What exactly is i? (And other weird questions about complex numbers)

by greenturtle3141, Apr 24, 2023, 5:05 AM

This is a topic that comes up quite a bit. This is of no surprise, of course, since imaginary numbers tend to be introduced in a confusing manner. Even math teachers often get tripped up by the numerous traps that surround this mysterious number $i$ . What is $i$ , anyway?

More likely than not, you were taught that $i$ is the square root of $-1$ . That is, $i = \sqrt{-1}$ . Of course, negative numbers don't have square roots, and the excuse that teachers tend to make is "well, just invent such a number $i$ that satisfies $i = \sqrt{-1}$ , and everything is ok". While this works for some time, it will not be long before the mathematically curious mind runs into some major issues.

Paradox 1: We start from the very true equation
$\frac{-1}{1} = \frac{1}{-1}.$ Taking the square root of both sides, we obtain
$\frac{i}{1} = \frac{1}{i}.$ Cross-multiplying gives $-1 = 1$ .

The astute reader may surely complain that the error is simply that the square root is multiplicative. While this is true, I'd say that this is merely a symptom of a larger problem. Indeed, why must square root suddenly not follow such a rule that worked perfectly for the reals? I purport that the "root cause" of such a paradox is, in fact, a misunderstanding of what a square root even is! What does it mean to take a square root in a complex plane?

Paradox 2: We know that $1 = e^{2i\pi}$ . Squaring both sides, we get $1 = e^{4i\pi}$ , so that $e^{2i\pi} = e^{4i\pi}$ . Taking the log gives $2i\pi =4i\pi$ . That is, $2=4$ .

Here, the issue is a failure of understanding clearly what taking the exponential or a logarithm of a complex number really means.

Now that I have laid out the issue, here is my outline for this post. My aims are:

Explain what $i$ "actually" is.
Lay out what $\sqrt{\cdot}$ and $e^{\cdot}$ and $\log(\cdot)$ really mean, in the complex plane, and in doing so, resolve the above paradoxes.
Along the way, I'll try to make a case for my perhaps controversial opinion that $i \neq \sqrt{-1}$ . I am calling it an "opinion" beacuse I admit that it is highly disputable. If it sounds like nonsense to you, you can totally skip that, and hopefully you'll learn some cool stuff in this post regardless!

The fallacy of $i := \sqrt{-1}$

"But isn't that the last bullet of your outline?" Not quite! My last goal was to argue against $i = \sqrt{-1}$ . Here, I will simply make an easy argument against $i := \sqrt{-1}$ , i.e. "the definition of $i$ is $\sqrt{-1}$ ".

As I mentioned before, negative numbers do not have square roots. So the attitude that people tend to have is that we will simply pretend that $\sqrt{-1}$ exists, and call it $i$ . But does that really make sense? You can't just take something that doesn't exist 'a priori' and then suddenly declare that it exists. This is like saying that unicorns have never been observed, but we will just pretend that they exist in order to benefit scientific study. Or, perhaps, this is like saying " $1/0$ does not exist, but we can just pretend that it exists and call it $\text{Cow}$ ". Math and logical reasoning simply does not work like that!

At its core, the assertion that $i$ is defined to be $\sqrt{-1}$ does not answer the question of what $i$ is, and neither does the assertion that $i$ is some magical "number" that, when squared, gives you $-1$ (but this does make me slightly happier, as we shall see soon).

So what, then, does it mean to come up with a "definition" for $i$ ? It means that we have to point at an existing mathematical object and call it $i$ . The mood here is similar to that of my real numbers post, where we came up with a definition for real numbers by essentially "playing God" with math. Starting from sticks and stones, we can bring into the mathematical world the concept of integers, rationals, and then the real numbers, by simply looking at the constructs and tools that we already have and using them an interesting way to "create new definitions". What we need to do for $i$ is essentially the same.

It turns out that, provided that we have the real numbers $\mathbb{R}$ and the ability to take ordered pairs $(x,y)$ of numbers, it is extremely easy to define what $i$ and complex numbers are.

