity, positive values stretching it and negative values shrinking it).
  2. Rotating it based on the imaginary part (this is the cool part).

So why does $e^{i \phi}$ perform a rotation? Well this comes out as a natural consequence of Euler's formula.

The simplest non-trivial example is of $\phi=\pi$, which gives us a rotation of $\pi$ radians, bringing the vector to $(-1,0)$ (this is a restatement of Euler's identity).

And in the general case, for any (real) value of $\phi$, Euler's identity gives us $e^{i \phi} = \cos\phi + i \sin\phi$, which has to be on the unit circle, since it's the vector $(\cos\phi, \sin\phi)$, the magnitude of which is:

$\|(\cos\phi, \sin\phi)\| = cos^2\phi + sin^2\phi = 1$

And this also helps us see that the effect of the imaginary part of the exponent is periodic, so if we want to, we can limit ourselves to only consider the range $\phi\in[0, 2\pi)$.

So to sum up, by performing a complex exponentiation, which is the combination of the scaling and rotation, we can get our vector anywhere on the complex plane, except for the origin (which is the limit when the real part tends to $-\infty$).

Edit: Btw, if you like infinities, you might find the concept of the Riemann Sphere cool. The idea here is to add another "point at infinity" to the complex plane, and wrap the whole plane up as a sphere, with the point at infinity being exactly opposite of the origin. So when considering complex exponentiation on the sphere, scaling acts like changing the latitude, while rotation acts as changing the longitude. And scaling by $\pm\infty$ become well-defined, such that scaling by $-\infty$ takes the vector to the point at zero, while scaling by $+\infty$ takes it to the point at infinity, on the other side of the sphere. And in both these extremes, the rotation has no effect, which is very similar to how when navigating on Earth, when you're at the poles (latitudes $180^\circ N$ and $180^\circ S$), the longitude loses meaning.

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A famous result in mathematics is Euler's Formula: $$e^{i\theta}=i\sin(\theta)+\cos(\theta)$$ Complex exponentials are used to parametize points in the complex plane so a point $$x+iy=re^{i\theta}=r(\cos(\theta)+i\sin(\theta))$$ So the function $e^{i\theta}$ can be viewed as a circle in the complex plane with radius $1$.

So complex exponentiation cannot be viewed as repeated multiplication anymore, and instead of stretching a point around the unit circle.

The reason e is the unique base that satisfies euler's formula has to do with the fact that the derivatives of $e^x$ and $cos(x),sin(x)$ cycle through the same values at a point.

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  • $\begingroup$ Note that you can rewrite $x^{i}$ as $e^{i \log x}$ $\endgroup$ – RoyPJ Nov 3 '17 at 15:02
  • $\begingroup$ I think he knows this stuff math.stackexchange.com/a/189836/369983 $\endgroup$ – Fawad Nov 3 '17 at 15:03
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    $\begingroup$ @RoyPJ That is a very dangerous thing to do, and it usually leads to inconsistent manipulations. $\endgroup$ – user228113 Nov 3 '17 at 15:06
  • $\begingroup$ @Fawad: Yes, I know this stuff... and apply it like voodoo magic. WHY does $e^{i\theta}$ act like a circle? $e$ to any power maintains the magnitude, it's the fun property of $e$, but it's that rotary behavior is what baffles me. $\endgroup$ – SF. Nov 3 '17 at 15:09
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    $\begingroup$ @G.Sassatelli how does this lead to inconsistency? As long as x is not zero I don't see a problem there. $\endgroup$ – RoyPJ Nov 3 '17 at 15:29
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Actually, I think your fundamental mistake is to think too fundamentally without thinking more fundamentally :P. The problem is deep rooted in how we define exponentiation.

This used to confuse me for some time :) I hope this answer helps you and anybody else that might stumble upon my answer!

A Simpler Problem

You seem comfortable calculating the value of $x^{1 \over b}$, but given a paper, a pen and infinite time, how would you evaluate:

$$3^{0.3}$$

How do you produce the digits? Most calculators reduce this expression to:

$$ e ^ {\log 3^{0.3}} = e ^ {0.3 \log 3}$$

And then we can use lookup tables to calculate $log3$ and indeed the exponential of the resulting number. The lookup tables are defined from limits and calculus though, so :

$$ e^x = \lim_{n\rightarrow \infty} \left( 1 + {x\over n} \right)^n $$

and:

$$\log x = \int_1^x {1 \over x} dx $$

So we can go ahead and work out this integral directly using a lookup as I mentioned, or by using something like Simpsons rule.

Notice that we've lost the original definition here, we are no longer defining:

$$ a^b = \underbrace{a * a * a * \ldots }_\text{n times} $$

but instead, we are now defining exponentiation by:

$$ a^b = e ^ {b \log a }$$

This represents a fundamental shift in how we should perceive exponentiation. Because the idea of multiplying something by itself 0.3 times is not well defined, and we shouldn't think of it in that way anymore :)

Extending to Complex Numbers

Now if we apply this definition to complex numbers, with the aide of Eulers amazing equation, that shows us multiplication of two complex numbers represents a multiplication of magnitudes and a rotation by their arguments, we can see that this equation is exactly what we would expect it to be:

$$ x ^ {i {\pi \over 4}} = e ^ {i \left( {\pi \over 4 }\log x \right)} = \cos\left( {\pi \over 4 }\log x \right) + i \sin \left( {\pi \over 4 }\log x \right)$$

This just uses our new definition of exponentiation, and it extends very naturally to complex numbers as you can hopefully see :)

Some Words

I think some of the others put this better than I could. As you learn more and more mathematics, you start to learn to explain things that you once could only explain with a certain pattern of thinking, in terms of new patterns that would have been absurd to show a new learner.

I indeed asked myself this same question some time ago, and after some careful thought, I ended up with this :)

Ps.

So I guess now you have to figure out how to calculate sin and cos ;)

Have a good one!

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  • $\begingroup$ As a side note, we can calculate $e^x$ using our old definition of repeated multiplication, I'm aware of that hole. But extending a definition doesn't mean that the old one doesn't still apply for the old domain :P so if n is an integer, you should be fine here. $\endgroup$ – user2662833 Nov 5 '17 at 3:47

protected by user21820 Nov 10 '17 at 11:26

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