PMTH333 COMPLEX ANALYSIS - University of New England

1 PMTH333 COMPLEX ANALYSISThe tutorial questions contained in this booklet have mostly been selected fromComplex Variables and applications , 6th Brown and Churchill, to relevant sections from this text are included with each tutorial. Ifyou have difficulty with particular questions, you should review these sections andattempt similar exercises from the worked solutions for the tutorial questions will appear at the coursewebsite. Resist the temptation to refer to the solution as soon as you encounter anydifficulty. You will learn and remember more from perseverance, successful or not,than from simply following through a worked at the University of New EnglandNovember, 20051 TUTORIAL 1 COMPLEX Variables and applications , Sect.

2 1-81. Letzbe any COMPLEX number. Prove that(a) Im(iz) = Rez,(b) Re(iz) = Solve the equationz2+z+ 1 = 0 by substitutingz=x+ Verify that(a)| z|=|z|,(b)z1z2= z1 z2,(c)(z1z2)= z1 z24. Sketch (or describe) the set of points determined by(a)|z 1 + i|= 1,(b)|2z i| 4,(c) Re( z i) = 25. Find the four roots of the equationz4+ 4 = Sketch (or describe) the closure of the sets(a) <Argz < (z6= 0),(b)|Rez|<|z|.2 TUTORIAL 2 COMPLEX Variables and applications , Sect. 9-141. Describe the natural domain of the functionf(z) =zz+ Letf(z) =x2 y2 2y+ i(2x 2xy), wherez=x+ iy. Usingx=z+ z2,y=z z2 ior otherwise, expressf(z) in terms Sketch the region onto which the sectorr 1, 0 4is mapped by thetransformations(a)w=z2,(b)w=z3,(c)w= Show that(a) limz 4z2(z 1)2= 4,(b) limz 11(z 1)3=.

3 5. Use the definitions to show that(a) limz 01z= ,(b) limz 1z= 3 COMPLEX Variables and applications , Sect. 15-191. Findf (z) wheref(z) =z 12z+1, (z6= 12).2. Show thatf (z) does not exist at any point if(a)f(z) = 2x+ iy2,(b)f(z) = exe Determine wheref (z) exists and find its value when(a)f(z) =x2+ iy2,(b)f(z) =zImz,(c)f(z) = e cos(lnr) + i e sin(lnr).4. Show that, forf(z) =u(r, ) + iv(r, ) wherez=rei , iff (z0) exists andz06= 0, thenf (z0) = iz0(u + iv ).3 TUTORIAL 4 COMPLEX Variables and applications , Sect. 20-251. Show that an analytic functionf(z) in a domainDwhich takes only realvalues for allz Dmust be a constant Find all values ofzsuch that ez= 3 Show that e z=ezand e zis not analytic Show that sin( z) is not analytic Find all roots of the equation coshz= Functions(a) Which of the following functions are harmonic on some domain inC?

4 Specify the (x,y) =yx2+y2f(x,y) = ex2 y2(b) Using Laplace s equation, show directly that Re1zis harmonic in a neigh-bourhood of any pointz0except the origin.(c) Let (x,y) = 6x2y2 x4 y4+y x+ is the real part of an analytic functionf(z), what is the imaginary part?Conversely, if is the imaginary part off(z), what is the real part? Are theanswers equal?(d) Find the harmonic conjugate of (x,y) = arctanxywhere 2<arctanxy 5 COMPLEX Variables and applications , Sect. 26-311. Show that(a) Log(1 i) =12ln 2 4i(b) Log(1 + i)2= 2 Log(1 + i),but Log( 1 + i)26= 2 Log( 1 + i)2. Find the principal value of (a) ii, (b) (1 i)4 Calculate the following:(a) 60e2 itdt,(b) 0e Evaluate 0excosxdxand 0exsinxdxby using 0excosxdx+ i 0exsinxdx= 0e(1+i) 6 COMPLEX Variables and applications , Sect.

5 32-381. Letz=z(t) be a smooth arc, andf(z) be analytic atz0=z(t0). Show thatddtf(z(t)) =f (z(t))z (t)att= Calculate Cf(z)dzwhereCis the arc fromz= 1 i toz= 1 + i along thecurvey=x3andf(z) ={1wheny <04ywheny > Show that Cdzz z0= 2 i whereCis given byz=z0+Rei , (R >0 fixed, changes from to .)4. Evaluate(a) i2ie zdz,(b) 3i(z 2) Apply the Cauchy-Goursat Theorem to show that Cdzz2+ 2z+ 2= 0whereCis the circle|z|= 1 oriented 7 COMPLEX Variables and applications , Sect. 39-421. Find Cg(z)dzwhereCis the circle|z i|= 2 in the positive sense andg(z) =1(z2+ 4) LetCbe the circle|z|= 3 in the positive sense. Show thatg(2) = 8 i andg(4 i) = 0 whereg(w) = C2z2 z 2z wdz,(|w|6= 3).3. LetCbe a simple closed contour oriented counter-clockwise.}

