- #1
futurebird
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I'm trying to understand the proof for this theorem, and I can't see what they did to get from one step to the next.
THEOREM: Suppose F(z) is an analytic function and that f(z) = F'(z) is continuous on a domain D. Then for a contour C lying in D with endpoints z1 and z2:
\int_{C}f(z)dz=F(z_{2}) - F(z_{1})
PROOF:
Using the definition of the integral, and assuming z'(t) is continuous
\int_{C}f(z)dz=\int_{C}F'(z)dz=\int^{b}_{a}F'(z(t))z'(t)dt
This next step is the one I don't get. They say they used the chain rule?
=\int^{b}_{a}\frac{d}{dt}\left[F(z(t))\right]dt
But HOW?
The rest makes sense:
=F(z(b))-F(z(a))
=F(z2)-F(z1)
Hence,
\int_{C}f(z)dz=F(z_{2}) - F(z_{1})
THEOREM: Suppose F(z) is an analytic function and that f(z) = F'(z) is continuous on a domain D. Then for a contour C lying in D with endpoints z1 and z2:
\int_{C}f(z)dz=F(z_{2}) - F(z_{1})
PROOF:
Using the definition of the integral, and assuming z'(t) is continuous
\int_{C}f(z)dz=\int_{C}F'(z)dz=\int^{b}_{a}F'(z(t))z'(t)dt
This next step is the one I don't get. They say they used the chain rule?
=\int^{b}_{a}\frac{d}{dt}\left[F(z(t))\right]dt
But HOW?
The rest makes sense:
=F(z(b))-F(z(a))
=F(z2)-F(z1)
Hence,
\int_{C}f(z)dz=F(z_{2}) - F(z_{1})
