Dividing Complex Numbers by Infinity [duplicate]

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My PreCalculus teacher recently reviewed the properties of limits with us before our test and stated that any real number divided by infinity equals zero. This got me thinking and I asked them whether a complex number (i.e. $$3+2i$$ or $$-4i$$) divided by infinity would equal zero.

This completely stumped them and I was unable to get an answer. After doing some theoretical calculation, knowing that $$i=\sqrt{-1}$$, I calculated that a complex number such as $$\frac{5i}{\infty}=0$$ since $$\frac{5}{\infty}\cdot \frac{\sqrt{-1}}{\infty} = 0\cdot 0 = 0,$$ using properties utilized with real numbers that would state that $$\frac{5x}{\infty} = 0$$ since $$\frac{5}{\infty}\cdot \frac{x}{\infty} = 0\cdot 0 = 0.$$ Is this theoretical calculation correct or is there more to the concept than this?

marked as duplicate by rschwieb, gen-z ready to perish, YuiTo Cheng, José Carlos Santos limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 26 at 8:57

• Possible duplicate of One divided by Infinity? The accepted answer talks complex numbers too, although it is moot. The main point is that throwing infinity in wrecks the field axioms, and it doesn't work. So whoever told you "a real number divided by infinity equals zero" is an underinformed person, or else did not know how to express the thing they actually wanted to express. – rschwieb May 25 at 23:42
• How do you define division by infinity at all? Within the realm of (finite) numbers, we define $a/b$ as the unique number $c$ such that $bc=a$. This cannot be extended to include division by $\infty$. – Hagen von Eitzen May 25 at 23:43
• I don't think it's a great idea to speak of dividing by infinity. You can't take an expression like $\frac 7{\infty}=0$ and deduce that $7=\infty \times 0$. Of course, it is true that if you take any real number and divide it by a very, very big real number you get something very near $0$. This is also true if you divide any complex number by a very big real number (or by a very big complex number). So, if that informal sense is what is meant, then I would agree that dividing any complex number by infinity yields $0$. When you study limits, you'll see better ways to speak of such things. – lulu May 25 at 23:45
• I simply tell people: “$\infty$ is a direction, not a number.” – gen-z ready to perish May 25 at 23:51
• Maybe the question is what happens when you divide a complex number by a very very very large real number? Something like limit of $\frac{i}{n}$ as n approaches infinity? – NoChance May 26 at 0:02

Infinity isn't a number. It's notation.

It's a shorthand for a precise definition.

The symbol $$\infty$$ doesn't denote any element of $$\Bbb{R}$$ or $$\Bbb{C}$$ (the real or complex numbers respectively), so when we say something like $$\frac{1}{\infty}=0$$, this is not division, it's shorthand for the statement $$\lim_{x\to\infty} \frac{1}{x} =0,$$ and once again we have another $$\infty$$ symbol, which means that this statement is also shorthand for the statement that

For every real number $$\epsilon > 0$$, there is some positive integer $$N$$ such that when $$|x| > N$$, $$\left|\frac{1}{x}\right| < \epsilon.$$

That said, we can consider whether this shorthand makes any sense for complex numbers.

Well, if $$a + bi$$ is some complex number, then $$|a+bi|=\sqrt{a^2+b^2}$$, so if we consider $$\frac{a+bi}{\infty} = 0,$$ this should be shorthand for the statement that

For every real number $$\epsilon > 0$$, there is some positive integer $$N$$ such that when $$|z| > N$$, $$\left|\frac{a+bi}{z}\right| < \epsilon.$$

This statement is still true. Given any $$\epsilon>0$$, we can find some integer $$N$$ large enough that $$\frac{|a+bi|}{N} < \epsilon$$, and then when $$|z| > N$$, we have $$\left|\frac{a+bi}{z}\right| = \frac{|a+bi|}{|z|} < \frac{|a+bi|}{N} < \epsilon.$$

Thus we can say that $$\frac{a+bi}{\infty}=0$$ for any complex number $$a+bi$$, but I want to emphasize again that this is not division. It's shorthand for a longer statement with a precise meaning.

• The OP did say this was a "precalculus" class. Unless they say otherwise, it must be assumed that they are not familiar with the notion of a limit, and certainly not with the rigorously defined limits. – grovkin May 26 at 8:29
• @grovkin That's why I gave careful statements of what was meant, without leaving it at just saying it's a limit. I realize that this answer may go slightly over the head of the average precalc student, but I hope that the gist of it will be conveyed clearly. – jgon May 29 at 12:27

For your purposes, you can treat $$i(=\sqrt{-1})$$ to be an unknown variable, $$x$$. While it is true that $$\infty$$ is a formal concept rather than a number, in order for that concept to be well-defined, it has to interact in a consistent way with all unknown variables (which are not equal to $$0$$ or $$\infty$$ because $$\infty$$ interacts in a special way with $$0$$ and $$\infty$$). And consistent means that equal things have to remain equal after the same operations are performed on them regardless of whether they are real or complex.

So $$\frac{2x}{\infty}$$ must be formally equal to $$\frac{2}{\infty}x$$ and it must be formally equal to $$2\frac{x}{\infty}$$. But $$\frac{2}{\infty} = 0$$, so $$\frac{x}{\infty}$$ must be formally equal $$0$$ in order for this treatment of $$\infty$$ to be consistent. But remember that $$x=i=\sqrt{-1}$$.

And now you can treat $$\frac{a+bi}{\infty}$$ as formally equal to $$\frac{a}{\infty}+ \frac{b}{\infty}i = 0+0\cdot i = 0$$.

So what do you I mean by "formally" equal? An object can be formally considered a number if the object's properties make all arithmetic operations on the object behave in the same way as these operations would behave on other numbers.

You already know some objects which you formally treat as numbers. Variables and functions are such objects. Imaginary numbers start out being formally treated as numbers. After some practice, it becomes matter-of-fact to treat complex numbers as actual numbers.

And while there are actual real-world concepts which can be easier expressed in terms of complex numbers, there is often a lot of initial resistance to treating them as such. But as long as they behave in the same way as real numbers do, it helps to think of them as numbers to simplify treatments of more complicated ideas.

Infinity is slightly different from complex numbers because it behaves as a number for most operations, but not all. But as long as you keep in mind the restrictions on when you can use simple arithmetic operations on $$\infty$$ and when you cannot, you can treat it as a number.