For context, I have been relearning a lot of math through the lovely website Brilliant.org. One of their sections covers complex numbers and tries to intuitively introduce Euler's Formula and complex exponentiation by pulling features from polar coordinates, trigonometry, real number exponentiation, and vector space transformations.

While I am now decently familiar with how complex exponentiation behaves (i.e. inducing rotation), I am slightly confused by the following.

$ 2^3 z$ can be viewed as stretching the complex number $z$ by $2^3$. This could be rewritten as $8z$. Therefore, Brilliant.org suggests that exponentiation of real numbers can be thought of as stretching a vector just like real number multiplication would. (check - understood)

Brilliant.org then demonstrates that multiplying $z_1$ by another complex number $z_2$ is equivalent to first stretching $z_1$ by the magnitude of $z_2$ and then rotating $z_1$ by the angle that $z_2$ creates with the real axis counterclockwise. (check - understood)

However, this is where I get confused. Why does, for example, $2^{2i}* z$ cause purely rotation of z but $2i*z$ does not (i.e. it causes stretching, too, in addition to rotation)?

To me, the fact that $2^{(2i+3)}$ causes both rotation and stretching makes perfect sense because we can rewrite this as $(2^3)*(2^{(2i)})$. As previously noted by Brilliant.org, exponentiation by real numbers can thought of as stretching.

Here is the crux of my issue:

I understand that the magnitude of the imaginary number in the exponent (for example, the $'2'$ in $e^{2i}$ ) can be thought of as a rate of speed...but why does this interpretation 'drop' when we are doing something like $2i * z$. i.e. Why is the $2$ in $2i*z$ not also treated like a rate of rotation but instead treated like a magnitude of stretching ?

My math skill is not particularly high level so if anyone can offer as much of an intuitive answer as possible, it would be greatly appreciated!

Edit 1: I guess another way of expressing this question is as follows:

Why does a duality exist between real number exponentiation and real number multiplication but a duality does not exist between imaginary number exponentiation and imaginary number multiplication (i.e. imaginary number multiplication can cause stretching in addition to rotation)?

Edit 2: While I accept that Euler's formula is a way of proving that exponentiation of purely imaginary numbers has a magnitude of 1 and therefore does not invoke stretching, that is not the sort of answer I am looking for. My question is aimed at identifying what was specified in Edit 1.

Edit 3: Here is a picture that helps clarify my point of confusion. Lack of Duality Between Exponentiation and Multiplication

Edit 4: The question that was asked in this post Which general physical transformation to the number space does exponentiation represent? is sort of the theme that I am going for. The answer that was given to this post, however, omits a reference to the complex numbers.

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    $\begingroup$ Do you mean $e^{2i}z$ where yo wrote 2^(2i)*z? $\endgroup$ – coffeemath Jun 16 '19 at 20:57
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    $\begingroup$ From what I understand about complex exponentiation, it could be any base. Conceptually, the question should still be the same. (Yes, I know that the rate of rotation changes depending on which base I use) $\endgroup$ – S.Cramer Jun 16 '19 at 20:59
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    $\begingroup$ Right, one could define $2^{2i}z$ as $e^{2\ln2 i}z$ which just has a different angle. $\endgroup$ – coffeemath Jun 16 '19 at 21:02
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    $\begingroup$ The magnitude of $e^{i\theta}=\cos\theta+i\sin\theta$ is $1$, so you're stretching by a factor of $1$, which is the same as not stretching. $\endgroup$ – saulspatz Jun 16 '19 at 21:05
  • $\begingroup$ hmmm, while I accept that is a quick way of proving this to be the case, I feel like this answer masks the truth that I am looking for. It's the failure in duality (which I mentioned in my edit at the bottom of the page) that is specifically confusing to me. $\endgroup$ – S.Cramer Jun 16 '19 at 21:09

To briefly address your specific question about "duality" first, there is no "duality" (not sure what is precisely meant by that term here) between complex exponentiation and complex multiplication, unless the arguments of both are real. This is because complex multiplication is commutative; complex exponentiation is not - neither is real exponentiation $(e^{ix}$ is not the same as $ix^{e}$). Take the second line:

$$c^{a+bi}\cdot z \leftrightarrow (a+bi)\cdot z $$ $$ \text{stretching and rotating} \leftrightarrow \text{stretching and rotating} $$

Yes, $(a+bi)\cdot z$ corresponds to a transformation consisting of some stretching and some rotation, but the stretching is due to both multiplying by $a$ and multiplying by $bi$.