Definition (Complex Numbers): The complex numbers $\mathbb{C}$ is literally just $\mathbb{R}^2$ , i.e. the set of all ordered pairs $(x,y)$ of real numbers. The value $i$ is defined to be the ordered pair $(0,1) \in \mathbb{C}$ . $\mathbb{C}$ is endowed with the following operations:

Addition: For complex numbers $(a,b)$ and $(c,d)$ , their sum is simply
$(a,b) + (c,d) := (a+c,b+d).$
Multiplication: For complex numbers $(a,b)$ and $(c,d)$ , their product is given by
$(a,b) \cdot (c,d) := (ac-bd,ad+bc).$

You might initially throw up at how contrived "multiplication" is, and indeed it appears to be unmotivated and ugly without viewing $i$ as $\sqrt{-1}$ , but just hold that thought for a minute so that we can appreciate how completely robust this definition is! There are no logical, technical, or rigorous issues here. We defined the complex numbers by simply pointing at something that already exists, named $\mathbb{R}$ . And, tacking on "operations" on $\mathbb{C}$ cannot be objected to, since morally we're simply defining functions $+,\cdot:\mathbb{C} \times \mathbb{C} \to \mathbb{C}$ .

Now, do note that we call these operations "addition" and "multiplication", but that doesn't immediately mean that they will necessarily satisfy properties that $+$ and $\times$ "typically" satisfy, like commutativity, associativity, and distributivity. These must be proven manually from this "ugly" definition, which is certainly a major disadvantage of this approach. Thus, I would absolutely not use this approach to introduce the concept of complex numbers to someone. It's just so much easier to play with things like $2+3\sqrt{-1}$ when doing algebraic manipulations.

...Speakiing of things like $2+3i$ , how exactly do we fit this into our rigorous framework? Remember, we've defined complex numbers to be ordered pair, and in particular we have $i := (0,1)$ . So if we write something like $2+3i$ , we're really writing $2+3(0,1)$ . Now, first issue: What exactly is $3 \cdot (0,1)$ ? We've never really defined it so it doesn't make sense. Never mind that though, suppose we got around that and somehow got the "obvious" result $3 \cdot (0,1) = (0,3)$ . Now we have to evaluate $2+(0,3)$ . What is that??? You can't add a real number to an ordered pair of numbers, that makes no sense. It's like mixing water and oil!

The solution to both of these problems is stupid, and we call it an "abuse of notation".

Abuse of Notation: We identify the $x$ -axis of $\mathbb{R}^2 = \mathbb{C}$ with the real numbers $\mathbb{R}$ . That is, whenever we see the ordered pair $(x,0)$ , we view it as the same exact thing as the number $x$ . Conversely, whenever we see the number $x$ , we view it as the exact same thing as $(x,0)$ .

Now now, this might sound absolutely horrifying (and it should!). Perhaps it might even sound completely un-rigorous! No worries though: The mindset to have behind "abuses of notation" is that we don't have to do them. They just make life a lot easier. Indeed, we can do complex number math just fine by writing them as ordered pairs all day long. But then easy equations like $i^4 = 1$ will start looking needlessly long, like $(0,1)^4 = (1,0)$ . It also makes algebraic manipulations with complex numbers much less intuitive. So, we cut back on the formality to make math easier.

Do not be pressured to do any of the following exercises. I write them here merely to point out that these are facts that aren't immediately obvious from our rigorous definition of complex numbers.

