First, sorry if my english does not correspond to the real/formal worlds that I should use for this domain in which i'm new. (logarithm in complex analysis and potential/flow)
I know I can only ask for one question but here it is a part of a whole problem and I also want to know if my reasoning is good so ...
I'm asked a small problem to study a potential defined by a function (I dont know what is $a$ ... ) $$ \Phi\left(z\right)=\ln\left(\frac{z-a}{z+a}\right) $$ in which I guess ln is defined on $\mathbb{C} \setminus \mathbb{R}^{-}$ by $$ \ln\left(z\right)=\ln\left(\left|z\right|\right)+i\text{arg}\left(z\right)=\ln\left(\rho\right)+i\theta $$ 1] Find the domain $D$ where $\Psi$ is holomorphic and explicit $\Phi'$.
2] Write the velocity field ( which is $\mathscr{C}^{\infty}$ ) of this flow. Is it possible to define it on an open set which contains strictly $D$ ?
3] Show that the flows lines are given by circular wedges which extremities are $a$ and $-a$.
4] With $\gamma^{+}$ a Jordan contour, discuss of the value of the integral $$ \int_{\gamma^{+}}\overline{V}\left(z\right) \text{d}z $$ where $\overline{V}$ is the conjugate of $V$.
- My attempt is to find where $\displaystyle z \mapsto \frac{z-a}{z+a}$ is a negative real. With $z=x+iy$ i've found that $x$ and $y$ verify $$ \frac{z-a}{z+a}=\frac{x^2+y^2-a^2+iya}{x^2+2ax+a^2y^2} $$ Hence $y$ needs to be $0$ in order to be a real ( or $a=i \alpha$ ) Hence it is a negative real iif $x^2+y^2 \leq a^2$ then it would be defined on $\mathbb{C}$ minus a circle centered in $(0,0)$ of radius $a$ and for $z \in D$
$$ \Phi'\left(z\right)=\frac{z+a-z+a}{\left(z+a\right)^2}\frac{z+a}{z-a}=\frac{2z}{\left(z+a\right)\left(z-a\right)} $$
- Should I calculate $$v_{x}=\frac{\partial \Psi}{\partial y}\left(x,y\right) \text{ and }v_{y}=-\frac{\partial \Psi}{\partial x}\left(x,y\right) $$
$$ \frac{z-a}{z+a}=K \Leftrightarrow z=a\frac{1+K}{1-K} $$ What can I conclude ? $z=cst \ \Rightarrow$ $x$ and $y \ \Rightarrow $ $\left|z\right|$ is constant ?
No idea of how to proceed. Any ideas ? Thanks !