Now, let's break down intuitively what different operations on the complex plane are (I know you said you already went over some of this in your question, but I think it will lead into the explanation for complex exponentiation nicely).

Complex addition is the same as vector addition: we add the components. Think of this intuitively by imagining each complex number as an arrow: adding two complex numbers is like sticking one of their arrows on the end of the other. Another way: think of adding a complex number not as a static operation, but as a transformation. Adding the number $(a+bi)$ is the same as shifting the origin of the complex plane onto the point $(-a-bi)$. Take a second to imagine why that's true. Thus $(a+bi)$ is a function with respect to addition, which maps every point in the complex plane to another point in the complex plane, (a+bi) away.

Complex multiplication corresponds to both a stretch and a rotation (usually). $$(a+bi)\cdot(c+di)=(ac-bd)+(ad+bc)$$

A better way to think about this is in terms of Euler's formula: represent your two complex numbers as polar coordinates, and multiplication becomes much clearer: $$r_1 e^{i\theta_1}\cdot r_2 e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}$$

So we can imagine complex multiplication of two numbers is taking the angles they make with the x axis, adding those two angles to get the angle of your new number, then multiplying the magnitudes of the two original numbers to get the magnitude of your new number. Think of complex multiplication by $(a+bi)$ as two, very dynamic transformations composted together: first stretching the entire complex plane by a factor of $\sqrt{a^2+b^2}$, then rotating the entire complex plane by a factor of $\tan^{-1}(\frac{b}{a})$. Thus $(a+bi)$ can also be thought of as a function with respect to multiplication: it maps every point in the complex plane to another point in the complex plane by combination of a rotation and a stretch.

What kind of function is "complex exponentiation"? We define it as follows: $a^{b+c\cdot i}$=$a^{b}\cdot a^{c\cdot i}$ where $a^{c \cdot i} = e^{c\cdot i \log a}$, etc. based on Euler's formula. $e^{c\cdot i \log a}$ is $e$ raised to a complex number and is thus a rotation. Note that we can imagine complex exponentiation again as a sort of dynamic transformation, squishing and mapping the complex plane to a new location.

Let's break this down into two cases. The first is the only one your question asks about.

1) The base of the exponent is real. As can be seen from above, every "complex exponentiation" is a transformation of the complex plane consisting of two transformations: first, a stretch by some factor, and then a rotation. This is very similar to complex multiplication, which begs the question, when do these two things behave in the same way? You called it "duality," I'm not going to call it that because that word means something specific in linear algebra, I'll just call this similarity "similarity."

Multiplying $x$ by $(a+bi)\rightarrow$ Stretch by $\sqrt{a^2+b^2}$, rotate by $\tan^{-1}(\frac{b}{a})$ radians.

Raising $x$ to the $(a+bi)$ power $\rightarrow$ Stretch by $x^a$, rotate by $b\cdot i \log x$ radians.

Note that the two are very dissimilar - exponentiation is dependent on the base whereas multiplication is not. This is a result of the effect that multiplication by a complex number is something called a linear transformation (https://en.wikipedia.org/wiki/Linear_map), raising something to a complex power is most certainly not.

2) The base of the exponent is complex. This gets a little more complicated, because raising a complex number to a real power corresponds in part to a rotation, so separating which parts are the rotation and which parts are the stretching is a little annoying and won't give much insight here. Complex exponentiation of complex numbers is really funky, giving rise to all sorts of weird fractal shapes when we consider which complex numbers get really large when we raise them to a complex power, and which ones don't - this is related to how the Mandelbrot set is formed (https://en.wikipedia.org/wiki/Mandelbrot_set). The point is that, again, the magnitude of the stretch and rotation is dependent on x, meaning that this is not a linear transformation.

If you want to gain an intuition for how complex exponentiation of complex numbers works, I recommend you play around with some functions with this grapher: http://davidbau.com/conformal/#z

So, to return to what I mentioned at the beginning, the reason why you're not seeing a "duality" in case 3 is that there's no "duality" in case 2 to begin with - yes, both complex exponentiation and complex multiplication correspond to both a rotation and a stretch, but the rotation and stretch for each behave in very different ways. Complex multiplication is a linear transform, complex exponentiation is not.

