Strange contradiction in the formulation of negative logarithms

Question

$\ln((-1)^2)=\ln(1^2)$

Use log properties to bring the two outside the expression, cancel twos and remove logs and $$-1=1.$$ But $e^{i\pi}=-1$. Is this a contradiction?

Answer 1

No. As $\log{x}$ is only defined for $x > 0$ in the real case, you must mean the complex logarithm. The complex logarithm has branches, so that you must talk about the region that you are working with. To find out more, read the "Branches of the complex logarithm section" of the wiki article.

Answer 2

"and remove logs "

just how does that work?

That is if $\ln x = \ln y\implies x =y$? how do we know that?

That assumes that $\ln:\mathbb R^{+}\to \mathbb R$ is an injective function and for the reals it is.

But for the complex numbers it does not.

Not only that but because $e^{ix} = e^{ix + i2\pi k}$ we don't have $\ln z$ is a single-value function but actually $\ln z$ is an equivalence class with many values. (All of which if you raise $e$ to, would result in $z$.)

$\log -1 = [-\pi i]$ where $[-\pi i]$ is the equivalence class of all $\{-\pi i+ 2k\pi i\}$ and $\log 1 = [0]$ where $[0]$ is the equivalence class of all $\{0 + 2k\pi i\}$.

So $\log (-1)^2 = \log 1^2 \implies$

$2\log -1 = 2 \log 1 \implies$

$2[-\pi i] = 1[0]\implies$

$[-2\pi i] = [0]$ which not at all a contradiction as $[-2\pi i] =\{-2\pi i + 2k\pi i\} = \{0 + 2k \pi i\}$.

Answer 3

Since you talk about $\ln(-1)$, you are not talking about the real logarithm; you are talking about the complex logarithm.

The logarithm (to base $b$) is the inverse function of the exponential (with base $b$). That inverses are single valued, so solving an equation requires more than applying an inverse function, is very likely not new to you.

• From $x^2 = 9$, we know the solution is $x \in \{-3,3\}$, but $\sqrt{9}$ is only $3$ -- we have to know about the other root.
• From $\sin(\theta) = \frac{1}{2}$, we know $\theta \in \left\{\frac{\pi}{6}, \frac{5\pi}{6} \right\} + 2 \pi k$, for any integer $k$, but $\sin^{-1}\left(\frac{1}{2} \right)$ is only $\frac{\pi}{6}$ (the particular solution given when we choose the family of solutions containing the first element listed in the set and then take $k=0$).
• From $\mathrm{e}^x = 1$, we know $x = 2 \pi \mathrm{i} k$, for any integer $k$, but $\ln 1$ is only $0$ (the particular solution given when $k=0$).

In particular, $3^2 = (-3)^2$, but this does not mean $3 = -3$. (The jargon is "the squaring function is not one-to-one" or "the squaring function is not injective".) If we say this in words, it might be clearer: Just because two things square to $9$ does not mean the two things are equal.

Similarly, $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = \sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{13\pi}{6}\right) = \cdots$. But this doesn't mean that all those angles are equal, they just have the same sines. (The sine function is not injective.)

So just because $\ln x = \ln y$ doesn't mean that $x = y$. (The complex exponential is not injective.) The real exponential is injective, so we can peel off the "$\ln$"s and get equal things, if we have real logarithms, but $\ln(-1)$ cannot be the real logarithm.

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But there's so much more going on here...

Using what I said we know in the bulleted list, $$ \mathrm{e}^{\mathrm{i}\pi + 2\pi \mathrm{i} k} = \mathrm{e}^{\mathrm{i}\pi} \mathrm{e}^{2\pi \mathrm{i} k} = \mathrm{e}^{\mathrm{i}\pi}\cdot 1 = \mathrm{e}^{\mathrm{i}\pi} = -1 $$ So the solution (set) of the equation $\mathrm{e}^x = -1$ is $x = \mathrm{i}\pi + 2\pi \mathrm{i} k$, for any integer $k$.

This means, whatever $\ln(-1)$ is, it is one of $\mathrm{i}\pi + 2\pi \mathrm{i} k$, for some integer $k$. (One way of picking a branch of the logarithm is to pick a fixed value of $k$. This is not the only way to do so.) A common choice is $k=0$, giving $\ln(-1) =\mathrm{i}\pi$.

