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