Exercise 8 define tan z sin z cos z where is this

This preview shows page 31 - 33 out of 164 pages.

Exercise 8.Definetanz=sinzcosz; where is this function defined and analytic?Solution.Not available.Exercise 9.Suppose that zn,zG=C- {z:z0}and zn=rneiθn,z=reiθwhere-π<θ,θn. Provethat if znz thenθnθand rnr.Solution.Not available.Exercise 10.Prove the following generalization of Proposition 2.20. Let G andΩbe open inCand supposef and h are functions defined on G, g:ΩCand suppose that f(G)Ω. Suppose that g and h areanalytic, gprime(ω)nequal0for anyω, that f is continuous, h is one-one, and that they satisfy h(z)=g(f(z))for z inG. Show that f is analytic. Give a formula for fprime(z).Solution.Not available.Exercise 11.Suppose that f:GCis a branch of the logarithm and that n is an integer. Prove thatzn=exp(n f(z))for all z in G.Solution.Not available.Exercise 12.Show that the real part of the function z12is always positive.Solution.We know that we can write z=reiθnequal0,-π<θ<πand Log(z)=ln(r)+iθ. Thusz12=eLogz12=e12Logz=e12(lnr+iθ)=e12lnre12θi=e12lnrparenleftbiggcosparenleftbiggθ2parenrightbigg+isinparenleftbiggθ2parenrightbiggparenrightbigg.27
Hence,Re(z)=e12lnrcosparenleftbiggθ2parenrightbigg>0,since e12lnr>0andcosparenleftBigθ2parenrightBig>0since-π<θ<π. Thus, the real part of the function z12is always positive.Exercise 13.Let G=C- {zR:z0}and let n be a positive integer. Find all analytic functionsf:GCsuch that z=(f(z))nfor all zG.Solution.Let Log(z)be the principal branch, thenlog(z)=Log(z)+2kπi for some kZ. Thus, we canwritef(z)=z1/n=elog(z)/n=e(Log(z)+2kπi)/n=eLog(z)/n·e2kπi/n.We know that the latter factor are the n-th roots of unity and depend only on k and n. They correspond tothe n distinct powers of the expressionζ=e2πi/n. Therefore, the branches of z1/non the set U are given byf(z)=ζk·eLog(z)/n,where k=0,...,n-1and therefore they are all constant multiples of each other.Exercise 14.Suppose f:GCis analytic and that G is connected. Show that if f(z)is real for all z inG then f is constant.Solution.First of all, we can write f:GCasf(z)=u(z)+iv(z)where u,v are real-valued functions. Since f:GCis analytic, that is f is continuously differentiable(Definition 2.3), we have that u and v have continuous partial derivatives. By Theorem 2.29, this impliesthat u,v satisfy the Cauchy-Riemann equations. That is,ux=vyanduy=-vx.(3.1)Since f(z)is realzG, this impliesv(z)0and therefore f(z)=u(z). So, since v(z)=0, we havevx=vy=0and by (3.1) we obtainux=uy=0and thus uprime(z)=0(see reasoning of equation 2.22 and 2.23 on page 41). Hence fprime(z)=0. Since G isconnected and f:GCis differentiable with fprime(z)=0zG, we have that f is constant.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture