1-pi-1-22-7-1-3-1-7-1-3-1-1-7-1-1-7-let-1-pi-t-1-1-7-t-1-1-7-t-1-t-7-1-t-7-1-t-7-hence-t-1-1-pi-1-i-request-all-math-profe

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)