3
$\begingroup$

This question came to mind when I saw a way of finding the value of $i^i$ which included transforming it to $e^{i\ln(i)}$ and taking $\ln(i)$ as $i\frac{\pi}{2}$

I understand how we can get $\ln(i)$=$i\frac{\pi}{2}$ geometrically but this got me thinking if $i$ is a positive number as the first thing I learned about $\ln$ since high school days is that it can't take negative numbers.

$\endgroup$
2

7 Answers 7

4
$\begingroup$

This answers quite a bit more than you asked, but hopefully it is useful in understand how logarithms of things that aren't positive reals can be defined and why it's not so simple.

Logarithms of complex numbers are a bit murky. To see why, let's look at a reasonable definition of a logarithm:

Define $y=\ln x$ so that $e^y=x$.

If you try to think of $\ln$ as a function that maps reals to reals, it should be pretty clear why you can only take logarithms of positive numbers, since $e^y>0$ for all real $y$.

However, once we start talking about logarithms in the complex plane, things get pretty messy. The first problem we encounter is something like

$$e^0=e^{2\pi i}=1.$$ So, judging by our definition, $\ln 1$ should be both $0$ and $2\pi i$, and in fact should be $2n\pi i$ for every integer $n$. In fact, since $e^{a+b}=e^ae^b$ is still a valid rule for complex exponentials, we have that $$y\text{ is a valid value of }\ln x \Leftrightarrow y+2\pi i\text{ is a valid value of }\ln x,$$ where we're using our definition above. One can show that, in fact, for every complex $x\neq 0$ there exists some $y$ so that $e^y=x$, and that any two such $y$ differ by an integer multiple of $2\pi i$.

How do we get around this? The standard way is to just pick your $y$ so the imaginary part is in a desired interval; this interval is usually $[0,2\pi i)$ or $(-\pi i,\pi i]$. This works if our only goal is to define a function, but we end up with nasty discontinuity issues. In sum, this works for some things and doesn't for others.

We also end up being led astray by examples such as $i^i$. Firstly, complex exponentiation is usually defined by the rule $$a^b=e^{b\ln a},$$ where we make some assumption that $\ln a$ is well-defined. This gets us into issues like the one you described; on one hand, $$i^i=e^{i\ln i}=e^{i\frac{i\pi}{2}}=e^{-\frac{\pi}{2}},$$ and on another, $$i^i=e^{i\ln i}=e^{i\frac{5i\pi}{2}}=e^{-\frac{5\pi}{2}}.$$ This seems bad, and mathematicians usually take great care to avoid these sorts of issues when dealing with complex exponentials.

In sum, the moral answer to your question is something like:

Logarithms of numbers that are not positive reals can be defined, but we have to give up some nice properties due to these logarithms having multiple values. In particular, complex exponentiation runs into some annoying problems.

Side note: complex exponentials and logarithms are usually formally defined using power series, I believe.

$\endgroup$
3
$\begingroup$

"Sign" only applies to real numbers as it depends on the truth value of the statements $x<0, x=0, x>0$. Ordering is implied by those statements.

The complex numbers have no linear ordering so you cannot say $z > w$ or $z <w$ for $z, w$ non-real complex numbers. Only equality applies.

And the purely imaginary numbers have the same "limitation". So $i$ cannot have a sign associated with it.

$\endgroup$
3
$\begingroup$

In the complex you can take the logarithm of negative real numbers.

$$\log(-1)=i\pi.$$

And $i$ has no sign.

$\endgroup$
2
$\begingroup$

$i=\sqrt{-1~}~$ is considered neither positive nor negative. The terms positive and negative are applied only on real numbers field. No one can place $~i$ on a number line that runs from negative to positive real numbers.

Although sometimes, parts of the imaginary numbers are referred to as positive imaginary and negative imaginary. For example, $~i~$ would be positive imaginary and $~-i~$ would be negative imaginary. The distinction is entirely conventional, and is not a discernible property of the numbers. Buy it’s not the same as “positive” and “negative” as applied to real numbers.



There is a basic difference of the definition of logarithm function in complex field and real field.

The real logarithm function $~\ln x~$ is defined as the inverse of the exponential function $~ y = \ln x~$ is the unique solution of the equation $~x = e^y~$. This works because $~e^x~$ is a one-to-one function; if $~x_1 \ne x_2~$, then $~e^{x_1}\ne e^{x_2}~$ . But this can't happens in the case for $~e^z~$; as $~e^z~$ is $~2πi~$-periodic, so all complex numbers of the form $~z + 2nπi~$, $($where $ n = 0 , ±1 , ±2 , ±3 , \cdots)$ are mapped by $~w = e^z~$ onto the same complex number as $~z~$.

For all nonzero complex number $~z~$, $~\text{Log}( z)~$ is defined by $~\text{Log}( z)=\ln(|z|)+i.\arg(z)~$, where $~\ln(z)~$ denotes real logarithm to the base '$e$'. Now $~|i|=1~$ and argument of $~i~$ has principal value $~\frac π2~$, and so the principal value of $~\text{Log}(i)~$ is $~\ln 1 + i.\frac π2 = \bf{i\frac π2}~$, as $~\ln 1 =0~$.
All the other values of $~\text{Log}(i)~$ are obtained by adding $~2nπi~$ with the principal value, where $~n~$ may be any integer.
Hence the set of all the values of $~\text{Log}(i)~$ is $~\{2n\pi +i\frac π2~ :~ n \in\mathbb Z\}~.$

$\endgroup$
1
$\begingroup$

What you are really asking (I assume if you dig down to it) is if there is a total ordering on $\mathbb{C}$. The answer to this is no. For instance, you can see this answer. Essentially, what you are asking for is a notion of positive and negative relative to $0$. But as the linked answer shows, no matter how you do this you will run into problems somewhere.

$\endgroup$
2
  • $\begingroup$ I'm not sure this is totally correct. Asaf Karagila shows a total order on $\mathbb{C}$ in (Question 1032257) and I think the order in the question you've linked is stronger than a total order. I believe what you mean to say is that $\mathbb{C}$ cannot be made into an 'ordered field'. $\endgroup$
    – Jam
    May 26, 2020 at 13:06
  • $\begingroup$ @Jam The OP did not state. The default assumption would be as a field, which it is not possible. Of course, you can give $\mathbb{C}$ the dictionary order you have on $\mathbb{R}^2$ but the point is this is just as a set. If you are just considering sets, every set can be given a total ordering (this is weaker than the Axiom of Choice). You could also give a total ordering to purely imaginary numbers. But the point is they do not preserve the field structure. Given the assumed level of the OP, I thought this was the most appropriate answer. $\endgroup$ May 27, 2020 at 3:02
1
$\begingroup$

The sign can be defined as: $$ sgn(x)=\begin{cases} \frac{x}{|x|}, & \text{if $x \neq 0$} \\ 0, & \text{if $x = 0$} \end{cases} $$ If you extend this to complex numbers, the sign can be any complex unit or zero. In that way $sgn(i)=i\neq0$ so it's not positive.

We usually don't define positive and negative numbers in the means of complex numbers. Some literature define it in the way I have shown above and then "$z$ is positive" implies that $\Im(z)=0$.

$\endgroup$
1
$\begingroup$

As other answers pointed out, there is no total ordering of $\mathbb C$, which is compatible with the one of the real numbers. If there was such an ordering, we could compare $i$ and $0$, so that either $i > 0$ or $i < 0$. In the first case, $i > 0$ implies that $i \cdot i = -1 > 0$, a contradiction. In the second case, $i < 0$ implies that $i \cdot i = -1 > 0$, too.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .