The field is always directed along the first coordinate axis. Therefore, its streamlines are lines parallel to that axis. Very easy to sketch. If you wish to connect this with the slope-from-the-ratio method, consider that the ratio $0/(x_1x_2)$ is zero. So, the slope of streamlines is everywhere zero.

Sure, the points with $x_1x_2=0$ are exceptional: they are stationary points of the field. You may want to mark them somehow to indicate this fact.

You'll have to accept or justify the definition that
$$(1) \quad \ln(a+b \cdot i)={{\ln(a^2+b^2)} \over 2}+\left({{\pi \cdot sign(b)} \over 2}-\tan^{-1}{ a \over b} \right) \cdot i$$
$$W=C \cdot \ln(z^2+a^2)$$
Where $C={m \over {2 \cdot \pi}}$

The stream function is given by the imaginary part of the complex potential. Assuming $z=x+y \cdot i$

$$z^2=x^2+2 \cdot x\cdot y \cdot i-y^2$$

Comparing with (1), we can see that the real and imaginary parts inside the logarithm are
$$a=x^2-y^2+a^2$$
and
$$b=2 \cdot x \cdot y \cdot i$$
The imaginary part of (1) is the stream function

$$\Psi=\left({{\pi \cdot sign(b)} \over 2}-\tan^{-1}{ a \over b} \right)$$

Substituting in and simplifying we get...
$$(2) \quad \Psi={C \over 2} \cdot \left(2 \cdot \tan^{-1} \left({{x^2-y^2+a^2} \over {2 \cdot x \cdot y}} \right)-\pi \cdot sign(x \cdot y) \right)$$

You'll want to justify this next part in more detail. The streamlines are where the stream function is constant. We'll get rid of the right hand side of (2) under justification that it's constant. Then since everything added to the left should be constant, substitute the value for a new constant. We'll get...

$$(3) \quad {{2 \cdot \Psi} \over C}+B=D=2 \cdot \tan^{-1} \left({{x^2-y^2+a^2} \over {2 \cdot x \cdot y}} \right)$$

$$(4) \quad tan^{-1}(D)=2 \cdot L=k={{x^2-y^2+a^2} \over {x \cdot y}}$$

finally,

$$(5) \quad x^2-y^2+a^2=k \cdot x \cdot y$$

## Best Answer

The velocity field $V(x)$ says what velocity fluid particles have at each particular point.

Streamline is a curve along which said particle flow. Denote its parametric form by $\gamma(t)$, where $t$ is time.

Since the velocity is the derivative of position, $V(\gamma(t))=\gamma'(t)$. This equation relates velocity field to streamline. The derivative $\gamma'(t)$ is also known as the tangent vector. It is usually drawn as an arrow beginning at $\gamma(t)$, not at $0$ - hence tangent, not just parallel to tangent. (There is a good reason for it, too: in the language of manifolds, $\gamma'(t)$ is an element of the tangent space to $\gamma$ at $\gamma(t)$).