I'm learning that imaginary numbers are just a way of representing a rotation. So, are imaginary numbers and multiple variable numbers $(x,y,z,...)$ just two different tools for representing numbers in multiple dimensions? Or are imaginary numbers still part of the same tool where $(x,y)$ represent points on a plane, and I'm misunderstanding their utility?

So it doesn't make sense to talk about calculus with complex numbers, since calculus is a tool to deal with functions of multiple variables?

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    $\begingroup$ I actually like this question, I've been asking it to my self many times. However I will let the intelligent people answer :) $\endgroup$
    – Zubzub
    Aug 10, 2016 at 14:13
  • $\begingroup$ This question may be of interest. $\endgroup$
    – user307169
    Aug 10, 2016 at 14:18
  • $\begingroup$ $(a,b)+(c,d)=(a+c,b+d)$ is in a sense natural, but $(a,b)\times (c,d)=(ac-bd,ad+bc)$ is a specific property of complex numbers $\endgroup$
    – Henry
    Aug 10, 2016 at 14:21

1 Answer 1


calculus is a tool to deal with functions of multiple variables

You probably mean "multivariate calculus", because ordinary calculus deals with single variables just fine.

There are several valid ways of looking at the complex numbers. One of the ways is to se $\mathbb C$ (the set of all complex numbers) as a vector space over $\mathbb R$. In that case, yes, you can look at $\mathbb C$ the same way as $\mathbb R^2$.

But this view misses two important points, one "positive" one not:

  • Plus : $\mathbb C$ is much more than just $\mathbb R^2$. Because you can multiply and divide complex numbers in a particular way, the complex numbers form what is called a field, making them very useful for all sorts of things. Furthermore, in complex numbers, you can define a complex derivative of a function similar to that in the real numbers, but having much nicer properties (see holomorphic functions). This, along with some other nice properties, is the reason that there is an entire subfield of analysis entirely devoted to functions of complex variables (complex analysis) So, you could say that $\mathbb C$ is a very rich version of $\mathbb R^2$.
  • Minus: This is only true for $2$. You cannot "enrich" $\mathbb R^3$ (this is a very complicated mathematical result called the Frobenius theorem) in a similar way, and you can only partially do it with $\mathbb R^4$ (see quaternions).
  • $\begingroup$ Great answer, thank you. That's exactly what I was looking for. $\endgroup$
    – BoddTaxter
    Aug 10, 2016 at 14:20

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