# Complex analysis and powers of complex values

There are some values: $$a^{2\alpha }, \quad (a^{\alpha})^{2}, \quad (a^{2})^{\alpha }.$$ How to prove that the sets of these values ​​are not equal?

Addition 1. For example, I used one simple task equal to $\left(z^{2}\right)^{\alpha}$.

Let's have two values: $z^{\frac{m}{n}}: z^{\frac{m}{n}}, \quad \left(z^{\frac{1}{n}}\right)^{m}$, where $n, m$ are not integer in general. Without loss of generality we can assume, that |z| = 1. $$z^{\frac{m}{n}} = e^{i\frac{m}{n}(\varphi + 2\pi k)} = cos\left(\frac{m}{n}\left( \varphi + 2 \pi k\right)\right) + isin\left(\frac{m}{n}\left( \varphi + 2 \pi k\right)\right).$$ $$\left(z^{m}\right)^{\frac{1}{n}} = cos\left(\frac{m}{n}\left( \varphi + 2 \pi k \right) + 2m\pi j\right) + isin\left(\frac{m}{n}\left( \varphi + 2 \pi k \right) + \frac{1}{n}2\pi j\right).$$ There are difference sets of values.

Addition 2. Previous example was bad. So I created new example: for $\alpha = x + iy$ $$\left(a^{\alpha}\right)^{2} = \left(e^{xln(r) - y(\varphi + 2 \pi k)}[cos(x (\varphi + 2 \pi k) + yln(r)) + isin(x (\varphi + 2 \pi k) + yln(r))]\right)^{2} =$$ $$= e^{2xln(r) - 2y(\varphi + 2 \pi k)}[cos(2x (\varphi + 2 \pi k) + 2yln(r)) + isin(2x (\varphi + 2 \pi k) + 2yln(r))],$$ $$\left(a^{2}\right)^{\alpha} = e^{2xln(r) - y(2\varphi + 2 \pi k)}[cos(x (2\varphi + 2 \pi k) + 2yln(r)) + isin(x (2\varphi + 2 \pi k) + 2yln(r))].$$

Can somebody generalize this result?

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They certainly look equal to me. Is there more information to the question? –  EuYu Sep 13 '12 at 17:49
Unfortunately, it isn't. But as the answer says that the values ​​are not equal. –  John Taylor Sep 13 '12 at 17:54
Are you sure you don't mean $a^{2\alpha}, a^{\alpha^2},$ and $a^{2^\alpha}$? –  Tarnation Sep 13 '12 at 17:57
Of course, I'm sure. –  John Taylor Sep 13 '12 at 17:59
What is the source you refer to? That is, where is that answer? –  AD. Sep 13 '12 at 18:03

Take $a=-1$ and $\alpha=\frac12$. Then $a^{2\alpha}=(-1)^{2\cdot1/2}=(-1)^1=-1$, but $(a^2)^\alpha = ((-1)^2)^{1/2}=1^{1/2}=1$ (or $\pm1$ when using the complex multivalued definition).

The following considerations all use the complex multivalued definition.

For real exponents, we can restrict to bases of the form $r\mathrm e^{\mathrm i\phi}$.

Then we have $a^{2\alpha}=\{\mathrm e^{\mathrm i\alpha(\phi+2\pi k)}:k\in\mathbb Z\}$ and $(a^\alpha)^2=\{\mathrm e^{2\mathrm i\left(\alpha(\phi+2\pi k)+2\pi j\right)}:j,k\in\mathbb Z\}=\{\mathrm e^{2\mathrm i\alpha(\phi+2\pi k)}:k\in\mathbb Z\}$ because the term proportional to $j$ is always an integer mutiple of $2\pi$.

Therefore we see that for all real $\alpha$ we have $a^{2\alpha}=(a^\alpha)^2$. Indeed, the same argument can be used to show that for all real $\alpha$ and integer $n$, we have $a^{n\alpha}=(a^\alpha)^n$.

For pure imaginary exponents $\alpha = \mathrm i\beta$, we have with $a=r\mathrm e^{\mathrm i\phi}$: $$a^{2\alpha} = a^{2\mathrm i\beta} = \mathrm e^{2\mathrm i\beta\left(\ln r + \mathrm i(\phi+2\pi k)\right)} = \mathrm e^{-2\beta(\phi+2k\pi)+2\mathrm i\beta\ln r}$$ and $$(a^\alpha)^2 = (a^{\mathrm i\beta})^2 = (\mathrm e^{-\beta(\phi+2k\pi)+\mathrm i\beta\ln r})^2 = \mathrm e^{-2\beta(\phi+2k\pi)+2\mathrm i\beta\ln r + 2\cdot2\pi\mathrm i j} = a^{2\alpha}$$ because again, the exponents only differ by an integer multiple of $2\pi\mathrm i$. Again, this can be extended to arbitrary $n\in \mathbb Z$.

