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












0












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






numerical-methods pi







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  • $begingroup$
    The numerical value is exactly $;(-3)^pi;$ ...
    $endgroup$
    – DonAntonio
    Jan 29 at 17:29






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    $$(-3)^{pi}=e^{pi ln (-3)}=e^{pi cdot (ln 3+mathrm ipi)}=e^{pi ln 3}cdot e^{mathrm i pi^2}$$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 29 at 17:31






  • 2




    $begingroup$
    The imaginary part is due to $(-1)^x$.
    $endgroup$
    – karakfa
    Jan 29 at 17:32
















0












$begingroup$


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






numerical-methods pi







share|cite|improve this question











$endgroup$












  • $begingroup$
    The numerical value is exactly $;(-3)^pi;$ ...
    $endgroup$
    – DonAntonio
    Jan 29 at 17:29






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    $$(-3)^{pi}=e^{pi ln (-3)}=e^{pi cdot (ln 3+mathrm ipi)}=e^{pi ln 3}cdot e^{mathrm i pi^2}$$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 29 at 17:31






  • 2




    $begingroup$
    The imaginary part is due to $(-1)^x$.
    $endgroup$
    – karakfa
    Jan 29 at 17:32














0












0








0





$begingroup$


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






numerical-methods pi







share|cite|improve this question











$endgroup$




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






numerical-methods pi




numerical-methods pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 17:38









Mohammad Zuhair Khan

1,6792625




1,6792625










asked Jan 29 at 17:27









KARAM JABERKARAM JABER

1033




1033












  • $begingroup$
    The numerical value is exactly $;(-3)^pi;$ ...
    $endgroup$
    – DonAntonio
    Jan 29 at 17:29






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    $$(-3)^{pi}=e^{pi ln (-3)}=e^{pi cdot (ln 3+mathrm ipi)}=e^{pi ln 3}cdot e^{mathrm i pi^2}$$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 29 at 17:31






  • 2




    $begingroup$
    The imaginary part is due to $(-1)^x$.
    $endgroup$
    – karakfa
    Jan 29 at 17:32


















  • $begingroup$
    The numerical value is exactly $;(-3)^pi;$ ...
    $endgroup$
    – DonAntonio
    Jan 29 at 17:29






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    $$(-3)^{pi}=e^{pi ln (-3)}=e^{pi cdot (ln 3+mathrm ipi)}=e^{pi ln 3}cdot e^{mathrm i pi^2}$$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 29 at 17:31






  • 2




    $begingroup$
    The imaginary part is due to $(-1)^x$.
    $endgroup$
    – karakfa
    Jan 29 at 17:32
















$begingroup$
The numerical value is exactly $;(-3)^pi;$ ...
$endgroup$
– DonAntonio
Jan 29 at 17:29




$begingroup$
The numerical value is exactly $;(-3)^pi;$ ...
$endgroup$
– DonAntonio
Jan 29 at 17:29




1




1




$begingroup$
Welcome to MSE. Please see this handy formatting reference to learn how to typeset math.
$endgroup$
– Théophile
Jan 29 at 17:29




$begingroup$
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




$begingroup$
@Théophile thanks man i wanted to just didnt know how!.
$endgroup$
– KARAM JABER
Jan 29 at 17:30




1




1




$begingroup$
$$(-3)^{pi}=e^{pi ln (-3)}=e^{pi cdot (ln 3+mathrm ipi)}=e^{pi ln 3}cdot e^{mathrm i pi^2}$$
$endgroup$
– Mohammad Zuhair Khan
Jan 29 at 17:31




$begingroup$
$$(-3)^{pi}=e^{pi ln (-3)}=e^{pi cdot (ln 3+mathrm ipi)}=e^{pi ln 3}cdot e^{mathrm i pi^2}$$
$endgroup$
– Mohammad Zuhair Khan
Jan 29 at 17:31




2




2




$begingroup$
The imaginary part is due to $(-1)^x$.
$endgroup$
– karakfa
Jan 29 at 17:32




$begingroup$
The imaginary part is due to $(-1)^x$.
$endgroup$
– karakfa
Jan 29 at 17:32










1 Answer
1






active

oldest

votes


















2












$begingroup$

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



$$(-3)^pi:= e^{piln(-3)}=e^{pi(ln|-3|+iarg(-3))}=e^{piln 3+ipi^2}=3^picdot e^{ipi^2}$$



where $e^{ipi^2}$ 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



$$e^{ipi^2}=cospi^2+isinpi^2$$



With WolframAlpha I get this numerical approximation:



$$(-3)^piapprox -28.47456 -i, 13.57354$$






share|cite|improve this answer











$endgroup$













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














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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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



$$(-3)^pi:= e^{piln(-3)}=e^{pi(ln|-3|+iarg(-3))}=e^{piln 3+ipi^2}=3^picdot e^{ipi^2}$$



where $e^{ipi^2}$ 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



$$e^{ipi^2}=cospi^2+isinpi^2$$



With WolframAlpha I get this numerical approximation:



$$(-3)^piapprox -28.47456 -i, 13.57354$$






share|cite|improve this answer











$endgroup$













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


















2












$begingroup$

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



$$(-3)^pi:= e^{piln(-3)}=e^{pi(ln|-3|+iarg(-3))}=e^{piln 3+ipi^2}=3^picdot e^{ipi^2}$$



where $e^{ipi^2}$ 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



$$e^{ipi^2}=cospi^2+isinpi^2$$



With WolframAlpha I get this numerical approximation:



$$(-3)^piapprox -28.47456 -i, 13.57354$$






share|cite|improve this answer











$endgroup$













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
















2












2








2





$begingroup$

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



$$(-3)^pi:= e^{piln(-3)}=e^{pi(ln|-3|+iarg(-3))}=e^{piln 3+ipi^2}=3^picdot e^{ipi^2}$$



where $e^{ipi^2}$ 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



$$e^{ipi^2}=cospi^2+isinpi^2$$



With WolframAlpha I get this numerical approximation:



$$(-3)^piapprox -28.47456 -i, 13.57354$$






share|cite|improve this answer











$endgroup$



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



$$(-3)^pi:= e^{piln(-3)}=e^{pi(ln|-3|+iarg(-3))}=e^{piln 3+ipi^2}=3^picdot e^{ipi^2}$$



where $e^{ipi^2}$ 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



$$e^{ipi^2}=cospi^2+isinpi^2$$



With WolframAlpha I get this numerical approximation:



$$(-3)^piapprox -28.47456 -i, 13.57354$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 29 at 18:00

























answered Jan 29 at 17:36









MasacrosoMasacroso

13.1k41747




13.1k41747












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




















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


















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






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




















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