Exercises:

(The key property of $i$ ) Prove that $i^2 = -1$ .
(Complex numbers can be written in the "standard" way) Let $a,b \in \mathbb{R}$ . Prove that $a+bi = (a,b)$ .
(Complex numbers scale in the obvious way) Let $a,b,c \in \mathbb{R}$ . Prove that $a(b+ci) = ab+aci$ .
(Multiplication behaves how we expect it to behave) Let $u,v,w \in \mathbb{C}$ . Prove that $uv = vu$ . Prove that $(uv)w = u(vw)$ .
(Multiplication interacts with addition the way we expect it to) Let $u,v,w \in \mathbb{C}$ . Prove that $u(v+w) = uv + uw$ .
Let $z$ be a complex number. Come up with a reasonable definition for " $-z$ ". Come up with a reasonable definition for " $1/z$ " when $z \neq 0$ .
Let $w,z$ be complex. Come up with a reasonable definition for " $w-z$ ". Come up with a reasonable definition for " $w/z$ " when $z \neq 0$ .
What does $\overline{z}$ mean?

Alright, we've defined $\mathbb{C}$ rigorously. We know that $i^2=-1$ . Now it's time to pop the question: Is it true that $\sqrt{-1} = i$ ? The answer is... that this question is ill-posed! As of right now, this function $\sqrt{\cdot}$ is only defined over non-negative real numbers. There is no sense in writing $\sqrt{-1}$ when $\sqrt{x}$ is literally not defined when you plug in $x=-1$ .

Now, there is not too much stopping you from making the definition $\sqrt{-1} := i$ . (Note that this is different from saying $i := \sqrt{-1}$ ; here, $i$ is the object that we are familiar with and $\sqrt{-1}$ is the object that does not yet exist, and $\sqrt{-1} := i$ says that we bring $\sqrt{-1}$ into existing by saying that it is exactly $i$ .) But before we do something so rash, we need to ask ourselves some key questions: Is this a useful definition, in that it satisfies "good" properties? Is it a natural definition?

"Square Root": One side of the coin

Before things truly go off the rails, let's discuss one way to define a square root on $\mathbb{C}$ .

Let's begin with re-examining how we came up with the square root over non-negative reals. For $x \geq 0$ , the goal is to find some real number $y$ for which $y^2 = x$ . As you are well aware, there are generally two such values of $y$ : One positive and one negative. Which do we choose to be the value of $\sqrt{x}$ ? We certainly cannot take it to be both such values; to say that $\sqrt{4} = 2 = -2$ would be ludicrous. For the square root to be a well-defined function, we are allowed to only select one possible value for each possible input!

Well, since positive numbers are in some sense more "pure" and "natural" than their negative counterparts, it is certainly sensible to choose the positive branch. This also ensures that the identity $\sqrt{xy} = \sqrt{x} \cdot \sqrt{y}$ holds for all $x,y \geq 0$ .

Extending the square root to $\mathbb{C}$ experiences a similar issue. You probably know, for example, that $\sqrt{2i}$ has "two sane values": $1+i$ and $-1-i$ . Which do we choose? Which one is a more natural choice? If you had to ask me, I would go with $1+i$ without too much hesitation. This is because $1+i$ and $\sqrt{2i}$ lie on the same "side" of the complex plane. To be precise, we can split the complex plane along the real line, to divide into two "halves". A complex number $z \in \mathbb{C} \setminus \mathbb{R}$ would then lie in one of these halves, and there would be exactly two values of $w$ that would satisfy $w^2 = z$ : One that lies in the "upper half" (i.e. has positive imaginary part), and one that lies in the "bottom half" (i.e. has negative imaginary part). It is logical to pick the value of $w$ that lies in the same half as $z$ .

This also makes sense from a perspective of polar form. A somewhat advanced reader may be aware that there is a little trick for dealing with complex numbers and multiplication/exponents: View it not as a number formed by a real part and an imaginary part. Rather, view all complex numbers as formed by a magnitude and an angle. (i.e. write in the form $re^{i\theta}$ ... which is a can of worms that we shall open up in the next section.) I won't go over how this works since that's not quite the main point of this post, but if you're interested then I highly recommend googling it (it's extremely useful).

Anyway, when viewing complex numbers in this way, it is dead simple to multiply complex numbers: You simply multiply their magnitudes and add their angles! Similarly, to square a complex number, you would square its magnitude and double its angle. So, the natural assumption then is that to take the square root of a complex number, you square root its magnitude and halve its angle... right?

