CONFORMAL TRANSFORMATION TO PROBLEMS IN HYDRODYNAMICS. 459 

 Let k > 0, and let be the surd conjugate to < &. Then 



= (wkfcw, 



which is positive for iv = 0, negative for w k, and positive for w = + oo. Hence 

 <&& has real roots, of which one (g) is between and k, and the other (g') is greater 

 than k. The values are 



g = k+^c (kc + JfC 2 )' 1 '*, g' = k + ^c+ (kc+^c^)' 1 ' (02) 



Now ^ is negative for w = 0, positive for w = k and for w = +00. Hence & has 

 one zero, which must be g. But 



wherein & > 0, has no zeros and is a conformal curve-factor. Its order at infinity is 

 zero, and therefore its angular range ought to be zero. This is readily verified, for if 

 X be the vector angle of ^ u when ^v is real and negative 



tan = cw' /2 w + & 



As w increases from zero the denominator of this expression diminishes from 

 + oo to a minimum value corresponding to w = />', namely 2k' 1 ', and then increases 

 again to infinity. Hence x diminishes from zero to a minimum value tan 1 (c 1/s /2i 1 ''), 

 and then increases again to zero. 



When w is negative the modulus of ^ M is {(g'w)/(gw)} ll \ 



When k < let it equal k''; then <$ takes the form 



(cw) ll > .......... (64) 



This can have no real zeros, since the two terms when both real are also both 

 positive. There can be no imaginary zeros in the relevant region because the vector 

 angle is x = tan" 1 {(cw) ll -/(w + k')}, which increases from zero (for w = 0) through 

 \-w (for w = k') to TT (for w = oo) and so gives an angular range TT, while the order 

 at infinity is unity. Thus ^ 15 is a conformal curve-factor. 



When w is negative 



which has real or imaginary factors according as c is greater or less than 4&'. In the 

 former case c may be put equal to 4F cosh 2 ft, so that ^ 15 takes the form 



......... (65) 



