I am looking for various examples on analytic function which maps to convex domain. Specially square functions are of our interest.
Hi, I tried to model this under MAPLE. If z is a complex number written as z=cos(theta)+I*sin(theta), then the real part of sqrt(1-z^4) takes the following form:
1/2*(2*((1-cos(theta)^4+6*cos(theta)^2*sin(theta)^2-sin(theta)^4)^2+(-4*cos(theta)^3*sin(theta)+4*cos(theta)*sin(theta)^3)^2)^(1/2)+2-2*cos(theta)^4+12*cos(theta)^2*sin(theta)^2-2*sin(theta)^4)^(1/2)
and this function of theta is not convex on [0,2*pi] as can be seen on the attached plot.
Nevertheless, my MAPLE calculus should be verified
3. The function is holomorphic in the interior of the unit disc, hence the maximum modulus principle applies.
2. Its maximum is not attained; same with minimum. However the sup value equals the vale of the continuous extention at z_1 and other three periodically distributed 8-th degree roots of unity, where it equals \sqrt{2}. The minimum of the continuous extension is attained at the remaining four roots: 1, i, -1, i and equals 0
1. The function on the segment of the straight line: { z = t, t \in (0: 1) } is a real valued function of a real variable $t$ with negative second derivatine. Thus, it cannot be convex. Moreover, for z_1 = e^{i \pi / 4} and z_2 = e^{- i \pi / 4} the value of the function is \sqrt{2}, wheras at the midle-point z_3 = 1/\sqrt{2} the value is less than 1 - thus the divided difference at these points is positive means: the function is meither convex nor concave (the continuity arguments have to be used, since the point z_1 and z_2 should be chosen inside the disc, but this can be done with arbitrary close points)
Regards
Remark. Obviously, the real part of a holomorphic non-constant function is never convex nor concave, since their Laplacean is zero and in almost all points the directional derivative wrt x and y are non-zero and of oposite sign. The examples proving directly the claim has been given for showing that no advanced analysis is necessary (in the case of the given function), elementary calculus is sifficient :)
Regards
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