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What all limitations do complex numbers have, for example in terms of operations that cannot be done on them (or some other limitations encountered working with real numbers)?(am aware it is possible for $\cos x$, $\sin x$ to have values greater than 1, and am aware of proof and reason for that) Where one can find or read about all motives for their existence, and what are all things (also including everything or anything in "real life") that are impossible to do or imagine without using complex numbers? Is everything that can be done on complex numbers must be done on real since R^2 is bijective to C? Sorry for maybe confusing question. Thank you for reading.

slomil
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  • "all applications of..." Such a list will not end. That is like asking for a list of all uses of numbers in general. There are as many uses as our imaginations allow and then some. – JMoravitz Nov 28 '23 at 13:07
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    There is some trouble defining $\ln z$ where $z$ is a complex number, but it can be made no more restrictive than defining $\ln$ on the real numbers. You can search "complex logarithm and branch cut" to see an explanation. – walkar Nov 28 '23 at 13:09
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    With regards to what operations are defined for them... that also will be a never-ending list. Looking at the operations you know and looking to see if they are defined for arbitrary complex is the way to go. A few notable exceptions, the familiar $\sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{b}$ fails for complex numbers, similarly $a^{bc}=(a^b)^c$ can be false... we do not have $z!$ for arbitrary complex $z$ defined, but we can instead talk about $\Gamma(z-1)$... otherwise complex numbers form a field. We don't lose nearly as much as when we go to something like quaternions. – JMoravitz Nov 28 '23 at 13:13
  • Thank you. I have edited my post. – slomil Nov 28 '23 at 13:14
  • Real numbers are ordered: for any reals $x$ and $y$ you have either $x \leq y$ or $y < x$. And this order plays nice with other operations, for instance $x \leq y$ is equivalent to $2 x \leq 2 y$ and to $-x \geq -y$. But for complex numbers we don't have a nice notion of order (you could of course define any arbitrary order you want, such as lexicographical order on $(Re, Im)$, but it wouldn't play nice with other operations). – Stef Nov 28 '23 at 13:33

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Ordering

Real numbers are ordered: for any reals $x$ and $y$ you have either $x \leq y$ or $y < x$. And this order plays nice with other operations, for instance $x \leq y$ is equivalent to $2 x + 3\leq 2 y + 3$ and to $-x \geq -y$.

But for complex numbers we don't have a nice notion of order.

You could of course define any arbitrary order you want, such as lexicographical order on the Cartesian or polar coordinates, but it wouldn't play nice with other operations.

Patching functions smoothly

A function is called smooth if it is differentiable and its derivative is differentiable and its second derivative is differentiable, and so on forever.

Smooth real functions can be patched together smoothly: for instance, if you have two smooth functions $g : (-\infty, -1] \to \mathbb R$ and $h : [+1, +\infty) \to \mathbb R$, then it is always possible to find a function $f : [-1, +1] \to \mathbb R$ such that the following function $\mathbb R \to \mathbb R$ is smooth: $$x \mapsto \begin{cases} g(x) &\mbox{ if } x \leq -1 \\ f(x) &\mbox{ if } -1 \leq x \leq +1 \\ h(x) &\mbox{ if } x \geq +1 \end{cases}$$

However, this is not possible in $\mathbb C$. Instead, we have the following property in $\mathbb C$:

If $D$ is a small open disc in the complex plane and $f : D \to \mathbb C$ is a smooth function, then there exists one unique way to complete function $f$ into a smooth function on all of $\mathbb C$.

Exponents, logarithm, non-unicity of argument

A logarithm $\log : (0, +\infty) \to \mathbb R$ is a continuous function, and we enjoy nice properties $(a^b)^c = a^{bc}$ and $\log (ab) = \log a + \log b$.

Exponents could be extended to the real numbers: the expression $a^b$ could make sense if $a \neq 0$ and $a, b \in \mathbb C$, using the definitions $r e^{i \theta} = r(\cos \theta + i \sin \theta)$ and $a^b = e^{b \log a}$. However, we'd lose a few properties, and any reasonable extension of the logarithm must be discontinuous.