So let's run the directed computation using what we know. \begin{align*} \ln\left( (-1)^2 \right) &= \ln\left( (1)^2 \right) \\ 2 \ln\left( -1 \right) &= 2 \ln\left( 1 \right) \\ 2 \pi \mathrm{i} &= 2 \cdot 0 \\ 2 \pi \mathrm{i} &= 0 \text{,} \end{align*} which looks just as silly. But, $\mathrm{e}^{2 \pi \mathrm{i}} = 1 = \mathrm{e}^0$. So what happened? Well maybe, on the left, we have $k = 1$ and on the right we have $k = 0$. Let's do that again, but keep full solution sets, not just pretend that both computations lie on the same branch of the logarithm. (This is actually what happened -- when we squared $-1$, we doubled its complex argument (its angle) from $\pi$ to $2\pi$, which means we left the $k=0$ branch and landed on the $k=1$ branch. More about this below.) \begin{align*} \ln\left( (-1)^2 \right) &= \ln\left( (1)^2 \right) \\ 2 \ln\left( -1 \right) &= 2 \ln\left( 1 \right) \\ 2 (\pi \mathrm{i} +2 \pi \mathrm{i} k) &= 2 (0 + 2 \pi \mathrm{i} m) \\ 2 \pi \mathrm{i} + 4 \pi \mathrm{i} k &= 4 \pi \mathrm{i} m \text{,} \end{align*} where $k$ and $m$ are integers. (We have to use both $k$ and $m$ because it would be an unreasonable coincidence that the sets on each side of the equal sign would happen to be in lock-step just as we write them down. It is more likely that one is a shifted and/or shuffled copy of the other.) But we can write the expression on the left as $(1+2k)2 \pi \mathrm{i}$ and the expression on the right as $(2m)2 \pi \mathrm{i}$. That is, the object on the left is odd multiples of $2 \pi \mathrm{i}$ and the object on the right is even multiples. These aren't even equal as sets. Even so, $\mathrm{e}^{(1+2k)2 \pi \mathrm{i}} = 1$ for any integer $k$ and $\mathrm{e}^{(2m)2 \pi \mathrm{i}} = 1$ for any integer $m$. So again, we got two things that when we exponentiate them, all we ever get is $1$ but they are not equal as things.

(By the way, we're showing, over and over, that the complex exponential is not injective.)

What's going on?

Let's look at what's going on in slow motion when we square $1$ and $-1$. First, we have to recognize that we are working in the complex numbers and we are about to take a logarithm, so writing in exponentials will be handy. Then we note ("$\forall k \in \mathbb{Z}$" just means "for any choice of $k$ from the integers")\begin{align*} \forall k \in \mathbb{Z}, \phantom{-{}}1 &= \mathrm{e}^{2\pi \mathrm{i} k} &\text{and} \\ \forall m \in \mathbb{Z}, -1 &= \mathrm{e}^{\mathrm{i} \pi + 2 \pi \mathrm{i} m} \text{.} \end{align*} Then \begin{align*} \forall k \in \mathbb{Z}, \phantom{-{}} 1^2 &= (\mathrm{e}^{2\pi \mathrm{i} k})^2 = \mathrm{e}^{4\pi \mathrm{i} k} &\text{and} \\ \forall m \in \mathbb{Z}, (-1)^2 &= (\mathrm{e}^{\mathrm{i} \pi + 2 \pi \mathrm{i} m})^2 = \mathrm{e}^{2 \pi \mathrm{i} + 4 \pi \mathrm{i} m} \text{.} \end{align*} We've already determined that the sets $4\pi \mathrm{i} k, k \in \mathbb{Z}$ and $2 \pi \mathrm{i} + 4 \pi \mathrm{i} m, m \in \mathbb{Z}$ are different. So, if we look in the exponents, we can see that these are different things, but when we evaluate the exponentials, we find out that these two sets are a partition of the set $2\pi \mathrm{i} k, k \in \mathbb{Z}$ and $1 = \mathrm{e}^{2\pi \mathrm{i} k}$ for any integer $k$, so all of these evaluate to $1$. (The exponential is very not injective.)

This stuff makes sense if we work on the Riemann surface for this function -- which lets us keep track of these exponents, rather than just on the complex plane, which forces us to evaluate these exponentials, getting $1$ all the time. This is a picture of that surface.

(from Wikipedia)

It is made by stacking an infinite number of copies of the complex plane, slicing all of them along the ray $[0,\infty)$, that is, the positive real axis. Then as you go around one in the clockwise sense, as you reach the cut, you glue the ray on your plane to the ray on the next plane down, so you can keep going. This makes a generalized right helicoid. If you take this helicoid and stretch it evenly vertically so that each point on one plane is $2\pi$ units above the same point on the plane below it, then the vertical coordinate is the exponent we've been using above (except missing the "$\mathrm{i}$").

Exponentiation smashes all the planes into one plane, so all the points above and below $1$, $\mathrm{e}^{2 \pi \mathrm{i} k}, k \in \mathbb{Z}$, get smashed into $1$.

When we square a complex number, we square its magnitude and double its argument. For us, this means we double its exponent, since we are only using number with magnitude $1$, so squaring them does nothing, but we are keeping the arguments in the exponents. Then all the things which smash to $1$ have coordinates $(1,0,2 \pi k), k \in \mathbb{Z}$ on the helicoid and all the things which smash to $-1$ are $(-1,0,\pi + 2 \pi m), m \in \mathbb{Z}$. When we square these, we imagine the material of the helicoid stretching along itself so that points at height $z$ get carried along to height $2z$. Then we find that \begin{align*} (1,0,2 \pi k) &\mapsto (1,0,4\pi k), k \in \mathbb{Z}, \text{ and} \\ (-1,0,\pi + 2 \pi m) &\mapsto (1,0,2\pi + 4 \pi m), m \in \mathbb{Z} \end{align*}

What am I going on about? In this "logarithm-land", squaring is injective. In particular, we have been showing that when we square anything that exponentiation smashes down to $1$ we cannot land in the same place as when we square anything that exponentiation smashes down to $-1$. The squares of the $1$ things and the squares of the $-1$ things are disjoint, but they all smash down to $1$. This means $\ln((-1)^2) \neq \ln(1^2)$ because in "logarithm-land", squaring is injective, so $(-1)^2 \neq 1^2$, so $-1 \neq 1$.