Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

And similarly, does $\operatorname{Log}(1-i)^2=2\operatorname{Log}(1-i)$?

If we were dealing with real numbers, it would hold. But I'm guessing that the fact that there are imaginary numbers involved changes things?

share|cite|improve this question
Research effort? – Jacob Wakem Jan 12 '14 at 17:49
Do you mean, how much effort did I put in to researching the answer before I asked here? If so, i googled around for about 5 minutes, then thought about it for some more till I asked. – Joe John Jan 12 '14 at 17:51
@JoeJohn : for your two specific examples, the answer is "yes", using the usual definition of "Log". But for other complex bases, the analogous equations are false. – Stefan Smith Jan 12 '14 at 19:52

The function $\mathrm{Log}$ is usually defined so as to be continuous on $D=\mathbb C\setminus\mathbb R_-$. Every point $z$ in $D$ can be uniquely written as $z=r\mathrm e^{\mathrm it}$ with $r$ and $t$ real, $r\gt0$ and $|t|\lt\pi$, then $\mathrm{Log}(z)=\log r+\mathrm it$.

For the identity $\mathrm{Log}(z^2)=2\,\mathrm{Log}(z)$ to hold for such a complex number $z=r\mathrm e^{\mathrm it}$, the condition is that the path $\gamma$ defined on $[1,2]$ by $\gamma(x)=\mathrm e^{\mathrm ixt}$ stays in $D$. One readily sees that this means that $|t|\lt\pi/2$.

If $z=1\pm\mathrm i$, then $r=\sqrt2$ and $t=\pm\pi/4$ hence $|t|\lt\pi/2$ and indeed $\mathrm{Log}(z^2)=2\,\mathrm{Log}(z)$ holds.

Likewise, for every integer $n\geqslant1$, the identity $\mathrm{Log}(z^n)=n\,\mathrm{Log}(z)$ holds if $z=r\mathrm e^{\mathrm it}$ with $|t|\lt\pi/n$. Still with $z=1\pm\mathrm i$, $\mathrm{Log}(z^3)=3\,\mathrm{Log}(z)$ holds, $\mathrm{Log}(z^4)$ is undefined, and, for every $n\geqslant5$, either $\mathrm{Log}(z^n)$ is undefined (this happens when $n$ is in $4+8\mathbb N$) or $\mathrm{Log}(z^n)\ne n\,\mathrm{Log}(z)$.

The identities $\mathrm{Log}(z^u)=u\,\mathrm{Log}(z)$ with $u$ complex are still another story.

share|cite|improve this answer

It is not easy to define a good notion of Logarithm in the complex numbers. The problem is essentially the following: $\exp: \mathbb{R} \rightarrow \mathbb{R}^+$ is bijective but this is not true for $\exp: \mathbb{C} \rightarrow \mathbb{C} - \{0\}$. Indeed, you have the well-known identity $\exp(z) = \exp(z + 2i\pi)$.

Since we want $\log$ to be an inverse of $\exp$, we have troubles to define it. You can deal with this in three different ways. First, you can define the logarithm to be a multi-valued function; that is you define $\log(z)$ to be the set of ALL $w \in \mathbb{C}$ such that $\exp(w) = z$. If $w_0$ is such a solution, then $\log(z)$ is the set $w_0 + 2\pi i \mathbb{Z}$. A second way to do things is to choose a preferred solution of $\log(z)$. Writing $z = \rho e^{i\theta}$, you define $\log(z)$ to be $\log(\rho) + i\theta$, choosing $\theta$ in $[0, 2\pi[$. In this way, $\log$ is a genuine function but it is not continuous since $\lim_{\theta \rightarrow 2\pi^-} (e^{i\theta})$ is $2i\pi$ whereas $\lim_{\theta \rightarrow 2\pi^+}$ is $0$.

The third way in to define $\log$ to be a genuine function with values in some space, called a Riemann surface. This is the modern solution used to view $\log$ as a univalued function, which is also continuous (and even holomorphic).

Concerning your question, what you can say of general is that of $\log (a^b)$ and $b \log(a)$ (using the second definition, with $b$ real) is that they differ from a multiple of $2i \pi$. In you case: $1 + i = \frac{1}{\sqrt{2}} e^{i \pi/4}$ and $(1+i)^2 = \frac12 e^{i \pi/2}$ so that $\log((1+i)^2) = \log(1/2) + i\pi/2 = 2\log(1+i)$. So this works with THIS definition of logarithm.

But remark that $1 = (-1)^2$ and that $\log(1) = 0$, whereas, in this definition, $\log(-1) = i\pi$.

share|cite|improve this answer
Thanks for the reply, really helpful explanations. – Joe John Jan 12 '14 at 17:58

In general, $\DeclareMathOperator{\Ln}{Ln}\DeclareMathOperator{\Arg}{Arg}\Ln(a^b)\neq b\Ln(a)$ in $\mathbb{C}$. See the following example:



So when dealing with complex number, we should be more careful.

share|cite|improve this answer
Of course everything you wrote is correct, but how does this answer the question if $\operatorname{Log}(1+i)^2 =2\operatorname{Log}(1+i)$? – user127.0.0.1 Jan 12 '14 at 17:41
In your example, your $b$ is complex, while OP's $b$ is real. Could you give an example where $a$ is complex and $b$ is real, but $\ln (a^b) \ne b \ln a$? – angryavian Jan 12 '14 at 17:42
What I want to say is, we can't apply what works in REAL numbers to COMPLEX numbers without consideration. As to this question, I think simple computation can show it doesn't hold. – Xucheng Zhang Jan 12 '14 at 17:46
Thanks a lot Xucheng, appreciate it. Yeah I was able to work it out from your example! – Joe John Jan 12 '14 at 17:49
@blf $Log(1+i)^2 \neq 2Log(1+i)^2$, it's the case. – Xucheng Zhang Jan 12 '14 at 17:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.