1550251515-Classical_Complex_Analysis__Gonzalez_

350 Chapter6

Fig. 6.7

By eliminating the parameter () we obtain the equation of the graph of
the Kasner circle in the form

1(-fzl = lfzl (6.11-2)

or, alternatively, letting ( = e + i17, and taking into account (6.7-8) and
(6.7-9),

[e -%(u,,, + vy)]^2 + [11-^1 Mv,,, -uy)]^2 =^1 / 4 ((u,,, - vy)^2 + (v,,, + uy)^2 ]

Hence the center of the circle is at the point

fz = %[(u,,, + vy) + i(v,,, - uy)]

and its radius is

lf.zl = %.j(u,,, - vy)^2 + (v,,, + uy)2

The Kasner circle reduces to the point fz when f is monogenic at z.

Example Let f(z) = z. Then fz = O, fz = 1, and ( = e-^2 i^6 (0::; ()::; 271').
Thus the Kasner circle is in this case a unit circle with center at the origin
described twice in the clockwise direction (Fig. 6. 7).

EXERCISES 6.2

1. (a) F'ind the directional derivative of the function f(z) = z^2 Imz at
   ·the point z = 1 in the direction () = -^1 / 4 71'.
   (b) Determine the point of the complex plane where the function in
   part (a) has an ordinary derivative.


2. (a) Find the directional derivative of the function f ( z) = ( z - 1 )^2 Re z
   at the point z = 0 in the direction () = 71' /3.
   (b) Determine the points of the complex plane where the function in
   part (a) has an ordinary derivative.