The only reasonable extension of the logarithm is something like $\log(z) = \log(|z|) + i \arg(z)$.

However, the argument $\arg (z)$ is neither unique nor continuous: if you take a full turn around the origin, you add $2\pi$ to the argument, and you're back at your starting point. Which means that as far as the argument is concerned, an angle of $2\pi$ and an angle of $0$ are the same thing... except of course $0 \neq 2\pi$, so we run into difficulties here.

For this reason, the logarithm cannot be continuous: no matter how we define the logarithm, if we continuously run around the origin, at some point the logarithm will have to make a $\pm 2\pi$ jump to get back on our feet. This is the same issue that Phileas Fogg encountered when his clock went out of sync by exactly 1 day while he was running around the world in Jules Verne's novel.

This also means that any definition of the logarithm will be somewhat arbitrary, as we have to decide where to put the discontinuity and which $2\pi$ offset to choose for the argument.

And most sadly, this also breaks our nice exponent properties: we have to be ready to accept that $\log(ab)$ might not be equal to $\log(a) + \log(b)$, and we have to accept that $(a^b)^c$ might not be equal to $a^{bc}$ if $b$ and $c$ are complex numbers.

For instance, if we tried to check whether $(e^{2\pi i})^{2\pi i} = e^{(2\pi i \times 2 \pi i)}$, we'd get: $(e^{2\pi i})^{2\pi i} = 1^{2 \pi i} = 1$, but $e^{(2\pi i \times 2 \pi i)} = e^{-4 \pi^2} \approx e^{-39.48} \approx 0.00000000000000000715$, so obviously something is wrong here.

Can you find out what is wrong in the previous paragraph? Hint: how is $1^{2 \pi i}$ defined?

Stef
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  • Thank you for answer. It is bit confusing to me and must reread it couple of times.. (am not sure how to determine what operations will not work as expected unless checking manually) – slomil Nov 28 '23 at 15:07
  • @slomil The first thing that does not work is that you can't have a continuous $\arg$ function that gives the polar angle of a complex number, because you need to go back to 0 after a full turn, but $2 \pi = 0$. And this issue extends to $\log$: you want $\log(r e^{i \theta}) = \log(r) + i \theta$, but $r e^{i \theta} = r e^{i (\theta + 2\pi)}$, and of course we cannot fix the inequality $\log(r) + i \theta \neq \log(r) + i (\theta + 2 \pi)$. So because of $r e^{i \theta} = r e^{i (\theta + 2\pi)}$, most properties of the logarithm break down. – Stef Nov 28 '23 at 15:16
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    @slomil Now, defining $e^z$ for complex $z$ is not a problem, but if you want something other than $e$ as the base, you need to use $\log$ in the definition; $a^b = e^{b \log a}$. So the properties of exponents break too. – Stef Nov 28 '23 at 15:18
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    @slomil Also, it's okay if you don't understand everything. Undegraduate students at university who follow a pure-math curriculum have to go through a course about complex analysis and they're all dazzled at the beginning; in particular, the consequences of extending the definition of derivatives $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ to the complex plane are not at all obvious (what is the meaning of $\lim$ when $h$ is allowed to be complex, and what are the implications of the existence of this limit? fun but relatively hard questions when you encounter them for the first time) – Stef Nov 28 '23 at 15:21
  • Btw, is there a definition for 1^i? (are there books about it) Is there definition for i^i for example?:) – slomil Nov 28 '23 at 18:37
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    @slomil You can define $a^b = e^{b \log(a)}$, so $1^i = e^{i \log(1)}$... But that relies on a definition of logarithm, which is only unique modulo $i 2\pi$. – Stef Nov 28 '23 at 18:45
  • Thank you very much @Stef .(do you perhaps have recommendation for a good book about it? In the meantime i will read the article on Wikipedia..:)) – slomil Nov 28 '23 at 20:32
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    @slomil Not right now but I can look for it – Stef Nov 28 '23 at 23:16