Section 31- 32, The Logarithmic function Flashcards
let $0 \not = z \in C$
If $e^w = z$ then
$w = ln|z| + i arg(z) = ln|z| + i(Arg(z) + 2 \pi i) , n \in Z$
Multiple valued logarithmic function log$z$ is defined as
log $z = ln |z| + i arg(z) = ln|z| + i(Arg(z) + 2 \pi i) , n \in Z$
Clearly $e^{log(z)}=$
z
for $z \not = 0, z \in C)
log($e^z$) = ?
$z + 2 n \pi i, n \in Z$
Log is a single valued function onwhat domain
How is Log z into log z
domain of Log z is $\mathbb{C}${0}
and
$log z = Log z + 2 n \pi i, n\in Z$
When and how does the complex logarithm turn into the real case
If $z = x + i 0, x>0$ then $|z| = x$ and Arg z = 0
so that Log z = ln x reduces to teh real logarithmic function
log$(z_1 z_2) = $
$log z_1 + log z_2$
log $\frac{z_1}{z_2} = $
$log z_1 - log z_2$
$z^n =$
$e^{nlogz} \qquad n \in Z$
$z^{1/n}=$
$e^{(1/n)logz} \qquad n\in Z$
log$(z_1 z_2) = $
NOT
$Log z_1 + Log z_2$
UNLESS
Re$(z_1) > 0$ and Re$(z_2)>0$