Let $\theta \in \mathbb { R }$. Let $g$ be the function from $] - 1,1 [$ to $\mathbb { C }$ defined by $$g ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { \mathrm { e } ^ { \mathrm { i } n \theta } } { n } x ^ { n }$$ a) Show that $g$ is of class $C ^ { 1 }$ on $] - 1,1 [$ and that, for all $x \in ] - 1,1 [$, $$g ^ { \prime } ( x ) = \frac { \mathrm { e } ^ { \mathrm { i } \theta } - x } { x ^ { 2 } - 2 x \cos \theta + 1 }$$ b) Show that, if $x \in ] - 1,1 [$, $$h ( x ) = - \frac { 1 } { 2 } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) + \mathrm { i } \arctan \left( \frac { x \sin \theta } { 1 - x \cos \theta } \right)$$ is well defined and that $h ( x ) = g ( x )$.
Let $\theta \in \mathbb { R }$. Let $g$ be the function from $] - 1,1 [$ to $\mathbb { C }$ defined by
$$g ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { \mathrm { e } ^ { \mathrm { i } n \theta } } { n } x ^ { n }$$
a) Show that $g$ is of class $C ^ { 1 }$ on $] - 1,1 [$ and that, for all $x \in ] - 1,1 [$,
$$g ^ { \prime } ( x ) = \frac { \mathrm { e } ^ { \mathrm { i } \theta } - x } { x ^ { 2 } - 2 x \cos \theta + 1 }$$
b) Show that, if $x \in ] - 1,1 [$,
$$h ( x ) = - \frac { 1 } { 2 } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) + \mathrm { i } \arctan \left( \frac { x \sin \theta } { 1 - x \cos \theta } \right)$$
is well defined and that $h ( x ) = g ( x )$.