Wellll there's a bit of a weird issue. If a complex number $z$ has magnitude $r$ and angle $\theta$ , then there are multiple angles $\varphi$ that, when doubled, would give you $\theta$ . This may sound incredibly false until you remember that angles are all taken mod $2\pi$ ... for instance, $2i$ has magnitude $2$ and angle $\pi/2$ . Now, $\pi/4$ is one obvious angle that gives $\pi/2$ when doubled. But another one is $\pi/4 + \pi = 5\pi/4$ . When doubled, you get $5\pi/2$ , which is the same as $\pi/2$ (up to a difference of $2\pi$ ). So there is a non-uniqueness issue! Fortunately this can be solved without too much trouble: Just give a very concrete procedure for halving the angle in a unique way. But what's the most natural procedure?

Let's try one "obvious" way.

Attempt 1

Start with a complex number $z$ .
Find its angle, $\theta$ , and restrict it to $[0,2\pi)$ .
Then the halved angle is $\theta/2$ .

Is this a mathematically natural function? Let's plot some complex numbers and take their square roots.

$[asy] real depth = 100; void exparrow(pair z, pen q){ draw(z+2*dir(45)*((z^(1+10/depth)-z))--z,p=q); draw(z+2*dir(-45)*((z^(1+10/depth)-z))--z,p=q); for(int i = 0; i < depth; ++i){ draw(z^(1+i/depth)--z^(1+(i+1)/depth),p=q); } dot(z^2,p=q+linewidth(5)); } size(7cm); draw((-4,0)--(4,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-4)--(0,4),arrow=Arrows(5),p=linewidth(1.5)); //i, 2*cis(60), 3*cis(135) exparrow(dir(45),red); exparrow(sqrt(2)*dir(30),orange); exparrow(sqrt(3)*dir(135/2),yellow); exparrow(-dir(-45),green); exparrow(-sqrt(2)*dir(-30),blue); exparrow(-sqrt(3)*dir(-135/2),purple); [/asy]$

At first sight, this might seem ok. However, I don't like it! This is because this lacks symmetry. As suggested by the diagram, we're always choosing the "clockwise" root, when there is no natural reason to do so. The issue lies in the first step: It is actually not natural to restrict to $[0,2\pi)$ . This is biased! In fact, it is more symmetrical to take $(-\pi,\pi)$ .

Attempt 2

Start with a complex number $z$ .
Find its angle, $\theta$ , and restrict it to $(-\pi,\pi)$ .
Then the halved angle is $\theta/2$ .

Now let's plot some square roots under this procedure.

$[asy] real depth = 100; void exparrow(pair z, pen q){ draw(z+2*dir(45)*((z^(1+10/depth)-z))--z,p=q); draw(z+2*dir(-45)*((z^(1+10/depth)-z))--z,p=q); for(int i = 0; i < depth; ++i){ draw(z^(1+i/depth)--z^(1+(i+1)/depth),p=q); } dot(z^2,p=q+linewidth(5)); } size(7cm); draw((-4,0)--(4,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-4)--(0,4),arrow=Arrows(5),p=linewidth(1.5)); //i, 2*cis(60), 3*cis(135) exparrow(dir(45),red); exparrow(sqrt(2)*dir(30),orange); exparrow(sqrt(3)*dir(135/2),yellow); exparrow(dir(-45),green); exparrow(sqrt(2)*dir(-30),blue); exparrow(sqrt(3)*dir(-135/2),purple); [/asy]$

I'd say this is much nicer! Look at the symmetry! It's so pretttyyyyy! ...but do you see the issue?

Yeah that's right, what's $\sqrt{-1}$ ? This procedure, in its pursuit of pure symmetry, has left out the case in which the angle is an odd multiple of $\pi$ . This is because when we restricted the angle to $(-\pi,\pi)$ , we did not include either $\pi$ or $-\pi$ . Which should we include?

