MILNE-THOMSON CIRCLE THEOREM Let )(zf be the complex potential for a flow having no rigid boundaries and such that there are no singularities within the circle az = . Then on introducing the solid

cylinder az = , with impermeable boundary, into the flow, the new complex potential

for the fluid outside the cylinder is given by )/()( 2 zafzfW += for az ≥ . Proof All singularities of )(zf occur in the region az > . Hence the singularities of

)/( 2 zaf occur in the region aza >/2 , i.e., az < . Thus the singularities of

)/( 2 zaf also lie in the region az < . It follows that in the region az > , the functions )(zf and )/()( 2 zafzf + both have the same analytical singularities. Thus both functions considered as complex potentials represent the same hydrodynamical distributions in the region az > . The proof of the theorem is now completed by considering what happens on the circular boundary az = . To this end, we write θ= iaez on the boundary where θ is

real. Then zaeza i == θ−/2 on the circular boundary. Thus, on the boundary az = ,

)()()/()( 2 zfzfzafzfW +=+= ,

which is entirely real. Hence on the boundary,

0Im ==ψ W .

This shows that the circular boundary is a streamline across which no fluid flows. Hence az = is a possible boundary for the new flow specified by the complex

potential )/()( 2 zafzfW += .