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Question Number 34634 by abdo mathsup 649 cc last updated on 09/May/18

l e t f (x) = l n (1 + i x) w i t h ∣ x ∣< 1

1) e x t r a c t R e (f (x)) a n d I m (f (x))

2) d e v e l o p p f (x) a t i n t e g r s e r i e .

Commented by math khazana by abdo last updated on 10/May/18

1) w e h a v e 1 + i x = \sqrt{1 + x^{2}} (\frac{1}{\sqrt{1 + x^{2}}} + i \frac{x}{\sqrt{1 + x^{2}}})

= r e^{i θ} \Rightarrow r = \sqrt{1 + x^{2}} a n d c o s θ = \frac{1}{\sqrt{1 + x^{2}}},

s i n θ = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow t a n θ = x \Rightarrow θ = a r c t a n x \Rightarrow

l n (1 + i x) = l n \sqrt{1 + x^{2}} + i a r c t a n x

= \frac{1}{2} l n (1 + x^{2}) + i a r c t a n x \Rightarrow

R e (f (x)) = \frac{1}{2} l n (1 + x^{2}) a n d I m (f (x)) = a r t a n x

2) w e h a v e {l n}^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n}

\Rightarrow l n (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} u^{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} u^{n}

\Rightarrow \frac{1}{2} l n (1 + x^{2}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n} x^{2 n} a l s o

{a r c t a n}^{^{'}} x = \frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} \Rightarrow

a r c t a n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} f i n a l l y

f (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{2 n} x^{2 n} + i \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} .

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

l n (1 + i x) = \frac{1}{2} l n (1^{2} + x^{2}) + {i t a n}^{- 1} (\frac{x}{1})

= \frac{1}{2} l n (1 + x^{2}) + {i t a n}^{- 1} (x)

l n (1 + t) = t - \frac{t^{2}}{2} + \frac{t^{3}}{3} - \frac{t^{4}}{4} + \dots .

{t a n}^{- 1} (x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \dots .

s o r e q u i r e d a n s i s

= \frac{1}{2} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{3} - \frac{x^{8}}{4} \dots) + i (x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \dots .)