I also enjoy brilliant.org's courses; if you're interested, I would recommend you check out their course on linear algebra next (https://brilliant.org/courses/linear-algebra/). This is the first answer I've actually posted. I would love feedback from anyone if they have it.

  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa Jun 19 '19 at 4:49
  • $\begingroup$ is another important point the fact that the "Stretching Event" caused by multiplication is different than the "Stretching Event" caused by real number exponentiation? For example, the "0.5" in 2.5 times x is a different sort of half than the "0.5" in x raised the 2.5. i.e. that 0.5 stretches the space differently depending on whether or not we are talking about exponentiation or multiplication. $\endgroup$ – S.Cramer Jun 19 '19 at 14:59
  • $\begingroup$ Yes, the arguments mean different things. Complex multiplication corresponds to a general stretching of your basis - imagine the plane as a lattice; stretch it out. Complex exponentiation corresponds to a stretching that is dependent on location. It's very rare that exponentiation and multiplication by the same argument correspond to the same transformation. My intuition is that this should only be true when the argument is 1. Go to davidbau.com/conformal/#z and compare the functions (2+i)z vs z^(2+i). $\endgroup$ – Laurel Turner Jun 19 '19 at 20:14

$2^{2i}z$ does cause a stretching of $z$. It causes a stretching by the magnitude of $2^{2i}$. And the magnitude of $2^{2i}$ is one. So, it stretches by a factor of $1$. If you choose to deny that stretching by a factor of $1$ is stretching, well, then that's where your problem is.

  • $\begingroup$ is another important point the fact that the "Stretching Event" caused by multiplication is different than the "Stretching Event" caused by real number exponentiation? For example, the "0.5" in 2.5 times x is a different sort of half than the "0.5" in x raised the 2.5. i.e. that 0.5 stretches the space differently depending on whether or not we are talking about exponentiation or multiplication. $\endgroup$ – S.Cramer Jun 19 '19 at 15:01

The failure of duality is that there was never really duality there in the first place.

It's true that in most cases, the vector from the origin to $c^{a+bi} z$ is rotated and stretched (or shrunk) relative to the vector from the origin to $z.$

But sometimes it is not stretched or shrunk: specifically, when $a = 0.$

It is also true that in most cases, the vector from the origin to $(a+bi) z$ is rotated and stretched (or shrunk) relative to the vector from the origin to $z.$

But sometimes that vector is not stretched or shrunk either: specifically, when $a^2 + b^2 = 1.$

The key observation to me is that when you write something like $c^{a+bi},$ you identify a point in the complex plane using the parameters $a$ and $b$ somewhat like polar coordinates. Parameter $a$ dictates the radius $r$, parameter $b$ dictates the angle $\theta.$

When you write $a + bi,$ however, the parameters $a$ and $b$ act like Cartesian coordinates $x$ and $y$ of a point in the complex plane.

Polar coordinates don't work like Cartesian coordinates, and vice versa. They both identify points in a plane, that's mostly what they have in common. So when you wrote that both $c^{a+bi} z$ and $(a+bi) z$ stretch a vector as well as rotating it, that was (mostly!) true, but the way that $a$ and $b$ contributed to making it be a rotation or a stretching was completely different in each case.

What really makes multiplication not stretch or shrink a vector, is you have to make sure the thing you multiply $z$ by is a complex number on the "unit circle." You can get this either by choosing $r=1$ in polar coordinates (which corresponds to $a=0$ in the formula $c^{a+bi}$) or by choosing $x^2 + y^2 = 1.$

  • $\begingroup$ is another important point the fact that the "Stretching Event" caused by multiplication is different than the "Stretching Event" caused by real number exponentiation? For example, the "0.5" in 2.5 times x is a different sort of half than the "0.5" in x raised the 2.5. i.e. that 0.5 stretches the space differently depending on whether or not we are talking about exponentiation or multiplication. $\endgroup$ – S.Cramer Jun 19 '19 at 15:01
  • $\begingroup$ Yes, that is a true point. The numbers get converted to “stretching factors” in very different ways. For example, if you put $-2$ in the exponent of your multiplier you shrink the vector, but if you just multiply by $-2$ the vector flips $180$ degrees around and gets longer. $\endgroup$ – David K Jun 19 '19 at 20:21

