# What is the numerical value of $(-3)^{\pi}$

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 $$(-3) \cdot(-3) \cdot (-3) \cdot(-3)^{\pi-3}$$?

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

$$e^{i\pi^2}=\cos\pi^2+i\sin\pi^2$$

With WolframAlpha I get this numerical approximation:

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

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