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using Plots, ComplexPhasePortrait, ApproxFun
gr();
Dr. Sheehan Olver
s.olver@imperial.ac.uk
This lecture we cover
Definition (Domain) A domain is a non-empty, open and connected set $D \subset {\mathbb C}$.
Definition (Homotopic) Two closed contours $\gamma_1 : [a,b] \rightarrow D$ and $\gamma_2 \rightarrow D$ in a domain $D$ are homotopic if they can be continuously deformed to one-another while remaining in $D$.
Theorem (Deformation of contours) Let $f(z)$ be holomorphic in a domain $D$. Let $\gamma_1$ and $\gamma_2$ be two homotopic contours. Then $$ \int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z) dz $$
Definition (Simply connected) A domain is simply connected if every closed contour is homotoic to a point.
Corollary (Deformation of contours on simply connected domains) Let $f(z)$ be holomorphic in a simply-connected domain $D$. If $\gamma_1$ and $\gamma_2$ are two contours in $D$ with the same endpoints, then
$$ \int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z) dz $$Demonstration We can test this experimentally: in the following, the integral (implemented as sum) over two different contours returns the same vaule (up to numerical precision)
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f = Fun( z -> exp(z), Arc(0.,1.,(0,π/2))) # Holomorphic!
f̃ = Fun( z -> exp(z), Segment(1,im)) # Holomorphic!
sum(f) , sum(f̃)
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Here is a depection of the two contours:
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plot(domain(f); ratio=1.0, legend=false, arrow=true)
plot!(domain(f̃), arrow=true)
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Contours are oriented: there is a notion of "left" and "right" inherited from $[a,b]$. For closed contours, there is as notion of positive/negative orientation:
Definition (Positive/negative orientation) Let $\gamma$ be a simple closed contour and $z$ in the interior of $\gamma$. We say that $\gamma$ is positively oriented if $${1 \over 2 \pi i} \oint_\gamma {d\zeta \over \zeta - z} = 1$$ It is negatively oriented if the reversed contour $\gamma_{\rm reversed}(t) = \gamma(b+a-t)$ for $t \in [a,b]$ is positively oriented, or equivalentally $${1 \over 2 \pi i} \oint_\gamma {d\zeta \over \zeta - z} = -1$$
Demonstration Here we see numerically that $1/z$ is positively oriented:
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sum(Fun(z -> 1/z, Circle()))/(2π*im)
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Cauchy's integral formula allows us to recover a function from knowlerdge of its values on a surrounding contour:
Theorem (Cauchy integral formula) Suppose $f$ is holomorphic inside and on a positively oriented, simple, closed contour $\gamma$. Then $$ f(z) = {1 \over 2 \pi i} \oint_\gamma {f(\zeta) \over \zeta - z} d \zeta$$
Demonstration This shows numerically that we can calculate $\E^{0.1}$ by integrating over a circle that surrounds $z = 0.1$:
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γ = Circle()
ζ = Fun(γ)
z = 0.1
sum(exp(ζ)/(ζ-z)) /(2π*im) - exp(z)
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Not only do we know $f$, we also know that $f$ is infinitely-differentiable, and we know all its values:
Corollary (Cauchy integral formula for derivatives) Suppose $f$ is holomorphic inside and on a positively oriented, simple, closed contour $\gamma$. Then $f$ is infinitely-differentiable at $z$ and $$ f^{(k)}(z) = {k! \over 2 \pi i} \oint_\gamma {f(\zeta) \over (\zeta - z)^{k+1}} d \zeta$$
Demonstration Here we show that we can recover $\E^z$ using any of the derivatives:
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k=10
factorial(1.0k)*sum(exp(ζ)/(ζ-z)^(k+1))/(2π*im) - exp(z)
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z = 0.1
k = 5
factorial(1.0k)*sum(exp(ζ)/(ζ-z)^(k+1))/(2π*im) - exp(z)
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expⁿ = (n,z) -> sum(z^k/factorial(1.0k) for k=0:n)
phaseplot(-20..20, -20..20, z -> expⁿ.(10,z))
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And now the $n$-term Taylor approximation to $1/(1-z)$:
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geometricⁿ = (n,z) -> sum(z^k for k=0:n)
phaseplot(-2..2, -2..2, z -> geometricⁿ.(20,z))
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And here the $n$-term Taylor approximation of $\sqrt z$ near $z_0$:
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function sqrtⁿ(n,z,z₀)
ret = sqrt(z₀)
c = 0.5/ret*(z-z₀)
for k=1:n
ret += c
c *= -(2k-1)/(2*(k+1)*z₀)*(z-z₀)
end
ret
end
z₀ = 0.3
n = 40
phaseplot(-2..2, -2..2, z -> sqrtⁿ.(n,z,z₀))
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