We can't include both of them, since then the angle's restricted value is not unique...
If we include neither of them, then $\sqrt{-1}$ is not defined...
If we include $\pi$ , so that the restricted angle is taken from $(-\pi,\pi]$ , then $\sqrt{-1} = i$ . The drawback is that we lose symmetry.
If we include $-\pi$ , so that $\sqrt{-1} = -i$ , then we run into a similar issue.

No matter how you spin it, something breaks! What now?

As far as I know, the two sane approaches are the 2nd and 3rd bullets.

Definition (Principal Square Root, "Totality Approach"): Let $z$ be a complex number, with magnitude $r$ and angle $\theta \in (-\pi,\pi]$ . Then $\sqrt{z}$ is the complex number with magnitude $\sqrt{r}$ and angle $\theta/2$ .

Definition (Principal Square Root, "Symmetry Approach"):

If $z$ is a negative real number, then $\sqrt{z}$ is not defined.
If $z$ is a non-negative complex number, with magnitude $r$ and angle $\theta \in (-\pi,\pi)$ , then $\sqrt{z}$ is the complex number with magnitude $\sqrt{r}$ and angle $\theta/2$ .

While a bunch of calculators (like Google's searchbar) and the Wikipedia article enjoy the "totality appoach" (that's totally a name I made up by the way), the "symmetry approach" is not without its proponents, because I exist. Maybe I'm alone in this "opinion", but I really do think that $\sqrt{-1}$ simply should be left undefined, because I really like symmetry. To have a value on the negative real line as a "special case" just seems so arbitrary to me.

Maybe you agree with this, and maybe you don't. Whatever the case, if you're still somehow reading this, I do have at least one other reason for my opinion. To properly explain it, we must start by addressing the elephant in the room...

What are complex exponents?

You've probably seen $e^{i\pi} = -1$ a hundred times by now. You've probably also asked yourself "what the heck does it mean to raise $e$ to an imaginary number" another hundred times. A common "proof" of this identity is to take the power series
$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots$ and then plug in $x = i\pi$ . The right side will be well-defined because we can always multiply and add complex numbers, so we can just take $e^{i\pi}$ to be whatever that right side comes out to (which is $-1$ ). Though, this still doesn't really address what it means to raise $e$ to a complex power!

So what does it mean? Unfortunately, I do not think these is any true intuitive way to think about it. After all, it's complete nonsense to try and multiply $e$ by itself $i$ times, or whatever. So... believe it or not, we literally define $e^z$ to be the power series!

Definition:Let $z$ be a complex number. Then $e^z$ is the complex number which we define to be the infinite sum
$e^z := \sum_{k=0}^\infty \frac{z^k}{k!}.$

This might sound absurd. But if you think really hard, it might not be so far-fetched! Think about it:

Exponents like $e^3$ are perfectly sane: It's just $e \times e \times e$ .
Exponents like $e^{2/3}$ are also fine. You first just take $e \times e$ , and then you find the cube root (which, as a real number, exists!).
...what the heck does something like $e^{\sqrt{2}}$ mean? How would you compute it?

Really, we run into a similar crisis when trying to define real exponents! Intuition doesn't really help much here either, since honestly, it's kinda nonsense to take $e$ and multiply it by itself $\sqrt{2}$ times. The way we resolve this is to take what we know - that we can have rational exponents - and use this to extend to real exponents by virtue of approximation. So, for example, $e^{\sqrt{2}}$ would be the value approached by $e^{1.4},e^{1.41},e^{1.414},\cdots$ .

So, to resolve complex exponents, we take what we know - that real exponents are a thing, and we have a power series $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$ - and use this to naturally extend to complex exponents. That is precisely what has happened here.

Now, as with how we defined the complex numbers, this definition of $e^z$ has some funny quirks:

The "crazy theorem" $e^{i\pi} = -1$ becomes almost completely trivial, once we recognize that $e^{i\pi}$ is nothing but $\cos \pi + i\sin\pi$ .
It is not obvious that $e^we^z = e^{w+z}$ . Remember, we defined $e^z$ as a power series, not as an exponent! So, we don't actually yet know that it satisfies the nice properties that real exponents do. Fortunately, this equality is true, and you can show this.
(Is it obvious that $e^z$ is well-defined? That is, can you find an easy argument for why the power series converges over complex numbers?)

