By Laywine C.F., Mullen G.L.

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**Additional resources for Discrete Mathematics Using Latin Squares**

**Example text**

Of the two arcs into which a and β divide a circle through them, each corresponds to a complete circle through λ and μ and these two circles are orthogonal. If — \π < arg(jS —α) ^ ^π, and if the circle through a and β is such that the arc of it which is described anti-clockwise from a to β contains an angle 0, then the interior of this circle corresponds to the two areas which lie inside one and outside another of the two orthogonal circles through λ and μ when arg k lies between θ — π and Θ. Some special cases will be found in Examples 3.

Hence, as z travels around the circle C and passes through zl9 w approaches wx and then recedes along a curve with the same tangent. On putting n = 2, zi = a, z2=-a, a2 dw Λ a2 — = 1—~, w = z+—9 ÛZ Z Z we obtain the special case considered above. If now instead of the circle C we take a slightly larger circle passing through z = a but slightly beyond z = —a and transform it, we get a figure with a cusp at w = 2a and a rounded end at w a little less than — 2a. f Glauert's Modification All Joukowski aerofoils have a cusp, whereas the trailing edge of an aeroplane wing is not a cusp.

To reduce jdx/yJX, where X is a quartic in x, to Legendre's standard form, use is made of a substitution due to Cayley. Write X = χχχ2 = (ax2 + 2bx + c){a'x2 + 2b'x + c')\ this we can always do if the coefficients in X are all real, which we are supposing. We then put Xt + u choosing λ9 μ so that αλμ + ο(λ + μ) + ο = 0 ) α'λμ + ν(λ + μ) + ο' = 0 fJ (3) are satisfied. For brevity, we shall omit the proof that real values of λ and μ can be found to satisfy (3) and illustrate the process by an example.