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Is this a real number? If it a complex number, then what is its $a+bi$ form?

$$(-3)^\sqrt{2}$$

Blue
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Angelo Mark
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    The main issue is not is it a real number or a complex number but how do you define $a^b$ when $a<0$ and $b>0$ are real numbers? – mathcounterexamples.net Sep 02 '19 at 04:41
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    $(-1)^{\sqrt{2}} = \left(e^{i\pi}\right)^{\sqrt{2}}$ – AgentS Sep 02 '19 at 04:43
  • We really should collectively look for the best thread explaining complex powers/exponentiation. A quick search gave me 1, 2 among others. No single satisfactory thread, but several scattered ones often based on one or the other familiar exponentiation rule breaking down when either the exponents are complex or the base is not a positive real number. – Jyrki Lahtonen Sep 02 '19 at 06:43

2 Answers2

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As a complex number, it has infinitely many values. We have $$(-3)^\sqrt{2}=e^{\sqrt{2}\log(-3)}=e^{\sqrt{2}(\ln{3}+(2n+1)\pi i)}=3^{\sqrt{2}}e^{\sqrt{2}(2n+1)\pi i}\\=3^{\sqrt{2}}\cos(\sqrt{2}(2n+1)\pi)+i3^{\sqrt{2}}\sin(\sqrt{2}(2n+1)\pi),\\n=0,\pm1,\pm2,\dots$$

saulspatz
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Exponentiation of a nonzero complex base $b$ with any rational number $r$ is always well-defined by letting $b^{p/q} = \frac{b^p}{b^q}$. For positive bases $b > 0$, we can extend the domain of $b^x$ for all real $x$ in two equivalent ways. The first is to notice that $b^x$ is continuous on $\mathbb{Q}$, and since the rationals are dense in $\mathbb{R}$ there is a unique way to extend the function to a continuous function on all of $\mathbb{R}$. The second is to fix the natural exponentiation function $e^x$ for any real-valued $x$, and then let $b^{x} = e^{x\ln{b}}$.

For a negative base $b$ (or more generally any non-positive complex number), neither of these definitions work since $b^x$ is no longer a continuous function on the rationals, and $\ln(x)$ is not defined. However, using the $complex$ logarithm $Log$, we can still use the second method to assign a value to $b^x$ for any complex numbers $b\neq 0$ and $x$ by again letting $b^{x} = e^{x\cdot Log{b}}$. However, the issue here is that the complex logarithm is not actually a function since it is multi-valued; for any number $z$ such that $e^z = w$, then $e^{z+2\pi i n} = w$ for any integer $n$ and so $z + 2\pi i n$ are all `logarithms' of $w$. Therefore, while we can assign a value to $(-3)^{\sqrt{2}}$, it is not unique; there are infinitely many such values we could have chosen.

Just to illustrate, notice that $e^{\ln{3} + \pi i} = 3e^{i\pi} = -3$ and so $\ln{3} + \pi i$ is a complex logarithm for $-3$. Thus, fixing this choice of logarithm, $$(-3)^{\sqrt{2}} = e^{\sqrt{2}\cdot (\ln{3} + \pi i)} = 3^{\sqrt{2}}e^{\pi\sqrt{2}i} = 3^{\sqrt{2}}(\cos(\pi\sqrt{2}) + i\sin(\pi\sqrt{2}))$$. However, we could have just as easily calculated that $$(-3)^{\sqrt{2}} = 3^{\sqrt{2}}(\cos(\pi(1+2n)\sqrt{2}) + i\sin(\pi(1+2n)\sqrt{2}))$$ for any integer $n$.

Andrew
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