Great. We now have a definition for $e^z$ , where $z$ is a complex number. But what is $w^z$ , where $w,z$ are both complex numbers? Well, we want to try to reduce this to just the case where $w = e$ , so here's a pretty "natural" approach.

Definition (Complex Exponents): $w^z := e^{z\log w}$

(Note: I do not like writing $\ln$ , I tend to prefer $\log$ . You might experience the same shift in preference when you go through college!)

Ok cool!

...wait a minute, we've turned one problem into another! What the heck is $\log w$ where $w$ is a complex number?

The woes of taking logs

Unfortunately, the answer is... not pretty at all.

For our definition $w^z := e^{z \log w}$ to have any hope of making sense, the property that " $\log$ " must satisfy is
$e^{\log z} = z.$ How in the world do we go about finding some complex number $w$ for which $e^w = z$ ? That sounds hard, and it is hard. Fortunately some smart people figured out how to do this... kinda. Let's be inspired by an interesting property that logs over real numbers satisfy, namely, that it is the integral of $1/x$ :
$\log x = \int_1^x \frac{1}{y}\,dy$ So, "naturally", we can try to define $\log z$ as
$\log z \stackrel?= \int_1^z \frac{1}{w}\,dx$ for $z$ a complex number. Now, this does not make sense as a definition, because what does it mean to integrate on a complex plane with these "complex endpoints"? Well, it turns out that you can integrate on a complex plane, but it's not as simple as "find an antiderivative and plug in the endpoints" (that sometimes works, but only in nice conditions). Rather, when we're in $\mathbb{C}$ , we take integrals over paths.

I'm not going to explain exactly what this means, but don't worry. You don't have to understand every little detail of this post. If you're still here, just hold out for a bit longer and believe me when I say that you can integrate over paths in the complex plane, and that that's basically how we do integrals. Then our attempted definition for $\log z$ would, more properly, be
$\log z := \int_\gamma \frac{1}{w}\,dx,$ where $\gamma$ is a path from $1$ to $z$ . Notice that I haven't quite specified what path to use: Is it a straight line? Is it a wavy line? Is it some weird curve?

Hm. Taking " $\log i$ " as an example, let's try a few paths in an attempt to find a good value for $\log i$ . One path from $1$ to $i$ is a straight line.

$[asy] size(7cm); draw((-2,0)--(2,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-2)--(0,2),arrow=Arrows(5),p=linewidth(1.5)); draw((1,0)--(0,1),p=purple,arrow=Arrow); dot("$1$",(1,0),S,p=red+linewidth(5)); dot("$i$",(0,1),W,p=blue+linewidth(5)); label((1.5,1.5),"$\gamma$",p=purple); [/asy]$

If we choose this straight line to be $\gamma$ , then it turns out that we get $\int_\gamma \frac{1}{w}\,dw = \frac{i\pi}{2}$ .

How about something curved, like a quarter-circle?

$[asy] size(7cm); draw((-2,0)--(2,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-2)--(0,2),arrow=Arrows(5),p=linewidth(1.5)); draw(arc((0,0),1,0,90,CCW),p=purple,arrow=Arrow); dot("$1$",(1,0),S,p=red+linewidth(5)); dot("$i$",(0,1),W,p=blue+linewidth(5)); label((1.5,1.5),"$\gamma$",p=purple); [/asy]$

Interestingly, we still get $\int_\gamma \frac{1}{w}\,dw = \frac{i\pi}{2}$ . Could this be a good value to assign to $\log i$ ...?

Ok what if we drew something crazy, like this?

