Part-1-S-ln-a-bi-a-b-R-i-2-1-is-S-C-b-0-Part-2-t-ln-ln-ln-k-k-R-k-gt-1-t-C-is-t-undefined-

Question Number 5091 by FilupSmith last updated on 11/Apr/16

Part 1

S = \ln (a + b i), a, b \in R

i^{2} = - 1

is S \in C ? (b \neq 0)

Part 2

t = \ln (\ln (\dots (\ln (k)))), k \in R, k > 1

∴ t \in C ?

is t undefined ?

Commented by prakash jain last updated on 11/Apr/16

S = x + i y

e^{x + i y} = a + b i

e^{x} (\cos y + i \sin y) = a + b i

Commented by 123456 last updated on 11/Apr/16

e^{x} \cos y = a

e^{x} \sin y = b

e^{2 x} = a^{2} + b^{2}

e^{x} = \sqrt{a^{2} + b^{2}}

x = \ln \sqrt{a^{2} + b^{2}}

\tan y = \frac{b}{a}

y = \arg (a + i b) + 2 π k

Commented by Yozzii last updated on 12/Apr/16

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

s u c c e s s i v e l n - a p p l i c a t i o n s t o k > 1, t h e r e s u l t

i s n e g a t i v e s o t h a t t h e l n - a p p l i c a t i o n s

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

r e s u l t . t w o u l d b e c o m e u n d e f i n e d i f

e v e r t_{n} = 0 i n t_{n + 1} = {l n t}_{n}, t_{1} = l n k, n ⩾ 1 .

S u p p o s e a f t e r k + 1 l n - a p p l i c a t i o n s

t_{k + 1} < 0 . \Rightarrow {l n t}_{k} < 0 \Rightarrow 0 < t_{k} < 1 .

{l n t}_{k} = (- {l n t}_{k}) (- 1 + 0) = (- {l n t}_{k}) (c o s π + i s i n π)

{l n t}_{k} = (- {l n t}_{k}) e^{π i} = t_{k + 1}

\Rightarrow t_{k + 2} = {l n t}_{k + 1} = l n ((- {l n t}_{k}) e^{π i})

t_{k + 2} = l n (- {l n t}_{k}) + π i

\Rightarrow t_{k + 2} \in C .

I f {t_{n}} c o n v e r g e s {w e}^{'} d n e e d t o s o l v e

u = e^{u} w h e r e f o r s u f f i c i e n t l y l a r g e n,

t_{n + 1} = t_{n + 2} = \dots . = u .

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