(−1)^π =? (−1)^((22)/7) =(−1)^(3+(1/7)) =(−1)^3 .(−1)^(1/7) =−(−1)^(1/7) let (−1)^π =t ⇒−(−1)^(1/7) =t ⇒(−1)^(1/7) =−t ⇒(−1)=(−t)^7 ⇒−1=−t^7 ⇒1=t^7 hence t=1 ∴(−1)^π =1 i request all math professionals to check this and if any error then pls comment.


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Question Number 110197 by redmiiuser last updated on 27/Aug/20

${(- 1)}^{π} = ?$ ${(- 1)}^{\frac{22}{7}}$ $= {(- 1)}^{3 + \frac{1}{7}}$ $= {(- 1)}^{3} . {(- 1)}^{\frac{1}{7}}$ $= - {(- 1)}^{\frac{1}{7}}$ $l e t {(- 1)}^{π} = t$ $\Rightarrow - {(- 1)}^{\frac{1}{7}} = t$ $\Rightarrow {(- 1)}^{\frac{1}{7}} = - t$ $\Rightarrow (- 1) = {(- t)}^{7}$ $\Rightarrow - 1 = - t^{7}$ $\Rightarrow 1 = t^{7}$ $h e n c e t = 1$ $∴ {(- 1)}^{π} = 1$ $i r e q u e s t a l l m a t h$ $p r o f e s s i o n a l s t o$ $c h e c k t h i s a n d i f$ $a n y e r r o r t h e n p l s$ $c o m m e n t .$
Commented by Dwaipayan Shikari last updated on 27/Aug/20

${(- 1)}^{π} = e^{π^{2} i} = c o s (π^{2}) + i s i n (π^{2})$ $I t h i n k i t i s a n i m a g i n a r y s o l u t i o n$
Commented by Aziztisffola last updated on 27/Aug/20

$why you put {(- 1)}^{π} = t and - {(- 1)}^{\frac{1}{7}} = t$ $you suppose that {(- 1)}^{π} = - {(- 1)}^{\frac{1}{7}}$ $but {(- 1)}^{π} \neq - {(- 1)}^{\frac{1}{7}} .$
Commented by Rasheed.Sindhi last updated on 27/Aug/20

$π i s a n i r r a t i o n a l n u m b e r a n d$ $i s n o t e q u a l t o \frac{22}{7} (r a t i o n a l)$ $Y o u {c a n}^{'} t u s e 22 / 7 i n t h e p l a c e$ $o f π .$ $(22 / 7 i s o n l y a p p r o x i m a t e d$ $v a l u e o f π)$
Commented by JDamian last updated on 27/Aug/20

${(- 1)}^{\frac{22}{7}} = {[{(- 1)}^{22}]}^{\frac{1}{7}} = {(1)}^{\frac{1}{7}} = 1^{\frac{1}{7}}$ $W h y d o y o u w r o n g l y a s s u m e π = \frac{22}{7} ?$
	Answered by mr W last updated on 27/Aug/20

	$- 1 = e^{π i}$ ${(- 1)}^{π} = e^{π^{2} i} = \cos π^{2} + i \sin π^{2}$
	Commented by redmiiuser last updated on 27/Aug/20

	$b u t s i r w h a t s t h e p r o b l e m$ $i n t h e a b o v e p r o c e s s$
	Commented by redmiiuser last updated on 27/Aug/20

	$w h y ?$
	Answered by 1549442205PVT last updated on 27/Aug/20

	$Set a = {(- 1)}^{π} \Rightarrow lna = π {lni}^{2} = 2 π \ln (i)$ $= 2 π [\ln 1 + i (\frac{π}{2} + 2 k π)] = i π^{2} (4 k + 1)$ $\Rightarrow a = {(- 1)}^{π} = e^{i π^{2} (4 k + 1)} (k \in Z)$ $(since \ln (i) = \ln (1) + i (\frac{π}{2} + 2 k π), k \in Z)$
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