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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?

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Research effort? –  Jacob Wakem Jan 12 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 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 at 19:52

3 Answers 3

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

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Thanks for the reply, really helpful explanations. –  Joe John Jan 12 at 17:58

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.

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

$Ln(e^i)=ln|e^i|+iArg(e^i)=i(1+2k\pi)$.

$iLn(e)=i[ln(e)+iArg(e)]=i-2k\pi$.

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

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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 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 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 at 17:46
    
Thanks a lot Xucheng, appreciate it. Yeah I was able to work it out from your example! –  Joe John Jan 12 at 17:49
    
@blf $Log(1+i)^2 \neq 2Log(1+i)^2$, it's the case. –  Xucheng Zhang Jan 12 at 17:50

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