# Fluid flow described by a complex potential

Given this complex potential $$\phi(z)=(\cos \alpha-i\sin \alpha)z$$ $\alpha>0$

• Find the equations of the streamlines
• Find the components $V_x$ y $V_y$ of the velocity vector at $(x,y)$.What angle does the velocity vector make with the positive x-axis.

I tried finding the components $u(x,y)$ and $v(x,y)$ from $\phi(z)$ $$\phi(z)=e^{-i\alpha}(x+iy)$$ So,$$u(x,y)=e^{-i\alpha}x$$$$v(x,y)=e^{-i\alpha}y$$ Making $u(x,y)=C$,where $C$ is a constant, to find the equipotential lines for the velocity.$$u(x,y)=C$$ $$e^{-i\alpha}x=C$$ $$x=K$$ Where $K$ is another constant. To find the velocity components $$V=\nabla u(x,y)$$ So,$$V=(e^{-i\alpha},0)$$ Is this process correct?

In potential flow one has $\vec v = \nabla \phi$ and for incompressible fluids $\nabla\cdot\vec v = 0$ one finds $\nabla^2 \phi = 0,$ which is solved by the harmonic functions. It turns out that a complex-smooth ("holomorphic" or "analytic") function is described by two such functions $p(x, y),$ $q(x, y)$ which are harmonic conjugates, as $f(x + i y) = p(x, y) + i~q(x, y),$ where the equations to make this thing complex-differentiable (and hence to make $p, q$ into harmonic conjugates) are those which make the Jacobian matrix into a scaled 2D rotation matrix, since complex multiplication is scaled rotation:$${\partial p\over\partial x} = {\partial q\over\partial y},~~~{\partial p\over\partial y} = -{\partial q\over\partial x}.$$ So if we were to identify, say, your $\phi$ as some $p$ then this prescribes some $q$ and therefore $f$ such that $\partial_x f = v_x - i~v_y,$ and one can immediately generalize this (since $x$ is only the most straightforward direction to take $d/dz$) to say, $${df\over dz} = v_x - i v_y.$$
You will therefore know that you are wrong because you wrote something like $u(x, y) = e^{-i\alpha}~x$, this cannot be correct because the $u$ you have written is not a real function; it has an imaginary component. On the other hand $v_x$ and $v_y$ and $p$ and $q$ above are all real. So whatever you have written it is not the above. Instead you may wish to multiply out $(\cos \alpha + i\sin\alpha)(x + i y)$ and see what the real and imaginary parts actually are.
• ahh. You're right $u(x,y)$ is not a real function. Thank you Commented Oct 3, 2017 at 2:39