$[asy] size(7cm); draw((-2,0)--(2,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-2)--(0,2),arrow=Arrows(5),p=linewidth(1.5)); draw((1,0)..(.5,1.5)..(-1,1)..(0.1,0.3)..(1.5,0.5)..(1,1)..(0.5,0.5)..(-.5,1)..(0,1),p=purple,arrow=Arrow); dot("$1$",(1,0),S,p=red+linewidth(5)); dot("$i$",(0,1),W,p=blue+linewidth(5)); label((1.5,1.5),"$\gamma$",p=purple+fontsize(20)); [/asy]$

Amazingly, we still get $\int_\gamma \frac{1}{w}\,dw = \frac{i\pi}{2}$ ! Wait, so if $\int_\gamma \frac{1}{w}\,dw$ is the same for every possible path $\gamma$ , then $\log i$ would be well-defined!

That's so cool! As a last sanity check, what if we try this path?

$[asy] real depth = 100; size(7cm); draw((-2,0)--(2,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-2)--(0,2),arrow=Arrows(5),p=linewidth(1.5)); pair z = 1.5^(1/depth)*dir(360/depth); for(int i = 0; i < depth; ++i){ draw(z^i--z^(i+1),p=rgb(1-i/depth,0,i/depth)); } draw((1.5,0){up}..(0,1){left},p=blue); dot("$1$",(1,0),S,p=red+linewidth(5)); dot("$i$",(0,1),W,p=blue+linewidth(5)); label((1.5,1.5),"$\gamma$",p=purple+fontsize(20)); [/asy]$

Then we get $\int_\gamma \frac{1}{w}\,dw = \frac{5i\pi}{2}$ . Coo- wait WHAT???

Oh no... this isn't what we've been getting before! ...did looping around $0$ break something?? What if we loop twice?

$[asy] real depth = 100; size(7cm); draw((-2,0)--(2,0),arrow=Arrows(5),p=linewidth(1.5)); draw((0,-2)--(0,2),arrow=Arrows(5),p=linewidth(1.5)); pair z = 0.6^(1/depth)*dir(720/depth); for(int i = 0; i < depth; ++i){ draw(z^i--z^(i+1),p=rgb(1-i/depth,0,i/depth)); } draw((0.6,0){up}..(0,1){left},p=blue); dot("$1$",(1,0),S,p=red+linewidth(5)); dot("$i$",(0,1),W,p=blue+linewidth(5)); label((1.5,1.5),"$\gamma$",p=purple+fontsize(20)); [/asy]$

It turns out that this path integral comes out to $\int_\gamma \frac{1}{w}\,dw = \frac{9i\pi}{2}$ . Blegh.

Hmm, perhaps we can figure out what's going on with some educated guessing (which will all be very correct because I'm writing it

Every time we loop around $0$ counter-clockwise, the value of the path integral increases by $2\pi i$ . This probably makes sense: The integrand $1/w$ induces smoe "spin" around $0$ , and looping around $0$ will "collect a full spin".
If, however, we do not loop around $0$ (or try to go around it "the other way"), then somehow by some magical mathematics, we keep getting the value $i\pi/2$ .

So, we got different answers for " $\log i$ " because of the loops that went around $0$ . Thus, in general, if we want $\log z$ to be well-defined, then we must restrict the domain of $\log z$ in such a way that it is impossible to draw loops around $0$ .

Finally, we are ready to reveal what a complex log is. (If you don't know what "open" means, don't worry about it.)

Definition (Complex Logarithms): Let $\Omega \subseteq \mathbb{C}$ be an open set satisfying the following conditions:

$1 \in \Omega$
$0 \not\in \Omega$
For any $z \in \Omega$ , there is a path starting from $1$ and ending at $z$ that always stays in $\Omega$ .
$\Omega$ does not "loop around $0$ ".

Then the complex logarithm whose domain is $\Omega$ is given by
$\log z := \int_\gamma \frac{1}{w}\,dw,$ where $\gamma$ is a path from $1$ to $z$ that stays in $\Omega$ .

...as usual, this has some weird but fascinating quirks.