6 Defineg(w) = Cz3+ 2z(z w) thatg(w) = 6 wwhenwis insideCandg(w) = 0 whenCis Show that iff(z) is analytic whithin and on a simple closed contourCandz0is not onC, then Cf (z)z z0dz= Cf(z)(z z0) 8 COMPLEX Variables and applications , Sect. 43-451. Show that ez= e n=0(z 1)nn!(z C).2. Find the MacLaurin series expansion of the functionf(z) =zz4+ 9=z9[11 +z49].3. Find the MacLaurin series expansion of the functionf(z) = sinz2and use itto show thatf(4)(0) =f(2n+1)(0) = 0,n= 0,1,2,..4. Derive the Taylor series representation11 z= n=0(z i)n(1 i)n+1,(|z i|< 2).5. Show that when 0<|z|<4,14z z2=14z+ n=0zn4n+ 9 COMPLEX Variables and applications , Sect. 46-511. Representf(z) =z+1z 1by(a) its MacLaurin series, and give the region of validity for the representation;(b) its Laurent series for the domain 1<|z|<.

7 2. Show that when 0<|z 1|<2,z(z 1)(z 3)= 12(z 1 3 n=0(z 1)n2n+ Differentiate11 z= n=0zn, (|z|<1), to obtain1(1 z)2= n=0(n+ 1)zn,(|z|<1)2(1 z)3= n=0(n+ 1)(n+ 2)zn,(|z|<1).4. Prove thatf(z) = coszz2 ( 2)2,whenz6= 2, 1 ,whenz= 2is Prove that iff(z) is analytic atz0andf(z0) =f (z0) = =f(m)(z0) = 0,the the functiong(z) = f(z)(z z0)m+1whenz6=z0,f(m+1)(z0)(m+1)!whenz=z0 ,is analytic 10 COMPLEX Variables and applications , Sect. 51, 53-551. Show that(a)ezz(z2+ 1)=1z+ 1 12z 56z2+ ,(0<|z|<1)(b)1ez 1=1z 12+112z 1720z3+ ,(0<|z|<1).2. Find the residues atz= 0 of the functions(a)1z+z2,(b)zcos1z,(c)z Evaluate the integral off(z) around the positively oriented circle|z|= 2 whenf(z) is(a)z51 z3,(b)11 +z2,(c) 11 COMPLEX Variables and applications , Sect.)

8 56-631. Find 0x2(x2+1)(x2+4) Find xsinaxx4+4dxwherea Show that 2 0d 1+12cos =4 ANALYSIS SOLUTIONSTUTORIAL PROBLEMS SET Letz=x+ iy. Then iz= y+ ixand therefore, Im iz=x= RezandRe iz= y= After the substitution we have(x+ iy)2+ (x+ iy) + 1 = (x2 y2+x+ 1) + (2xy+y) i = 0which gives two real equationsx2 y2+x+ 1 = 02xy+y= 0 From the second equation we find thaty= 0 orx= 12. In the first case,the first equation won t have a real solution (x= Rezmust be real!). In thesecond case, the first equation becomes14 y2 12+ 1 = This gives the two solutionsz= 12 (a) Letz=x+ iy. By definition| z|= x2+ ( y)2= x2+y2=|z|.(b)(x1+ iy1)(x2+ iy2) =x1x2 y1y2+ (x1y2+x2y1) i=x1x2 y1y2 (x1y2+x2y1) i = (x1 iy1)(x2 iy2)= z1 z2(c) This can be done analogously to (b) or as follows (by reduction to (b)):The assertion is equivalent to(z1z2) z2= to (b) the left hand side equals(z1z2)z2= z1which proves (c).

9 4. (a) a circle centred at 1 i of radius 1 (the distance ofzto 1 i equals 1).(b) The equations is equivalent to|z i2| 2. This is a closed disc (discincluding the boundaring circle) centred ati2of radius 2.(c) This equations is equivalent tox= 2, wherex= Rez. Thus the point setis a vertical line that intersects the real axis atx= Fromz4+ 4 = we getz2= 2 i. This can be solved by passing toz=x+ findx2 y2= 02xy= 2 Hence,x= yandx2= 1. Therefore, the four solutions are 1+i, 1 i, 1+i, 1 (a) The given set is the COMPLEX plane with the real half line ( ,0] closure is the entire COMPLEX plane, since any point of the real half line islimiting point for the given set.(b) The given formula is equivalent toy2>0 which in turn is equivalent toy6= 0.)

10 This is the COMPLEX plane with deleted real axis. Again, the closure isthe entire COMPLEX PROBLEMS SET The natural domain of this function consists of all COMPLEX points for whichthe denominatorz+ z= 2xdoes not vanish. Thus the natural domain is thecomplex plane with deleted imaginary (z) = z2+ 2 iz3. (a) The functionw=z2squares the absolute value and doubles the the image will consist of all point whose absolute value is betwee0 and 1 and whose argument is between 0 and 2. This is the (closed) upperright quadrant.(b) This function cubes the absolute values and triples the argument. Thus,the image is the sectorr 1 and 0 2 4.(c) Here the image is the closed upper half (a)According to (5) in Sect. 14 we havelimz 4z2(z 1)2= limz 04/z2(1/z 1)2= limz 04(1 z)2 Since the latter function is continuous at 0 the limit is the value of the functionat 0 which is 4.