Let $z_o$ and $z_d$ be two complex number, lets call the first one operator and second one operand, and let $z'_d$ be the transformed version( state after it being operated by $z_o$ ) of $z_d$ i.e. $$ \begin{align} z_d' &= z_o \times z_d \\ &= |z_o| e^{i arg(z_o)} \times |z_d| e^{i arg(z_d)} \\ & = |z_o z_d| e^{i(arg(z_o) + arg(z_d))}\end{align}$$

From here all your results get derived:

  1. If $z_o$ is purely real, then $ arg(z_o) = 0, \pi$ when $z_o$ are positive and negative respectively. In the case when $z_o$ is positive

$$ z'_d = |z_o z_d| e^{i(arg(0 + arg(z_d))} = |z_o z_d| e^{i(arg(z_d))} = |z_o| z_d$$ the number $z_d$ is scaled by factor of $|z_d|$

For the case when $z_o$ is negative real number, $$ z'_d = |z_o z_d| e^{i(\pi + arg(z_d))} = -|z_o| z_d $$ In this case, the number is rotated $\pi$ radians, you can its flipped and scaled by the factor of $|z_o|$, but no one is wrong here, because only positive real numbers can be expressed in the terms of real exponents, in that case, what I've said conform with the website.

  1. If $z_o$ is complex number, $$ z_d' = |z_o z_d| e^{i(arg(z_o) + arg(z_d))}$$ as you can see here, the magnitude of $z_d$ is scaled by $|z_o|$ and in addition to that argument of complex number has been changed by the addition of argument of operator, so here the number is rotated and scaled.

Now, coming to your question, for multiplication of $2^{2i}$ with $z$ , if I denote $z'$ as the transformed version, then $$ z' = 2^{2i} z = |2^{2i}| e^{arg(2^{2i})} z = (1) e^{arg(2^{2i})} z $$ As you can see here, its magnitude is one, while in the case when $z_o = 2i$, $|2i| = 2, arg(2i) = \cfrac{\pi}{2} $, so here you will rotation and scaling.

Coming to your crux-points, If $z_o$ is $e^{2i}$, its implication is $|z_o| = 1$ which implies it can't scale the complex number, and it can only rotated $z_d$ by $2$ radians. When $z_o = 2i, |z_o| = 2,arg(2i) = \cfrac{\pi}{2}$, so it is scaled two times and rotated half pi radians.


Think of ot this way. $2iz=2(iz)$, so you have two steps here: $z\mapsto iz$, which is a pure rotation and $(iz)\mapsto 2(iz)$, which is a pure stretching. Of course, you can also write $2iz=i(2z)$ in which case you'll do the stretching first, but the result will be the same.

Now, when you multiply by $2$ in the exponent, something different is going on. Instead of $xyz=(xy)z$, you have $x^{yz}=(x^y)^z$, so $e^{2i}z=(e^i)^2z=e^i(e^iz)$, i.e., you apply the rotation given by $e^i$ twice, not follow or precede it by stretching two times. That's why the rotation remains a rotation but the angle doubles.

Of course, you can write it also as $(e^2)^iz$, but this way you will have to say that you apply the stretching ($z\mapsto e^2z$) imaginary ($i$) number of times. There is no intuitive physical meaning of applying some transformation $i$ times but, since you want to maintain the usual laws for powers as much as you can, you are forced to say that stretching done $i$ times becomes a rotation. Strange, but since there is no contradiction with common sense (common sense just says nothing about applying a transformation imaginary number of times), possible.


Write $z=r_1e^{i\theta_1}$. When multiplying $w=r_2e^{i\theta_2}$ by $z$, we stretch by a factor of $r_1$, and rotate by an angle of $\theta_1$. This is easy to see, since $zw=r_1r_2e^{i(\theta_1 +\theta_2)}$.

If $z$ is purely imaginary, so that we can take $\theta _1=\dfrac{\pi}2$, then we get rotation by $\dfrac{\pi}2$ and stretching by $r_1$.

Now for multiplication by $z^{ri}$, we have $z^{ri}=e^{ri\ln z}$. Depending on the value of $\ln z$, we have various possibilities. But we do know there will be no stretching. For, the complex number $e^{ri\ln z}$ has magnitude $1$.


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