Math 210A Homework 1

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1 Math 0A Homework Edward Burkard Exercise. a) State the defiitio of a aalytic fuctio. b) What are the relatioships betwee aalytic fuctios ad the Cauchy-Riema equatios? Solutio. a) A fuctio f : G C is called aalytic if it is cotiuously differetiable o G. b) A fuctio is aalytic o G if ad oly if it satisfies the Cauchy-Riema equatios o G. Exercise. Cosider the fuctios: fz) = z, fz) = x, fz) = y, fz) = x, fz) = e z+, fz) = e si z z + 3 z, fz) =, ad fz) = cos z z 3 + respectively. Determie the poits at which they are differetiable. Determie their regios of aalyticity. Solutio. fz) = z ) We have that fz) = z = x + y. Thus u = x + y ad v = 0. Takig the ecessary partials, we see: u x = x v y = 0 u y = y v x = 0. Sice the oly value at which the Cauchy-Riema equatios are satisfied is 0, f is differetiable oly at the origi. Thus f is aalytic owhere. fz) = x ) I this fuctio u = x ad v = 0, thus the partials we have are: u x = x v y = 0 u y = 0 v x = 0. So this fuctio is differetiable o the lie ir C. However, f is aalytic owhere. fz) = y ) I this fuctio u = y ad v = 0, thus the partials we have are: u x = 0 v y = 0 u y = y v x = 0. So this fuctio is differetiable o the lie R C, but aalytic owhere. fz) = x) I this fuctio u = x ad v = 0, thus the partials we have are: u x = v y = 0 u y = 0 v x = 0. So this fuctio is ot differetiable or aalytic aywhere o C. fz) = e z+ ) Let s first try to separate the real ad imagiary parts of f: Thus ad fz) = e z+ = e x+iy+ = e x+)+iy = e x+ cos y + i si y) = e x+ cos y + ie x+ si y. The first partials of these are: u = e x+ cos y v = e x+ si y. u x = e x+ cos y v y = e x+ cos y u y = e x+ si y v x = e x+ si y. Sice u y = v x ad u x = v y for all x + iy, f is differetiable o all of C, hece aalytic o all of C.

2 ) fz) = e z First ote that fz) = h gz) where hz) = e z ad gz) =. First let s fid out where g ad h are aalytic: z Sice gz) = e z = e x+iy = e x cos y + ie x si y, we have: u x = e x cos y v y = e x cos y u y = e x si y v x = e x si y. So we see that g is aalytic o all of C. Now do the same for h: The partials are: hz) = z = x iy x + y = x x + y i y x + y. u x = x +y v x +y ) y = x y x +y ) u y = xy v x +y ) x = xy. x +y ) Thus, whe the partials exist whe z 0), they are equal. Therefore h is aalytic o C \ {0}. Sice we have that gc) C \ {0}, by the chai rule f is aalytic o C \ {0}. differetiable o C \ {0}. Of course, f is also fz) = si z cos z ) To do this we will use the followig fact: Lemma. Suppose that f ad g are aalytic o a regio G. Let G G be the set of poits where g vaishes. The f g is aalytic o G \ G. ad Let gz) = si z ad hz) = cos z. We have: gz) = eiz e iz i hz) = eiz + e iz, i so g ad h are aalytic o all of C. The set of zeros of h is give by G = {x + iy C x = π + π, y = 0, N}. ) fz) = z+3 z 3 + So fz) is aalytic o the set C \ G. Lettig gz) = z + 3 ad hz) = z 3 +, we have fz) = gz) hz). As both g ad h are their ow power series represetatios, the { fuctios are aalytic o all of C. The set of values o which hz) = 0 are the cube roots of, i.e. G = + i 3,, i } 3. Thus f is aalytic o C \ G. Exercise 3. a) What are the relatioships betwee aalytic fuctios ad harmoic fuctios? b) Fid a harmoic fuctio that is a polyomial of degree four. Solutio. a) If f = u + iv with u, v C C) is aalytic, the u ad v are harmoic. I fact, a fuctio beig aalytic actually implies that it is C, thus give ay aalytic fuctio f = u + iv, both u ad v are harmoic. O the other had, if a real-valued fuctio u is harmoic, the if its harmoic cojugate, call it v, exists, the fuctio f = u + iv is aalytic.

3 3 b) Claim: The fourth degree polyomial: ux, y) = x 3 y xy 3, is harmoic. The partial derivatives are: u xx = 6xy ad u yy = 6xy ad so: Thus u is harmoic. u = 6xy 6xy = 0. Exercise 4. State Abel s Theorem. Theorem. Let a z a) have radius of covergece ad suppose that a coverges to A. The: lim r a r = A. Exercise 5. Fid the radius of covergece of the followig series: p z p = ay iteger), z Proof. p z ) The coefficiets of this power series is give by: So by the ratio test: R = lim Thus the radius of covergece is. a a + a = p.!,!z, ad z. = lim p + ) p =. z! ) Here the coefficiets of the power series are a =. So agai, usig the ratio test:! R = lim a = lim! = lim + =. a + Thus the radius of covergece is. +)!!z ) This time the coefficiets of the power series are a =!. By the ratio test: R = lim a = lim! + )! = lim + = 0. a + Thus the radius of covergece is 0. z! ) This is ot a power series i the traditioal sese, so let s rewrite it i the form of this power series are give by: {, if = k! for some k N b = 0, otherwise. b z where the coefficiets Clearly we caot use the ratio test here sice there are multitude of zero terms i betwee each ozero oe. Thus we will use Hadamard s formula: = lim sup b = lim sup b =, R so that the radius of covergece is. = Exercise 6. Determie the values of z so that =0 ) z is coverget. + z

4 4 Proof. This series will coverge if z z + <. This is true if z < z +. Replacig z = x + iy we get: x + y < x + ) + y = x + x + + y squarig both sides we have: which is true if ad oly if: or x + y < x + x + + y 0 < x + x >. So the series coverges if z {x + iy C x > }. Exercise 7. Show that the radius of covergece of ) z+) is, ad discuss the covergece for z =,, i, i. Proof. Frist write this i the form of a traditioal power series We will use Hadamard s formula, so we eed: b z where the b are as follows: = {b } = = {0,, 0, 0, 0,, 0, 0, 0, 0, 0,, 0,...}. 3 { b } = = {0,, 0, 0, 0, 6, 0, 0, 0, 0, 0,, 0,...} 3 Thus solvig for R: Thus we see that R =. lim sup { b } = lim +) = R, l R = lim l = lim + ) l + ) = 0. Now we will discuss the covergece: z = ) At z = we have: ) z+) = ) which, by the alteratig series test coverges. z = ) First ote that + ). The at z = we have: which coverges. ) z+) = ) )+) = ) z = i) Pluggig i z = i:

5 5 ) z+) = ) = ) = ) i+) +) ) +3) = = ) + ) = This is a series with a =. Clearly each a 0. It is also clear that lim a = 0. Now we must verify that {a } is decreasig. a a + = ) + ) + ) = = Thus this is a decreasig sequece, ad so, by the alteratig series test, the series coverges. z = i) Sice + ) we have that ) i)+) = ) i+) which we already kow coverges. Thus the series coverges at z = i. Exercise 8. Expad z + 3 z + Proof. i powers of z ). What is the radius of covergece? z + 3 z + = z ) + 5) + z ) = z ) + 5 ) ) z = z ) + 5 ) ) z ) =0 = ) z ) =0 =0 ) z ) = ) z ) + 5 ) z ) = =0 = 5 [ + ) + 5 ) ] z ) = = 5 ) z ) 4 =

6 6 This series has radius of covergece: R = lim ) ) =.