1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation

= ( G - E) sin 20 + 2F cos 20

and

d2.

d0 2 (p^2 ) = 2( G - E) cos 20 - 4F sin 20 = h( 0)

Equating the first derivative to zero, we have

2F

tan20 = --

E-G

369

(6.16-5)

(6.16-6)

(6.16-7)

provided that E of. G. If 01 is a solution of (6.16-7), all other solutions

are given by 01 +^1 / 2 mr. Hence there are four essentially different direction

angles, 01, 02 = 01 +! 7r, 03 = 01 + 7r, and 04 = 01 + 37r /2, for which p may

have either a local maximum or a local minimum.

Let

D = y'(E-G)^2 +4F^2

and suppose that 01 is that solution of (6.16-7) for which

. 20 2F
sm 1= D and

By substitution in (6.16-6) we have

E-G

cos20 1 = n-

h(01) = -2 (E - G)2 + 4F2 = -2D

D

£'{6.16-8)

and similarly, h(0 3 ) = -2D. Hence p attains a maximum in the two

opposite directions 01 and 01 + 7r.
For 02 = 01 + % 7r we have cos 202 = -cos 201 and sin 202 = -sin 201.
Hence

h(02) = 2 (E - G)2 + 4F2 = 2D

D

and similarly, h(0 4 ) = 2D. Thus p attains a minimum in the two opposite
directions 02 and 02 + 7r. These directions are perpendicular to those giving
a maximum. The lines (not the rays) determined by 01 and 02 are called

the principal directions of f at z.

If E = G, F of. O, (6.16-5) equated to zero gives cos 20 = 0. Hence

a solution is 01 =. 7r / 4, and the other essentially different solutions are
02 = 37r/4, 03 = 57r/4, and 04 = 77r/4. In this case we have

h(0 1 ) = h(03) = -4F

and