What conditions on the Laurent coefficients correspond toz0being(i) a pole of orderm?(ii) a removable singularity?(b) Letf(z) =eizz2+ 25and let Γ be the contour shown in Figure 1, consisting of the line segment[-R, R] and the semicircle in the upper half-plane of radiusR.By integratingfaround Γ, or otherwise, prove thatZ∞-∞cosxx2+ 25dx=π5e-5.MATH 2101CONTINUED4

5.(a) Letfbe holomorphic in a discD={z:|z-z0|< r}and suppose that thereis a sequence of pointszninDwith the properties:zn6=z0, zn→z0asn→ ∞,and such thatf(zn) = 0 for alln. Show thatf(z) = 0 for allzinD.[You may use without proof the fact thatf(z) can be expressed as a convergentpower series inD.](b) State Rouch´e’s Theorem. Show that the polynomialp(z) =z6+ 8z4+ 5z+ 1has precisely 3 zeros in the annulus{1/2<|z|<2}.MATH 2101END OF PAPER5

Question 1(a) The function is holomorphic atz0if it is complex-differentiable there, in thesense that the limitlimz→z0f(z)-f(z0)z-z0(1)exists.BW/Definition2 marksIffis holomorphic atz0, with derivativef0(z0), then, since the limit is inde-pendent of the way in whichzapproachesz0, we havelimh→0f(z0+h)-f(z0)h=f0(z0) = limh→0f(z0+ih)-f(z0)ihforh∈R.The LHS isfx(z0) whereas the last term is-ify(z0). Hencefx(z0) =f0(z0) =-ify(z0)and sofx+ify= 0 as required.BW4 marks(b) If arg(f(z)) =α, say, it means thatv=Cufor some real constantC(oru= 0ifα=π/2). The real form of the Cauchy–Riemann equations isux=vy, uy=-vx.Combining withv=Cu, we obtainux=vy=Cuy, uy=-vx=-Cux.Thus (1 +C2)ux= 0 soux= 0.Fromuy=-Cux, it follows thatuy= 0,and then by the Cauchy–Riemann equations,vx=vy= 0 as well.Thatis,f0(z) = 0.The caseα=π/2,u= 0 is simpler, for the CR equationsimmediately imply thatvx=vy= 0, sof0(z) = 0.Seen Similar4 marksIfz0=x0+iy0is any point, thenf(x0,0)-f(0,0) =Zx00fx(t,0) dt= 0becausefx= 0. Thenf(x0, y0)-f(x0,0) =Zy00fy(0, t) dt= 0becausefy= 0. Hencef(x0, y0) =f(x0,0) =f(0,0)sofis constant.MATH 2101END OF PAPER6

BW5 marks(c) The condition for conformality isf0(z0)6= 0.BW3 marksThis part will unfortunately generate many different solutions. One way is torealiseHas the set of points closer to 2 than to 0,H={z:|z-2|<|z|}.So the mapf(z) = (z-2)/zwill do the job. For full credit, some discussion isrequired of the fact that an inverse exists (implying the map is bijectiveH→D) and verification thatf0(z)6= 0 forz∈H, or invoking that conformality isa general property of M¨obius transformations.

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