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Let $z$, $\lambda$, $\mu$ be complex numbers. Find a case where $(z^\lambda)^\mu$ is not equal to $z^{\lambda\mu}$.

In our book, $a^b = \exp( b \cdot \operatorname{Log}(a) )$.

$\operatorname{Log}(a) = \ln |a| + i \operatorname{Arg}(a)$.

$\operatorname{Arg}(a)$ is a value in $(-\pi,\pi]$.

Thank you for your help.

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2 Answers 2

Let $z=-1$, $\lambda=2$, and $\mu=\frac{1}{2}$.

Try figuring out both quantities $(z^\lambda)^\mu$ and $z^{\lambda\mu}$ yourself, then check your work below once you're done by moving your cursor over the gray area.


We know that $\operatorname{Arg}(-1)=\pi$, and $\ln|-1|=\ln(1)=0$, so $$z^\lambda=(-1)^2=\exp(2\cdot\operatorname{Log}(-1))=\exp(2\cdot (0+i\pi))=\exp(2\pi i)=1,$$
so $$(z^\lambda)^\mu=((-1)^2)^{1/2}=1^{1/2}=\exp(\tfrac{1}{2}\cdot\operatorname{Log}(1))=\exp(\tfrac{1} {2}\cdot(0+i0))=\exp(0)=1.$$ However,
$$z^{\lambda\mu}=(-1)^{2\cdot (1/2)}=(-1)^1=\exp(1\cdot\operatorname{Log}(-1))=\exp(1\cdot(0+i\pi))=\exp(\pi i)=-1.$$

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6  
This is a really eerie example because only real numbers are involved. Who would doubt that $((-1)^2)^\frac {1}{2}=1^\frac {1}{2}=1$ ? And who would doubt that $(-1)^{2 \cdot \frac {1}{2}}=(-1)^1=-1$ ? So the unescapable conclusion is that $((-1)^2)^\frac {1}{2}\neq (-1)^{2 \cdot \frac {1}{2}}$ And the really perturbing thing is that we only used positive integral exponents for $-1$, and a rational exponent for the positive number $1$, which seems completely harmless. Caveat exponentiator! –  Georges Elencwajg Feb 1 '12 at 23:38
    
I don't have a mathematics comment here; I just wanted to say that the mouse-over-grey effect is cool. I haven't seen it used here before. –  Mark Beadles Feb 1 '12 at 23:49
1  
Thanks @Mark! I quite often forget it exists myself. Some helpful formatting commands, including for the "spoiler" feature, are listed here. –  Zev Chonoles Feb 1 '12 at 23:54
    
@Georges: +1, especially for Caveat exponentiator! –  Zev Chonoles Feb 1 '12 at 23:57
    
@ZevChonolesReally nice touch with the hint and grey. Too bad I could only vote once - s.b. +1 for the answer and ++ for the format. –  Andrew Feb 2 '12 at 0:32

According to the definition, $(z^\lambda)^\mu = \exp(\mu \log(z^\lambda))$ while $z^{\lambda \mu} = \exp(\lambda \mu \log(z))$. Now $\exp(a) = \exp(b)$ if and only if $a - b$ is an integer multiple of $2 \pi i$, so you want $\mu \log(z^\lambda) - \lambda \mu \log(z) = \mu (\log(z^\lambda) - \lambda \log(z))$ not to be an integer multiple of $2 \pi i$. Now $\log(z^\lambda) = \log(\exp(\lambda \log(z))) = \lambda \log(z) + 2 \pi i k$ where $k$ is the integer that puts the imaginary part of this in the interval $(-\pi, \pi]$, and then $\log(z^\lambda - \lambda \log(z) = 2 \pi i k$. So what you need is that $k$ is a nonzero integer and $\mu k$ is not an integer.

You could use any $z$ that is not $0$ or $1$. If $w = \log(z)$, choose a complex number $\lambda$ such that $\text{Im}(\lambda w)$ is outside the interval $(-\pi, \pi]$, find the integer $k$ such that $\text{Im}(\lambda w) - 2 \pi k$ is in $(-\pi, \pi]$, and then any $\mu$ such that $\mu k$ is not an integer.

Zev's example has $w = \pi i$ and $k = 1$ so his $\mu = 1/2$ works, but so would any non-integer $\mu$. Another example is $z = e$ with $\lambda = 2 \pi i$ and any non-integer $\mu$.

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I tried to up vote you, but don't have enough of a reputation to do so yet. Thank you very much for your thorough explanation. –  user24205 Feb 2 '12 at 2:09
    
You are able to accept an answer, I think. –  Robert Israel Feb 2 '12 at 7:13

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