wingtipti
2022-01-23
Answered

We know that negative number to the power of any integers or some fractions will always have a solution. Is it possible for us to solve $(-2)}^{\frac{1}{3}$ or $(-2)}^{e$ , by modifying/extending our current number system?

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sineurosi0f

Answered 2022-01-24
Author has **14** answers

Consider the value of the "number" $(-2)}^{e$ ; this is equivalent to considering

$x={(-2)}^{e}$

$\mathrm{ln}x=e\mathrm{ln}(-2)$

$\frac{\mathrm{ln}x}{e}=\mathrm{ln}(-2)$

$e}^{\frac{\mathrm{ln}x}{e}=-2$

$\frac{12}{e}}^{\frac{\mathrm{ln}x}{e}=-1$

$e}^{\frac{\mathrm{ln}x}{e}+\mathrm{ln}\frac{1}{2}=-1$

Given the equation${e}^{a+bi}={e}^{a}(\mathrm{cos}b+i\mathrm{sin}b)$ , we only need to find x such that $\frac{\mathrm{ln}x}{e}=i\pi +2ni\pi -\mathrm{ln}\frac{1}{2}$ or $\mathrm{ln}x=ei\pi +2\ne i\pi +e\mathrm{ln}2$ which is

$x={e}^{ei\pi +2\ne i\pi +e\mathrm{ln}2}={2}^{e}(\mathrm{cos}(e\pi +2\ne \pi )+i\mathrm{sin}(e\pi +2\ne \pi )$

which is an infinite set of unique complex values in a circle of radius$2}^{e$

Given the equation

which is an infinite set of unique complex values in a circle of radius

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