A={z∈C: 2<∣z∣<4} fine log(A) where log is complex logaritmique


Question and Answers Forum

All Questions Topic List
Logarithms Questions
Previous in All Question Next in All Question
Previous in Logarithms Next in Logarithms
Question Number 169918 by pticantor last updated on 12/May/22

$A = {z \in C : 2 <∣ z ∣< 4}$ $f i n e l o g (A)$ $w h e r e l o g i s c o m p l e x l o g a r i t m i q u e$
	Answered by pticantor last updated on 12/May/22

	$b e s o i n d^{'} a i d e p l e a s e!!$
	Answered by Mathspace last updated on 12/May/22

	$w h a t m e a n l o g A . . .$
	Commented bypticantor last updated on 13/May/22

	$l o g A I s i m a g e o f A b y t h e {l o g a r i t h m i c}^{'} f o n c t i o n$ $a n d A i s t h e r i n g$ $o r$ $f : A \to C$ $z ↣ l o g (z)$ $f i n d l o g (A)$
	Answered by Mathspace last updated on 13/May/22
	$if z=re^(iθ) ⇒∣z∣=r⇒2<r<4 and logz=logr+i(θ+2kπ) ⇒ logA={logr+i(θ+2kπ) /k∈Z} une determination principale est logr+iθ −π<θ<π$
	$i f z = {r e}^{i θ} \Rightarrow∣ z ∣= r \Rightarrow 2 < r < 4$ $a n d l o g z = l o g r + i (θ + 2 k π) \Rightarrow$ $l o g A = {l o g r + i (θ + 2 k π) / k \in Z}$ $u n e d e t e r m i n a t i o n p r i n c i p a l e e s t$ $l o g r + i θ - π < θ < π$
	Commented bypticantor last updated on 13/May/22

	$j e c r o i s q u e v o u s a v e z j u s t e d o n n e r$ $l a d e f i n i t i o n d u l o g a r i t h m e c o m p l e x!$ $m a i s o n m e d e m a n d e l a n a t u r e d e l o g (A) e t l e$ $t r a c a g e d e l a c o u r b e$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum