Solutios for Math 4 Assigmet #2 A2. For each of the followig fuctios f : C C, fid where f(z is complex differetiable ad where f(z is aalytic. You must justify your aswer. (a f(z = e x2 y 2 (cos(2xy + i si(2xy (b f(z = z 2 + z 2 z 2 if z (c f(z = z 2 if z = x 3 if z 0 (d f(z = z 0 if z = 0 Here z = x + yi. Solutio. (a Clearly, e x2 y 2 cos(2xy ad e x2 y 2 si(2xy have cotiuous partial derivatives everywhere o R 2 ad are hece totally differetiable everywhere. Sice x + i = (cos(2xy + i si(2xy + e x2 y 2 x + i (e x2 y 2 (cos(2xy + i si(2xy x + i e x2 y 2 (cos(2xy + i si(2xy = e x2 y 2 (2x 2yi(cos(2xy + i si(2xy + e x2 y 2 ( 2y si(2xy 2xi si(2xy + 2yi cos(2xy 2x cos(2xy = 0 everywhere o C, f is complex differetiable ad aalytic everywhere. (b Clearly, f(z = z 2 + z 2 = 2(x 2 y 2 has cotiuous partial derivatives everywhere ad are hece totally differetiable o C. Sice x + i f = 4x 4yi http://www.math.ualberta.ca/ xiche/math47f/hw2sol.pdf
2 oly vaishes at (0, 0, f(z is oly complex differetiable at (0, 0. Sice f(z is ot complex differetiable i ay disk cetered at (0, 0, f(z is ot aalytic at (0, 0. Therefore, f(z is owhere aalytic. (c Clearly, f(z = z + for all z C. Sice z + is a polyomial i z, it is complex differetiable ad aalytic everywhere. (d For z 0, x + i f = x + i x 3 x2 + y 2 = 2x4 + 3x 2 y 2 (x 2 + y 2 3/2 i x 3 y (x 2 + y 2 3/2 is cotiuous ad oly vaish at x = 0. So f(z is complex differetiable at {z : Re(z = 0, z 0}. At z = 0, sice we have 0 x 3 x 2 + y 2 x f(x, y f(0, 0 0(x 0 0(y 0 z 0 (x, y (0, 0 So f is totally differetiable at z = 0 with x = 0 = 0 ( 0 0 x + i = 0. x 3 = z 0 x 2 + y = 0 2 Therefore, f(z is complex differetiable at 0. I coclusio, f(z is oly complex differetiable o the imagiary axis {z : Re(z = 0}. But for every z with Re(z = 0, f(z is ot complex differetiable i ay disk at z. So f(z is owhere aalytic. A2.2 Show that uder the chage of coordiates x = r cos θ ad y = r si θ, r = cos θ x + si θ ad θ = r si θ x + r cos θ Derive the Cauchy-Riema equatios uder polar coordiates.
Solutio. For every differetiable fuctio f(x, y, r θ (r cos θ (r si θ = + r x r = cos θ + si θ x = (r cos θ θ by chai rule. Therefore, x + (r si θ θ = r si θ + r cos θ x r = cos θ x + si θ ad θ It follows that [ ] [ r cos θ si θ = r si θ r cos θ θ Therefore, x + i = r ] [ x ] ( r cos θ r si θ θ So we obtai the CR equatios: r cos θ u u si θ r θ r si θ u u + cos θ r θ or equivaletly, = r si θ x + r cos θ [ ] [ x cos θ si θ = r si θ r cos θ [ r cos θ si θ = r + i r r si θ cos θ ] [ r ] [ r θ θ ( r si θ r + cos θ θ v v = r si θ + cos θ r θ v v = r cos θ + si θ r θ r u r = v θ u v = r θ r A2.3 Let f : D C be a complex fuctio o a domai (coected ad ope set D C. Show that if both f(z ad f(z are aalytic o D, the f(z is costat o D. ] ] 3
4 Proof. Let f(z = u(x, y + iv(x, y for u = Re(f ad v = Im(f. Sice both f(z ad f(z are aalytic o D, we have x + i f = 0 x + i (u + iv = 0 x + i f = 0 x + i (u iv = 0 x + i u = x + i v = 0 u x = v x = 0 f (z = u x + iv x = 0 o D. Ad sice D is coected, f(z is costat o D. A2.4 For a differetiable map f : U R m o a ope set U R, the Jacobia of f is the m matrix [ ] i if f is give by x j m f(x, x 2,..., x = (f (x, x 2,..., x, f 2 (x, x 2,..., x,. f m (x, x 2,..., x. Let f : D C be a complex fuctio o a ope set D C. Show that if f(z is complex differetiable at z 0 D, the det J f (z 0 = f (z 0 2 Proof. Let f(z = u(x, y + iv(x, y for u = Re(f ad v = Im(f. The [ ] ( ( ( ( u/x u/ u v u v J f = det(j v/x v/ f = x x Whe f(z is complex differetiable at z 0, u x = v, v x = u ad f (z 0 = u x + iv x
5 at z 0. Therefore, ( ( ( ( u v u v det J f (z 0 = x x ( 2 ( 2 u u = + x = f (z 0 2. A2.5 For a sequece {a : = 0,, 2,...} of complex umbers, show that a + a ad the equality holds if the it a + exists. You may assume that a 0 for all. Proof. We let b = l a l a for ad b 0 = l a 0. The l a = l a = b ( a = exp b a + ( = exp b So to show a + a it suffices to show that Suppose that b L = b. a b
6 The for every u > L, there exists N such that b u for all > N. Fixig u ad N, we have That is, b = N b m + m=n+ N ( N b m + b N b m + b u b m u for all u > L. Therefore, b L = b. ( N Replacig b by b, we have ( b ( b b b I coclusio, we have b Takig exp, we obtai a + b b b a u = u b b a Whe a + = a + a = a + a + a.
7 we have a + = a = a = a + a.