This implies that there are actually infinitely many ways to define a logarithm, one for each possible selection of the domain $\Omega$ .
There is no logarithm whoes domain is $\mathbb{C} \setminus \{0\}$ (where it is natural to exclude $0$ because $\log 0$ isn't even a thing for real numbers). Indeed, $\Omega = \mathbb{C} \setminus \{0\}$ does not satisfy the conditions because it allows you to spin around $0$ while staying in $\Omega$ .
On the other hand, the slit plane $\Omega = \mathbb{C} \setminus (-\infty,0]$ DOES satisfy the conditions! If you can't cross the negative reals, then you can never loop around $0$ . So there is a log on this domain $\Omega$ .

Definition (Principal Logarithm): The logarithm whose domain is the slit plane is the principal logarithm.

The slit plane is the biggest symmetrical domain we can hope to define a logarithm over, where by "a logarithm", we mean "an antiderivative of $1/z$ . So you cannot reasonably take the log of negative numbers if you hope to preserve the key property with the integral. Hm... does that sound familiar?

Back to Square Roots

What is $\sqrt{z}$ ? It's just $z^{1/2}$ . But what does that mean? Well, by definition of exponentiation, I suppose it's just $e^{\frac12\log z}$ . But what is $\log z$ ? The most natural log to take here is the principal one. So, here is the second argument I wish to make for my opinion: Since the principal log is not defined on negative numbers, neither should the square root.

Resolving the paradoxes

It turns out that $\log(re^{i\theta}) = \log(r) + i\theta$ when we take the principal logarithm and restrict $\theta$ properly, as we might expect from log. However, it is not true in general that $\log(wz) = \log w + \log z$ . We can break this because we can spin around $0$ enough to make things bad. For example, if we take $z = e^{\frac{2i\pi}{3}}$ , then $\log(z^2) = \log(e^{\frac{4i\pi}{3}}) = -\frac{2\pi i}{3}$ , whereas $2\log z = \frac{4\pi i}{3}$ , so we do not have $\log(z^2) = 2\log z$ .

This also explains why $\sqrt{wz} \neq \sqrt{w}\sqrt{z}$ in general. Spiritually, trying to take a square root can be reduced to trying to take a log, and since the log doesn't satisfy $\log(wz) = \log w + \log z$ , we cannot expect that $e^{\frac12\log(wz)} = e^{\frac12\log w + \frac12\log z}$ , i.e. $\sqrt{wz} = \sqrt{w}\sqrt{z}$ .

This fully resolves the first paradox in a very overkill way, but what about the second paradox? For that, we can reason that the identity $\log(re^{i\theta}) = \log(r) + i\theta$ should only hold for $\theta$ restricted in $(-\pi,\pi)$ (or $(-\pi,\pi]$ , if you prefer that philosophy). This is because if we imagine the slit plane, we can only "go up to $\pi$ counter-clockwise or $\pi$ clockwise around $0$ ".

Final Notes

My definition for the principal log is different from what most people would define it as. Indeed, most people happily include the negative reals, so that in some sense its behavior on the negative reals is akin to how it behaves on the upper half of the complex plane, similar to how people deal with square roots of negative reals.
You might be wondering why we would ever use a log other than the principal log (however you wish to define that). It turns out that there actually are some proofs in complex analysis that use a domain for a log that isn't the slit plane!
There are other things I don't like about defining the square root of the negative numbers. In particular, it won't be continuous if we do that.
How can we, say, reconcile the undefinedness of $\sqrt{-1}$ with convenient facts such as the quadratic formula? Well, we can just interpret the square root in the quadratic formula as "a square root", not the "most natural" square root.
There is another really weird approach for dealing with complex numbers and exponents. Basically, you make a cut along the negative reals, and then stack infinitely many complex planes on top of each other, glued along the slits, to make an infinite spiral. Google "Riemann surface".

Whether or not you agree with my opinion, I hope this was a fun read! (I once again spent way too much effort on a blog post.)

This post has been edited 4 times. Last edited by greenturtle3141, Mar 11, 2025, 1:12 AM

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