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

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

Solution 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 $\log$).

Solution 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.