Complex velocity of line vortex, simplification I am considering the potential flow where I have a uniform flow past wing and two line vortices, one at the origin and one at a position $(x_1,y_1)$ relative to a wing of chord $D$.
I'm using the following potentials
$$\phi_f(x,y)=Ux+Vy$$
$$\phi_{v1}(x,y)=\frac{\Gamma_1}{2\pi}\tan^{-1}\left(\frac{y}{x}\right)=\frac{\Gamma_1}{2\pi}\theta$$
$$\phi_{v2}(x,y)=\frac{\Gamma_2}{2\pi}\tan^{-1}\left(\frac{y-y_1}{x-x_1}\right)=\frac{\Gamma_2}{2\pi}\theta_1$$
I've tried to calculate the velocity potential $W(z)=\phi+i\psi$ where $\phi_x=\psi_y$, and $\phi_y=-\psi_x$ and I'm getting stuck when I try to simplify the equations.
Just considering the 1st vortex for now my text uses an example where and it says $\theta-i\ln(r)=\ln(z)$ where $z=x+iy$. Basically, I've been able to show that 
$$Z=\frac{y}{x}=-i\frac{z-z^*}{z+z^*}$$
where $z^*$ is the complex conjugate. I was able to rewrite the arc tan part of the equation
$$\tan^{-1}(Z)=\frac{i}{2}\ln\left(\frac{1-iZ}{1+iZ}\right)$$ and $$\ln(r)=\frac{i}{2}\ln\left(zz^*\right)$$ so that
$$\theta-i\ln(r)=\frac{i}{2}\ln\left(\frac{1-iZ}{1+iZ}\right)-\frac{1}{2}\ln\left(zz^*\right).$$
Could I get some help figuring this out? I haven't been able to get any further.
 A: Playing around with the algebra with complex numbers in polar form led to this solution.
Define the complex number $z=x+iy$ in polar form, along with its conjugate.
$$z= re^{i\theta},\; z^*=re^{-i\theta}$$
The product and quotient of the complex number and its conjugate respectively are.
$$\therefore zz^*=r^2,\; \ln\left(\frac{z}{z^*}\right)=2i\theta$$
The complex potential for a line vortex with the potential and streamfunction is
$$w(z)=\phi+i\psi$$
$$\phi_v=\frac{\Gamma}{2\pi}\theta,\; \psi_v=-i\frac{\Gamma}{2\pi}\ln\left(r\right)$$
$$w_v(z)=\frac{\Gamma}{2\pi}\left(\theta-i\ln(r)\right)$$
Follow through with some algebra
$$w_v(z)=\frac{\Gamma}{2\pi}\left(\frac{2i}{2i}\theta-\frac{2}{2}i\ln(r)\right)$$
$$w_v(z)=\frac{\Gamma}{2\pi}\left(\frac{1}{2i}\ln\left(\frac{z}{z^*}\right)-\frac{1}{2}i\ln(zz^*)\right)$$
$$w_v(z)=\frac{\Gamma}{4\pi i}\left(\ln(z)-\ln(z^*)+\ln(z)+\ln(z^*)\right)$$
$$w_v(z)=\frac{\Gamma}{4\pi i}(2\ln(z))=\frac{\Gamma}{2\pi i}\ln(z)$$
The methodology shown here is valid even when the vortex is at $(x_1,y_1)$, or represented as $(r_1,\theta_1)$.
