show-that-pi-ie-1-2-0-

Question Number 82041 by M±th+et£s last updated on 17/Feb/20

s h o w t h a t

π^{i e} + \frac{1}{2} = 0

Commented by mr W last updated on 17/Feb/20

c e r t a i n l y {i t}^{'} s n o t y o u r f a u l t t h a t i t i s

n o t e q u a l t o z e r o . i t i s s o, n o b o d y i s

t o b l a m e f o r i t .

Commented by mr W last updated on 17/Feb/20

π^{i e} + \frac{1}{2} = e^{i (e \ln π)} + \frac{1}{2} = \cos (e \ln π) + \frac{1}{2} + i \sin (e \ln π)

\approx - 0.499553 + 0.029890 i \neq 0

Commented by mr W last updated on 17/Feb/20

s o m e t h i n g i s w r o n g w i t h t h e q u e s t i o n .

Commented by M±th+et£s last updated on 17/Feb/20

t h a n k s f o r a l l a b o u t t h e s o l u t i o n d a n d i a m

s o s o r r y i t s m y f a u l t i t {d o e s n}^{'} t e q u a l z e r o

Answered by MJS last updated on 17/Feb/20

a^{i b} =

[a > 0]

= \cos (b \ln a) + i \sin (b \ln a)

{\begin{cases} \frac{1}{2} + \cos (b \ln a) = 0 \\ \sin (b \ln a) = 0 \end{cases}

{\begin{cases} b \ln a = 2 k_{1} π \pm \frac{2 π}{3}; k_{1} \in Z \\ b \ln a = k_{2} π; k_{2} \in Z \end{cases}

\Rightarrow k_{2} = 2 k_{1} \pm \frac{2}{3} \Rightarrow k_{1} \in Z xor k_{2} \in Z \Rightarrow no solution

\Rightarrow \forall a \in R^{+} \forall b \in R : a^{i b} + \frac{1}{2} \neq 0

z = \cos (b \ln a) + i \sin (b \ln a)

\Rightarrow abs z = 1 \land \arg z = b \ln a

\Rightarrow π^{ie} = e^{ie \ln π} \approx - . 999553 + . 0298898 i

\Rightarrow π^{ie} + \frac{1}{2} \neq 0

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