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Question Number 187359 by Humble last updated on 16/Feb/23

w h a t a r e t h e t w o c o m p l e x s o l u t i o n t o

X^{- x} + {(- X)}^{x} = 0 i n a d d i t i o n t o \pm 1 ?

Answered by Frix last updated on 16/Feb/23

x = r i; let r > 0 [due to symmetry]

{(r i)}^{- r i} + {(- r i)}^{r i} = 0

{2 e}^{\frac{π r}{2}} \cos (r \ln r) = 0

\cos (r \ln r) = 0

r \ln r = (n - \frac{1}{2}) π; n \in Z

Since the minimum of r \ln r is

- \frac{1}{e} \approx - . 367879 \Rightarrow n \approx . 382900

and r \in R

\Rightarrow

r \ln r = (n - \frac{1}{2}) π; n \in N^{★}

We get infinite solutions

n = 1 r \approx 2.10729948

n = 2 r \approx 3.64417367

n = 3 r \approx 4.92574568

\dots

Commented by Humble last updated on 16/Feb/23

n i c e a p p r o a c h . o n e c a n a l s o g o w i t h

L a m b e r t W f u n c t i o n

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