Nor for arbitrary complex exponents, first note that $a^{\alpha+\beta}=a^\alpha + a^\beta$ (this follows directly from the properties of the exponential function and the fact that complex number multiplication is commutative). Therefore especially $a^{x+\mathrm iy} = a^xa^{\mathrm iy}$.

Furthermore, for the multivalued definition of the power (unlike the principal value definition), we have $(ab)^c = a^cb^c$, as can be seen as follows: $$a^c b^c = \mathrm e^{c(\ln a + 2\pi\mathrm i j)}\mathrm e^{c(\ln b + 2\pi\mathrm i k)} = \mathrm e^{c(\ln a + \ln b + 2\pi\mathrm i (j+k))} = \mathrm e^{c(\ln (ab) + 2\pi\mathrm i (j+k))} = (ab)^c$$

Therefore with $\alpha = x+\mathrm iy$ and the previous results, we have always $$(a^\alpha)^2 = (a^xa^{\mathrm iy})^2 = (a^x)^2(a^{\mathrm iy})^2 = a^{2x}a^{2\mathrm iy} = a^{2x+2\mathrm iy} = a^{2\alpha}$$

Again, the same argumentation applies for any integer instead of $2$.

# Conclusion

For the multivalued definition of the complex power, for any complex $a$ and $\alpha$ and every integer $n$ we have $${a^\alpha}^n = a^{n\alpha}$$ However we generally (but not always) have $${a^n}^\alpha \ne a^{n\alpha}$$ One obvious exception to the second inequality is where $\alpha\in\mathbb Z$.

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Maybe there are some complex values that create a differense in the sets of values? –  John Taylor Sep 13 '12 at 18:13
Let's have two values: $z^{\frac{m}{n}}: z^{\frac{m}{n}}, \quad \left(z^{\frac{1}{n}}\right)^{m}$. Without loss of generality we can assume, that |z| = 1. $$z^{\frac{m}{n}} = e^{i\frac{m}{n}(\varphi + 2\pi k)} = cos\left(\frac{m}{n}\left( \varphi + 2 \pi k\right)\right) + isin\left(\frac{m}{n}\left( \varphi + 2 \pi k\right)\right).$$ $$\left(z^{\frac{1}{n}}\right)^{m} = cos\left(\frac{m}{n}\left( \varphi + 2 \pi k \right) + 2m\pi j\right) + isin\left(\frac{m}{n}\left( \varphi + 2 \pi k \right) + 2m\pi j\right).$$ There are difference sets of values. –  John Taylor Sep 13 '12 at 20:48
@Maxim_Ovchinnikov: $\frac{m}{n}(\phi+2\pi k)+2m\pi j = \frac{m}{n}(\phi+2\pi (k+nj)) = \frac{m}{n}(\phi+2\pi k')$. Since $k$ may be any integer, the sets are the same. However, if $\alpha$ is irrational, the sets should be indeed different. –  celtschk Sep 14 '12 at 9:05
Can you show that? –  John Taylor Sep 14 '12 at 13:20
Yes, I understood, that that example was bad. I created new example: for $\alpha = x + iy$ $$\left(a^{\alpha}\right)^{2} = \left(e^{xln(r) - y(\varphi + 2 \pi k)}[cos(x (\varphi + 2 \pi k) + yln(r)) + isin(x (\varphi + 2 \pi k) + yln(r))]\right)^{2} =$$ $$= e^{2xln(r) - 2y(\varphi + 2 \pi k)}[cos(2x (\varphi + 2 \pi k) + 2yln(r)) + isin(2x (\varphi + 2 \pi k) + 2yln(r))],$$ $$\left(a^{2}\right)^{\alpha} = e^{2xln(r) - y(2\varphi + 2 \pi k)}[cos(x (2\varphi + 2 \pi k) + 2yln(r)) + isin(x (2\varphi + 2 \pi k) + 2yln(r))].$$ –  John Taylor Sep 14 '12 at 13:27