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I'm currently studying a Fluid Dynamics module and mock exam question has me completely stumped, I have been given the complex potential and shown it to be in the form given, however when trying to separate the complex potential into velocity potential and stream function I cannot seem to do the manipulation. The question is as follows:

$w(z)=\frac{m}{2\pi}log[z^2+a^2]$

Show that the streamlines are given by $x^2-y^2+a^2=kxy$

where k is a constant.

Any help would be appreciated! Thanks in advance.

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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,

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

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  • $\begingroup$ Thank you for the quick response! Looking at my scribbling I was on the right track however my manipulation let me down towards the end! Here's to hoping for a nicer question on this in tomorrows exam! $\endgroup$ – user2662468 Jun 10 '15 at 16:35

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