ANSWERS 591ll. We evaluate lim f(• ) - / (O) = lim l•l'-o = lim •• = lim z = 0. Follow the
z - O z-0 z - 0 .z-0 z-0 z z-0
hint for the rest.13. f(••lt2- !C.-% 1 il = i^3 t - 1 -^13 = t-i-l - 1 = i. The minimum modulus of points on the
line y = 1 -x is {j- (prove this!). But f' (z) = 3z^2 , and the only solutions
to the equation 3z^2 = i have moduli equal to 1 (prove), which is less than
{j-(prove this also).Section 3.2. T h e Cauc hy-Riemann Equations: page 110la. u(x, y) = -y, v(x, y) = x + 4; u., =Vy= 0, and Uy= -v., = - L The
partials are continuous everywhere, so f' (z) = u., +iv.,= i for all z.
le. u,, = Vy = -2 (y + 1) and Uy = -v., = -2x. The partials are continuous
everywhere, so f' (z) = u., +iv., = - 2 (y + 1) + i 2x for all z.
le. f is differentiable only at z = i, and f' ( i) = 0.
lg. U:r: =Vy = 2 x, Uy = 2y, and v., = 2y. The conditions necessary for Theorem
(3.4) are satisfied if and only if y = 0, and for z = (x, 0), f' (z) = 2x.- a = 1 and b = 2.
- f' (z) = f" (z) = e"' cosy+ ie"' sin y by Theorem (3.4).
7a. u,, = -eYsinx, Vy= eYsinx, Uy= eYcosx, -v., = - eYcosx. The Cauchy-
Riemann equations hold if and only if both sinx = 0 and cosx = 0, which
is impossible.9a. u., = sinhxsiny =Vy and Uy= coshxsiny = -v,,. The partials are contin-
uous everywhere, so f is entire.lla. f is differentiable only at points on the coordinate axes. f is nowhere ana.-
lytic.
Uc. f is differentiable and analytic inside quadrants I and Ill.
- The form of the definition is identical, but the meaning is more subtle in
the complex case. For starters, the limit must exist when z --+ zo from any
direction in the complex case. The real case is limited to two directions.
15. Since f = u + iv is analytic, u and v must satisfy the Cauchy- Riemann
equations. Since f is not constant, this means the functions u and - v do
not satisfy the Cauchy-Riemann equations. Explain why this is the case,
and then use T heorem (3.3) to conclude that g = u - iv is not analytic.