Complex potentials and gradient I am currently working on a question on complex potentials. I do have a complex potential $w$ defined as
$$w = \mathrm{Log}\bigg(z^2+\frac{a^4}{z^2}\bigg)$$ with $a>0$. $w$ ist the potential of a vector field $\vec{v}(x, y)$ built from the superposition of some basic vector field.
I did already decompose $w$ to $w=-2\mathrm{Log}(z)+\sum_{k=0}^{3}\mathrm{Log}(z-z_k)$ with $z_k$ as the roots of $z^4+a^4=0$. My solution is
$$\vec{v}=-\frac{2}{z}+\sum_{k=0}^{3}\frac{1}{z-z_k}$$
The solution manual lists as solution everything with conjugated complex numbers. Every $z$ is a $\overline{z}$ and I don't know why. Can anybody please help me?
Thanks.
 A: For two-dimensional potential flow, the velocity field $(u,v)$ has zero curl and must be the gradient of a potential $\phi:$
$$u = \frac{\partial \phi}{\partial x}, \,\,\,\,\, v = \frac{\partial \phi}{\partial y}$$
Typically we assume that the velocity components are infinitely differentiable with respect to the coordinates $(x,y)$.  The complex potential is an  analytic function $f(z) = \phi(x,y) + i \psi(x,y)$ of the complex variable $z = x+iy$, where the real part $\phi$ is the velocity potential and the imaginary part $\psi$ is called the streamfunction.
With $f$ analytic, we can apply the Cauchy-Riemann equations to obtain
$$u = \frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y}, \\ v= \frac{\partial \phi}{\partial y} = -\frac{\partial \psi}{\partial x},   $$
and the complex derivative is given by
$$\frac{df}{dz} = \frac{\partial \phi}{\partial x} + i\frac{\partial \psi}{\partial x} = \frac{\partial \psi}{\partial y} - i\frac{\partial \phi}{\partial y} = \frac{\partial \phi}{\partial x} - i\frac{\partial \phi}{\partial y} = u - iv.   $$
Notice that the derivative $\frac{df}{dz}$ gives us the so-called complex velocity $u - iv$ and not $u + iv$, where the real and imaginary parts are the velocity components with correct signs.
In order to get $u + iv$ we have to change the sign of $i$ everywhere in $f(z)$ and take the derivative with respect to $\bar{z}$. 
The ordinary conjugate of the complex number $f(z)$ is $\overline{f(z)} = \phi(x,y) - i\psi(x,y)$.   Define the conjugate complex function $\bar{f}$ as
$$\bar{f}(z) = \overline{f(\bar{z})} = \phi(x,-y) - i \psi(x,-y),$$
so
$$\bar{f}(\bar{z}) = \overline{f(\bar{\bar{z}})} = \overline{f(z)}= \phi(x,y) - i \psi(x,y).$$
Thus, 
$$\frac{d}{d\bar{z}}\bar{f}(\bar{z}) = \frac{\partial \phi}{\partial x} - i\frac{\partial \phi}{\partial(-y)} = \frac{\partial \phi}{\partial x} + i\frac{\partial \phi}{\partial y} = u +iv.$$
The bottom line is -- to get $u +iv$ -- replace $i$ with conjugate $-i$ and $z$ with conjugate $\bar{z}$ everywhere in $f(z)$ and then take the derivative with respect to the conjugate $\bar{z}$.
As an example, suppose the complex potential is $f(z) = -iz^2$.
Then
$$u - iv = \frac{df}{dz} = -2iz = -2i(x+iy) = 2y -i(2x)\\ \implies u = 2y, \,\,\,\,\, v = 2x $$
and
$$u + iv = \frac{d\bar{f}}{d\bar{z}} = \frac{d}{d\bar{z}}(i\bar{z}^2) = 2i\bar{z} = 2i(x-iy) = 2y +i(2x)\\ \implies u = 2y, \,\,\,\,\, v = 2x $$
A: Formally, the solutions of $z^4+a^4=0$ are
$$\left\{z_0= -(-1)^{1/4} a\right\},\left\{z_1= (-1)^{1/4}
   a\right\},\left\{z_2= -(-1)^{3/4} a\right\},\left\{z_3= (-1)^{3/4}
   a\right\}$$ Developed as complex numbers, they are
$$\left\{z_0= -\frac{(1+i) }{\sqrt{2}}a\right\},\left\{z_1= \frac{(1+i)
   }{\sqrt{2}}a\right\},\left\{z_2= \frac{(1-i) }{\sqrt{2}}a\right\},\left\{z_3=
   -\frac{(1-i) }{\sqrt{2}}a\right\}$$
