- #1
blahblah8724
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I don't understand this estimation lemma example :(
We are given the 'curve',
And part of the example is showing that the contour integral over the top semicircle [itex]C_R[/itex] tends to zero.
Apparently we use the estimation lemma and the fact that, [itex] |z^2 +1| \geq |z|^2 - 1 [/itex] to show,
[itex]\left| \int_{C_R} \frac{e^{iz}}{z^2 + 1} dz \right| \leq \int_0^\pi \frac{e^{-Rsin(t)}}{R^2 - 1} dt \leq \frac{2\pi R}{R^2 - 1} \to 0[/itex] as [itex] R \to \infty [/itex]
However I don't understand the part where [itex] e^{iz} [/itex] 'goes to' [itex] e^{-Rsin(t)} [/itex], is this some sort of parameterisation on the curve [itex] C_R [/itex]?
Help would be much appreciated!
Thanks
We are given the 'curve',
And part of the example is showing that the contour integral over the top semicircle [itex]C_R[/itex] tends to zero.
Apparently we use the estimation lemma and the fact that, [itex] |z^2 +1| \geq |z|^2 - 1 [/itex] to show,
[itex]\left| \int_{C_R} \frac{e^{iz}}{z^2 + 1} dz \right| \leq \int_0^\pi \frac{e^{-Rsin(t)}}{R^2 - 1} dt \leq \frac{2\pi R}{R^2 - 1} \to 0[/itex] as [itex] R \to \infty [/itex]
However I don't understand the part where [itex] e^{iz} [/itex] 'goes to' [itex] e^{-Rsin(t)} [/itex], is this some sort of parameterisation on the curve [itex] C_R [/itex]?
Help would be much appreciated!
Thanks