y-x-i-x-R-i-2-1-is-y-C-

Question Number 4555 by FilupSmith last updated on 07/Feb/16

y = x^{i}

x \in R

i^{2} = - 1

is y \in C ?

Commented by Yozzii last updated on 07/Feb/16

A l l k n o w n n u m b e r s a r e m e m b e r s o f

C s i n c e C = R \cup {w h o l l y c o m p l e x n u m b e r s} .

C i s u n b o u n d e d a n d o p e n .

Commented by FilupSmith last updated on 07/Feb/16

Then, more spesifically, i s y \in R ?

Commented by Yozzii last updated on 07/Feb/16

y c o u l d b e r e a l o r c o m p l e x .

x \in R

I f x > 0, x = e^{l n x} \Rightarrow x^{i} = e^{i l n x} = y

x^{i} = c o s (l n x) + i s i n (l n x)

I f y = x^{i} \in R \Rightarrow s i n (l n x) = 0

\Rightarrow l n x = n π \Rightarrow x = e^{n π} (n \in Z) .

I f y = x^{i} \in {w h o l l y c o m p l e x} \Rightarrow c o s (l n x) = 0

∴ l n x = 2 n π \pm 0.5 π \Rightarrow x = e^{2 n π \pm 0.5} (n \in Z) .

F o r x > 0, x \neq e^{n π} a n d x \neq e^{2 n π \pm 0.5}, y \in C .

I f x < 0 l e t x = - a, a > 0 (a \in R)

\Rightarrow x = - a = a (c o s π + i s i n π) = {a e}^{π i}

x = e^{l n ({a e}^{i π})} = e^{l n a + i π}

\Rightarrow x^{i} = e^{i l n a + i^{2} π} = e^{i l n a - π} = e^{- π} (c o s (l n a) + i s i n (l n a))

S o, s i m i l a r l y f o r x < 0, y \in R o r y \in C .

I {d o n}^{'} t t h i n k y i s d e f i n e d f o r x = 0 .

Commented by Yozzii last updated on 07/Feb/16

{L e t}^{'} s c o n s i d e r t h e g e n e r a l f o r m

y = {(a + b i)}^{c + i d} w h e r e i^{2} = - 1,

a, b, c, d \in R, a n d i f a = b, t h e n a \neq 0 f o r

i f a = b = 0, y i s u n d e f i n e d s i n c e a r g (0)

i s u n d e f i n e d .

I n m o d u l u s - p o l a r f o r m,

a + b i = {r e}^{i θ}

w h e r e r = \sqrt{a^{2} + b^{2}}, r > 0, a n d θ, - π < θ ⩽ π, i s t h e

p r i n c i p a l a r g u m e n t o f t h e c o m p l e x n u m b e r

a + b i . e i s {E u l e r}^{'} s c o n s t a n t .

∴ y = {({r e}^{i θ})}^{c + i d} = r^{c + i d} e^{i θ (c + i d)}

y = r^{c} {(r^{d})}^{i} e^{c i θ - θ d} = r^{c} e^{- θ d} {(r^{d} e^{c θ})}^{i} = {n x}^{i}

w h e r e n = r^{c} e^{- θ d} > 0, x = r^{d} e^{c θ} > 0,

n, x \in R .

y = n {(e^{l n x})}^{i} = {n e}^{i l n x} = n (c o s (l n x) + i s i n (l n x))

y = n c o s (l n x) + i n s i n (l n x)

F r o m t h i s w e k n o w t h a t ∣ y ∣= n = r^{c} e^{- θ d}

a n d a r g (y) = l n x = l n (r^{d} e^{c θ}) = c θ + d l n r .

(1) I f y i s r e a l \Rightarrow n s i n (l n x) = 0

\Rightarrow l n x = t π \Rightarrow x = e^{t π}, t \in Z .

∴ y = n c o s (l n x) = r^{c} e^{- θ d} c o s ({l n e}^{t π})

y = {(a^{2} + b^{2})}^{\frac{c}{2}} e^{- d θ} c o s t π

y = {(- 1)}^{t} e^{- d θ} {(a^{2} + b^{2})}^{c / 2}, t \in Z .

{S i n c e x = r^{d} e^{c θ} a n d x = e^{t π} \Rightarrow e^{t π} = r^{d} e^{c θ}

∴ t π = d l n r + c θ

t = \frac{d l n r + c θ}{π} \in Z

I s i t n e c e s s a r y f o r t \in Z t h a t r = 1 ?}

(2) I f y i s w h o l l y c o m p l e x \Rightarrow n c o s (l n x) = 0

\Rightarrow l n x = 2 t π \pm 0.5 π \Rightarrow x = e^{2 t π \pm 0.5 π}, t \in Z

∴ y = n i s i n (2 t π \pm 0.5 π)

y = n i (s i n 2 t π c o s 0.5 π \pm s i n 0.5 π c o s 2 t π)

y = \pm n i = \pm r^{c} e^{- θ d} i

x = r^{d} e^{c θ} ∴ r^{d} e^{c θ} = e^{2 t π \pm 0.5 π}

d l n r + c θ = 2 t π \pm 0.5 π

t = \frac{d l n r + c θ \pm 0.5 π}{2 π} \in Z

(3) F o r r^{d} e^{c θ} \neq e^{(2 t \pm 0.5) π}, e^{q π} w h e r e

t, q \in Z, I m (y) \neq 0 a n d R e (y) \neq 0 a n d h a s t h e f o r m

y = r^{c} e^{- d θ} (c o s (c θ + d l n r) + i s i n (c θ + d l n r))

R e (y) = {(a^{2} + b^{2})}^{\frac{c}{2}} e^{- d θ} c o s (c θ + \frac{1}{2} d l n (a^{2} + b^{2}))

I m (y) = {(a^{2} + b^{2})}^{\frac{c}{2}} e^{- d θ} s i n (c θ + \frac{1}{2} d l n (a^{2} + b^{2}))

Commented by FilupSmith last updated on 07/Feb/16

Wow! This is amazing and very interesting!

I love it!