why is $i$ not like $0$?

Question

Consider the following proof: $$e^{i(\theta+2n\pi)}=e^{i(\theta)}$$ where $n$ is a non-zero integer $$ i(\theta+2n\pi)=i(\theta) $$ $$ \theta+2n\pi=\theta $$ $$ 2n\pi=0 $$ From the answers that I have read so far, it seems the error is introduced in the 2nd line, after taking the logarithm of both sides, as the logarithm in this case would be a multivariate function.

My question is: Could it be that an error is introduced in the 3rd line, after dividing by $i$? What if $i$ is like $0$, in that $$ 2\pi \times 0 = 0 \times 0 $$ i.e. what if $i$ had similar properties to $0$ in that $i \times a = i \times b$ for two distinct numbers $a$ and $b$ (except that it would be more restricted, in that the difference between $a$ and $b$ would be an integer multiple of $2\pi$)? Even if it is said that there is no proof that $i$ is as such, then what is the proof that it isn't? Why should $i$ be assumed to behave like non-zero real numbers when multiplying and dividing?

Answer 1

No. The only mistake you made is in the second line. In general you have $$ e^{z+2\pi i} = e^z, $$ because $e^{z+2\pi i} = e^z\cdot e^{2\pi i} = e^z\cdot 1 = e^z$.

Answer 2

The division by $i$ is not a problem. The problem is asserting that because

$$ \exp(i(\theta + 2\pi n)) = \exp(i\theta) \text{ for all } n $$

it follows that $$ i(\theta + 2\pi n) = i\theta \text{ for all } n . $$

The first statement is just the periodicity. You can see the same false argument with no mention of complex numbers if you think about $\sin$ (without the $i$) instead of $\exp$.

The same false argument comes up whenever a function is not injective. The fact that $ 1^2 = (-1)^2$ does not imply $1 = -1$.

Answer 3

The powre rule that you are using does not apply in all cases.

Note that $e^{2\pi i } =1$ and you can multiply a number by one without changing the answer.

For example we have $1^4=1^5$ but $4\ne 5$

Also you have $i^4= i^8$ while $4\ne 8$

Or $(-1)^4= (-1)^2$ while $4\ne 2$

Thus one has to be careful when usign these rules.

Answer 4

Your actual question seems to be about "Why can we divide by $i$?" and "Why should i be assumed to behave like non-zero real numbers when multiplying and dividing?"

We define division by complex numbers $(a+bi) / (c+di)$ to be $\dfrac{(ac+bd)+(bc-ad)i}{c^2+d^2}$.

This happens to agree with our usual definitions of multiplication and division and has all of the nice properties that we expect.

In particular, this operation is well defined for every $(a+bi)$ and $(c+di)$ with the exception of when $c=d=0$, and this implies then that we can cancel common factors of nonzero complex numbers that are common to both sides of an equation. That is, if $zu=zv$ with $z$ a nonzero complex number this implies that $u=v$.

Answer 5

you have to realize that what you are dealing with here is not the standard equality you have in the real numbers.

when talking about the argument of complex numbers you have to note that since we have a period of $2 \pi$ in our angles you may have infinitely many pairs of numbers with distance from the origin $\mathrm r$ and angles $\theta_1,\theta_2\ldots\theta_n$ showing the same vector in the complex plane.

a more accurate way of talking about the equality above would be using the modulus:

$$\theta + 2\pi n \equiv \theta \pmod {2\pi} , 0\leqslant \theta \leqslant 2\pi $$

thats what gets most students, in short, powers don't work quite the same in the complex plane.

just like these two vectors here: $$ e^{i\pi},e^{i3\pi} $$

which point to the same number, $-1$.

Answer 6

Yes, the error is in the implication from the first to the second line. It doesn't follow at all that because $$e^x=e^y,$$ then $x=y,$ where $x,y$ are nontrivial complex numbers. This is simply because the complex exponential is not injective, so many complex values (indeed infinitely many) get mapped to the same value. Hence, we simply can't invert like you'd want in order to make your deduction.

Having settled that, you say

My question is: Could it be that an error is introduced in the 3rd line, after dividing by $i$?

Not at all. Dividing by $i$ is not a crime since $i\ne 0.$

What if $i$ is like $0$, in that $$ 2\pi \times 0 = 0 \times 0 $$ i.e. what if $i$ had similar properties to $0$ in that $i \times a = i \times b$ for two distinct numbers $a$ and $b$ (except that it would be more restricted, in that the difference between $a$ and $b$ would be an integer multiple of $2\pi$)?

The complex unit $i$ is not at all like $0.$ If anything, it is more like the real units $\pm1,$ than like $0.$ We do not have that $ia=ib$ for $a\ne b$ since otherwise we would have to say that two complex numbers in rectangular form with different coordinates are equal, which is not how we define equality of complex numbers.

Even if it is said that there is no proof that $i$ is as such, then what is the proof that it isn't? Why should $i$ be assumed to behave like non-zero real numbers when multiplying and dividing?

A proof has been given above. To add more assurance, we know what $0$ is and how it behaves, and we know what $i$ is and how it behaves. Their distinguishing behaviors are not in the least similar, so they can't be the same. Your question is like asking why we can assume $-1,$ for example behaves like a nonzero number. Well, the reason is because we define it to be so. We want the symbols $-1$ and $0$ to mean different things. Similarly, $i$ and $0$ mean different things.