As the title suggest what is the numerical value of $(-3)^{\pi}$?

could we derive an answer using numerical analysis something along the lines of well if its basically $(-3) \cdot(-3) \cdot (-3) \cdot(-3)^{\pi-3}$?

  • $\begingroup$ The numerical value is exactly $\;(-3)^\pi\;$ ... $\endgroup$ – DonAntonio Jan 29 at 17:29
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    $\begingroup$ Welcome to MSE. Please see this handy formatting reference to learn how to typeset math. $\endgroup$ – Théophile Jan 29 at 17:29
  • $\begingroup$ @Théophile thanks man i wanted to just didnt know how!. $\endgroup$ – KARAM JABER Jan 29 at 17:30
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    $\begingroup$ $$(-3)^{\pi}=e^{\pi \ln (-3)}=e^{\pi \cdot (\ln 3+\mathrm i\pi)}=e^{\pi \ln 3}\cdot e^{\mathrm i \pi^2}$$ $\endgroup$ – Mohammad Zuhair Khan Jan 29 at 17:31
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    $\begingroup$ The imaginary part is due to $(-1)^x$. $\endgroup$ – karakfa Jan 29 at 17:32

Mathematicians generally define powers of negative real numbers using the principal value of the complex logarithm, that is

$$(-3)^\pi:= e^{\pi\ln(-3)}=e^{\pi(\ln|-3|+i\arg(-3))}=e^{\pi\ln 3+i\pi^2}=3^\pi\cdot e^{i\pi^2}$$

where $e^{i\pi^2}$ is the complex number defined by a vector on the complex plane of length $1$ such that the angle with the real line is $\pi^2$, that is


With WolframAlpha I get this numerical approximation:

$$(-3)^\pi\approx -28.47456 -i\, 13.57354$$

  • $\begingroup$ $\cos \pi^2 \approx \frac12(\pi*(\pi-3))^2-1$ $\endgroup$ – karakfa Jan 29 at 17:45

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