What did Johann Bernoulli wrong in his proof of $\ln z=\ln (-z)$?

Question

Some people say, Johann Bernoulli has proven $\ln z=\ln (-z)$ in the following way $$\ln ((-z)^2 )=\ln(z^2)\;\;\;\Rightarrow\;\;\;2\ln(-z)=2\ln z\;\;\;\Rightarrow\;\;\;\ln (-z)=\ln z$$ While the statement is not true, I'm unable to figure out what exactly went wrong:

• Since $(-z)^2=z^2$ for all $z\in\mathbb{C}$, it holds $\ln ((-z)^2 )=\ln(z^2)$, too.
• By definition of $a^b=e^{b\ln a}$ for all $a\in\mathbb{C}^{-},b\in\mathbb{C}$ it follows $\ln (e^{2\ln z})=\ln (e^{2\ln (-z)})$ for all $z\in\mathbb{C}\setminus\mathbb{R}$
• Further, for $z=re^{i\varphi}$ we've got $$\ln (e^{2\ln z})=\ln (r^2)+i\pi +i\text{arg}_0(e^{i(2\varphi -\pi)})=2\ln r+2i\varphi$$ and $$\ln z=\ln r+i\varphi$$ So, it seems like $2 \ln z=\ln (z^2)$ holds, too (at least for all $z\in\mathbb{C}\setminus\mathbb{R}$).

So, what is wrong here?

PS: Let's define $$\mathbb{C}^{-}:=\left\{z\in\mathbb{C} : \text{Re }z>0\vee \text{Im}\ne 0\right\}$$

Answer 1

Let's see where things go wrong. I'm going to assume my logarithm has a branch cut along the negative $x$-axis as per usual (and that we're looking at the principal argument where $\theta$ is between $-\pi$ and $\pi$ for symmetry). If $z = re^{i\theta}$

$$\log(z) \stackrel{\text{def}}{=} \log r+i\theta.$$

With a branch cut, $-z$ is a little ambiguous. If $\theta < 0$, then $-z = re^{i\theta+i\pi}$. If $\theta > 0$, then $-z = re^{i\theta-i\pi}$. Let's consider each case separately.

Case 1: $\theta < 0$, then

$$\log(-z) = \log(re^{i\theta+i\pi}) = \log r + i(\theta+\pi).$$

Note that this is most definitely not the same as $\log(z)$.

Case 2: $\theta > 0$, then

$$\log(-z) = \log(re^{i\theta-i\pi}) = \log r + i(\theta-\pi).$$

Similarly, this is not the same as $\log(z)$.

The question then becomes: where's the faulty assumption in his proof?

Clearly $(-z)^2 = z^2$ is true but complex logarithms no longer obey $\log(ab) = \log(a)+\log(b)$ in the usual way. If $a = r_1e^{i\theta_1}$ and $b = r_2e^{i\theta_2}$ (let's say neither of them are pure imaginary), then $\log(ab) = \log(r_1r_2 e^{i\theta_1+i\theta_2})$. What this evaluates is heavily predicated on what $\theta_1$ and $\theta_2$ are.

Suppose $-\pi < \theta_1+\theta_2 < \pi$, then indeed $\log(ab) = \log(r_1r_2)+i(\theta_1+\theta_2)$ which agrees exactly with $\log(a)+\log(b)$.

However suppose that $\theta_1+\theta_2>\pi$, then similar to above above, we would have that $\log(ab) = \log(r_1r_2)+i(\theta_1+\theta_2-2\pi) \neq \log(a)+\log(b)$.

If instead $\theta_1+\theta_2 < -\pi$, then $\log(ab) = \log(r_1r_2)+i(\theta_1+\theta_2+2\pi) \neq \log(a)+\log(b)$.

If you restrict to the case that $a = b = -z$, the above analysis would read that $\log((-z)^2) = 2\log(-z)$ if (and only if) $\theta < \left|\frac{\pi}{2}\right|$ as Did mentions below. So really the discrepancy is due to the periodic nature of $e^{i\theta}$ (and hence the multivalued nature of $\log$).

Answer 2

Of course the identity $\mathrm{Log}(z^2)=2\ln(r)+2\mathrm i\varphi$ when $z=r\mathrm e^{\mathrm i\varphi}$, holds only when $-\frac\pi2\lt\varphi\leqslant\frac\pi2$ although the identity $\mathrm{Log}(z)=\ln(r)+\mathrm i\varphi$ holds for every $-\pi\lt\varphi\leqslant\pi$ (recall that by definition the principal value $\mathrm{Log}(z)$ is the logarithm whose imaginary part lies in the interval $(−π,π]$).

For a little practice, choose $z=-1+\mathrm i\sqrt3$, compute $r$, $\varphi$, $\ln(r)$, $\mathrm{Log}(z)$ and $\mathrm{Log}(z^2)$, and compare.

Answer 3

Let's start from your first display: $$ \log((-z)^2 )=\log(z^2) \implies 2\log(-z)=2\log z\implies \log (-z)=\log z $$ (I find anachronistic using “ln”, which I find awful anyway).

The “error” is in the first implication, which requires, assuming $z>0$, that

1. the logarithm exists for negative numbers;

2. it satisfies the same property for exponents as the logarithm for positive numbers.

Actually, there is no error: Bernoulli's argument is no more than a motivation for a definition. After having defined the logarithm for positive numbers, Bernoulli wants to extend it by setting $\log z=\log(-z)$ for $z<0$ or, in other words, $$ \log z=\log|z|\qquad(z\ne0). $$

Do the common properties hold? Yes, because $|xy|=|x|\,|y|$, so also $|z^n|=|z|^n$ (at least when $n$ is an integer).

This is, in a way, no different from extending the Gamma function to negative numbers by using the property that $\Gamma(t+1)=t\Gamma(t)$ for $t>0$, which can be proved from the definition $$ \Gamma(t)=\int_0^{\infty}x^{t-1}e^{-x}\,dx $$ (for $t>0$). Thus we can define, for $-1<t<0$, $$ \Gamma(t)=\frac{\Gamma(t+1)}{t} $$ and go backwards, using this definition for getting $\Gamma(t)$ when $-2<t<-1$, and so on.

What's the difference between these two cases? That the $\Gamma$ function can be shown to be extendable to the whole complex plane (except at the nonpositive integers): $$ \Gamma(t)= \frac{e^{-\gamma t}}{t}\prod_{n\ge1}\left(1+\frac{t}{n}\right)^{-1}e^{t/n} $$ (where $\gamma$ is the Euler-Mascheroni number).

Bernoulli's extension of the logarithm to the negative numbers, instead, has no relationship with the exponential function any more. It's just one possible extension, but it turns out to be useless. The reasons for its uselessness have been given in other answers.

Answer 4

the key point is that $z^2=y^2 \implies z=y$ is false for $z,y \in \mathbb C$ just like it is for $z,y \in \mathbb R$

which then falls into the classic take square root of both sides without adjusting signs where needed