Week 6 Worksheet - Solutions

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Cauchy's Integral Formula

    1. We have f(z)=|z+1|2=(z+1)(z+1)=x2+y2+2x+1 and calculate (1u)(x,y)=2x+2(2u)(x,y)=2y as well as (1v)(x,y)=0=(2v)(x,y) for all x,y. The Cauchy-Riemann equations are only satisfed at z=1 so f is not holomorphic on any domain whatsoever.
    2. Note that f(z)=2+z+z when |z|=1. It just so happens that 1z=z|z|2=z when |z|=1 so let's take g(z)=2+z+1z.
    3. Now γ|z+1|2dz=γ2+zdz+γ1zdz and, since γ is closed, the existence of an antiderivative for 2+z on C leaves us with γ|z+1|2dz=2πi(12πiγ1zdz) to be evaluated. The term in brackets is the Cauchy integral formula applied to f(z)=1 and b=0.
  1. Since f is entire it is represented by a power series n=0anzn centered at 0 with infinite radius of convergence. Moreover, for every R>0 we have an=12πiγRf(z)zn+1dz where γR(t)=Reit on [0,2π]. Thus an=12π02πf(Reit)(Reit)ndt and the estimation lemma, together with the hypothesis, gives |an|12π2πKRkRn=KRkRn which, for n>k, converges to zero as R. Thus all but finitely many of the coefficients an are equal to zero and f is a polynomial.

Taylor's Theorem

  1. We can differentiate power series. By induction f(k)(z)=n=kann(n1)(n(k1))(zb)nk and plugging in 0 gives f(k)(b)=akk(k1)21=k!ak.
    1. First we verify that (sin(z))2=(exp(iz)exp(iz)2i)2=exp(2iz)2+exp(2iz)4=12cos(2z)2 so that (sin(z))2=1212n=0(1)n(2n)!(2z)2n=z213z4+245z6 with infinite radius of convergence.
    2. Write 12z+1=11(2z)=n=0(2z)n=n=0(2)nzn whenever |2z|<1 using the geometric series. The radius of convergence is 1/2.
    3. We can plug z2 into the power series of the exponential function exp(z2)=n=01n!(z2)n=n=01n!z2n giving a Taylor series with infinite radius of convergence.
  2. We know that Log(w) is holomorphic on C(,0]. Thus f(z)=Log(1+z) is holomorphic on C(,1]. It therefore has a Taylor on B(0,1). First a0=f(0)=Log(1)=0. Now f(z)=(Log(1+z))=11+z from the chain rule. But we know f(z)=n=0(1)nzn on B(0,1) from the geometric series. Thus f(z)=n=0(1)nn+1zn+1=n=1(1)n+1nzn=z12z2+13z314z4+ on B(0,1).
  3. We can write 11+z2=12(11iz+11(iz))=12n=0(iz)n+12n=0(iz)n=n=0in+(i)n2zn=1z2+z4z6+z8z10+ whenever |iz|<1 and |iz|<1 using the geometric series. Thus the series is valid on B(0,1).
  4. Whenever B(b,R)C{c,d} we will have a Taylor series for f on B(b,R). The largest ball centered at b and entirely contained within C{c,d} is the ball of radius min{|bc|,|bd|}.

Applications

  1. Plug into the quadratic formula (2+2i)±(2+2i)24(2)i2i=(2+2i)±16i2i=(2+2i)±4(12+i12)2i to get (1+i)i+2(1+i)i=(1+i)(1+2)i(1+i)i2(1+i)i=(1+i)(12)i as the roots.
  2. If p has degree one then p(z)=a1z+a0=a1(za0a1) as desired. We proceed by induction. Fix a polynomial p(z)=anzn+an1zn1++a2z2+a1z+a0 of degree n. By the fundamental theorem of algebra p(cn)=0 for some cnC. Polynomial long division therefore lets us write p(z)=(zcn)q(z)+d where q(z) is a polynomial of degree n=1. Plugging in z=cn gives d=0 so p(z)=(zcn)q(z). By induction q(z)=b(zcn1)(zc2)(zc1) and therefore p has the desired form.
  3. Suppose there is aC not equal to any p(z). Then the function f(z)=1/(p(z)a) is defined on all of C and holomorphic there. Moreover, as in the proof of the fundamental theorem of algebra, the function is bounded. Liouville's theorem then implies f and therefore p is constant. This is a contradiction, as no polynomial of degree at least 1 is constant.