Properties and applications of complex logarithms and exponentials.

Question

We have these two identities where $a>0$ real, $n$ integer and $z$ complex : \begin{align} (a^z)^n=a^{nz} \tag{1} \\ (a^n)^z=a^{nz} \tag{2} \end{align}

$(1)$ is true. Is $(2)$ also true?

Assuming that $(2)$ is true, consider for $a,b>0$ real, $n,m$ integer and $z_1,z_2$ complex, and the steps below must be in this order:

\begin{align} & a^n = b^m \\ & (a^n)^{z_1} = (b^m)^{z_2}\\ & a^{nz_1} = a^{mz_2} \tag{3} \end{align}

What will happen if we take the natural $\log$ of the last equation above? Since $z_1, z_2$ are complex, the properties of $\log$ are now different.

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I saw in this [post][1] the following formula, for some integer $k$: \begin{align} \log_{b_1}({a_1}^d) = \dfrac{d \ln(a_1) + 2 \pi i k}{\ln(b_1)} \end{align}

and tried to apply on $(3)$ considering $b_1=e$, the natural $\log$:

\begin{align} &a^{nz_1} = b^{mz_2}\\ &\ln a^{nz_1} = \ln b^{mz_2} \\ & nz_1 \ln a + 2i \pi k= mz_2 \ln b. \tag{4} \end{align}

But I got some weird results working with the above, so something is wrong. I highlighted the questions.

Answer 1

I agree that my answer is complicated, but that is because the subject is complicated and even seemingly innocuous expressions like $1^z$ are actually complicated since while we generally and implicitly take $1^z=1$, so constant, it is legitimate to take $1^z=\exp({2\pi iz})$ too and the last is a completely different object, an entire integral function.

So coming to your question about 4:

Line 1: $a^{nz_1}=b^{mz_2}$ - I do not know what each side means for you, since I do not know what you mean by $a^{nz}$ - see above even for $a=1$; if it means the canonical interpretation, $a^{nz}=exp(n\log a z)$, where $exp, \log$ are given unambiguously by the power series mentioned in my answer, the line is correct; also the line can be correct in other circumstances, but it can also be incorrect with differe3nt choices for each side

Line 2: Only as sets in general so the set {$\ln (a^{nz_1}$} equal to the set {$\ln b^{mz_2}$}, not necessarily as individual values

Line 3: same as line 2, as sets only not as individual values, so RHS should be understood as a set too

Answer 2

For 1, you actually need $a \neq 0$ since for $a=0, a^z$ doesn't really make sense unless $z$ is a positive integer (just think how you would make sense of $0^{-1}$),

For $a>0$ there is a canonical choice of $\log a$ (the unique one that's real and for example given by $\log 1 =0$, $\log a = -\Sigma{\frac{b^n}{n}}, b=1-a, 0<a<1$, $\log a = -\log \frac{1}{a}, a>1$), so both sides are $\exp(nz\log a)$, where $\exp{z} = \Sigma{\frac{z^n}{n!}}$ is the standard entire exponential function.

For $a<0$ it gets tricky since there is no canonical choice of $\log a$, so you need to fix one such (e.g. $\log a = \log|a| + \pi i$) and then both sides are $\exp(nz\log |a| + n\pi i z)$

For 2, same thing, for $a>0$ with the canonical choices of $\log a, \log {a^n} =n\log a$ as above, both sides are $\exp(nz\log a)$, while for $a=0$ either expression doesn't generally make sense and for $a<0$, if you fix a choice as above and use it consistently in the sense that $\log {a^n}$ is the pointwise choice $n\log a$, so for example $\log {(-1)^2}= 2\pi i$, not the canonical choice $0$, both sides are still $\exp(nz\log |a| + n\pi i z)$

Note that in both 1 and 2 you need to fix the same $\log a$ for $a<0$ on both sides as otherwise the expressions won't generally be equal unless $z$ is an integer too, while in addition in 2, you need to use the choice consistently as noted above; for $a>0$ we have that by "default"

Since $\exp$ is periodic the expressions in 1 and 2 are well defined and unique (up to the choice of $\log a, a <0$ and consistency of its use as noted) so you can say that both 1 and 2 are true if interpreted appropriately. Taking logarithms though is very tricky, since being the inverse of periodic $\exp$, $\log$ is multivalued, so expression involving logarithms tend to be equal only set-wise, not pointwise - e.g $(-1)^2=1$ means $exp(2\pi i) = \exp (0)$, perfectly legitimate as both sides are $1$; taking logarithms it just means that the set {$2\pi i + 2k\pi i$} is equal to the set {$2k\pi i$} which again is perfectly legitimate but doesn't tell you anything about individual values; so, in particular, it is very easy to get incorrect expressions taking logarithms pointwise rather than set-wise

The bottom line is that when dealing with complex powers (or even real powers of negative numbers) it is wise to express everything in terms of the standard exponential function defined by the Taylor series above and forget about the meaning of powers in the real case since that can lead one into error; when taking logarithms everything should be thought in terms of sets rather than pointwise, unless you have well defined analytic functions on both sides, so for example defining the $\log$ by cutting the negative imaginary axis, so the argument lives in $(-\frac{\pi}{2}, \frac{3\pi}{2})$, and considering the function $z^2$ there, one can define a consistent logarithm of $z^2$, call it Log, on that domain for which Log${(-1)^2}$ has a defined pointiwse value, Log${1}$ has a defined pointwise value but they are not necessarily the usual ones and equal - one such choice would fix Log$1=0$ and then for any $z_0$ in the domain, take a smooth path $\gamma$ joining $1$ to $z_0$ and define Log$z_0^2 = \int_{\gamma}\frac{2}{z}$ which doesn't depend on the choice of $\gamma$ in the plane minus the negative imaginary axis; but you can start at $-1$, define Log$(-1)^2 = 2\pi i$ and repeat the procedure above, except that now Log$z_0^2 = \int_{\gamma}{\frac{2}{z}} + 2\pi i$ for any path joining $-1$ to $z_0$. Then you will have well defined number Log${z^2}$ for each $z$ in the domain above, satisfying the required $\exp {Log ({z^2})} = z^2$ but not necesarily Log$(-1)^2=$ Log$1$