Question

use the maclaurin series of f(z)= [ (cos ^2) of z] - [ (sin ^2) of z ]
to compute integral on Curve for ( [ (cos ^2) of z] - [ (sin ^2) of z ] ) / ( z^53)
C :|z|=2 oriented positively

Answer 1

Answer on Question #49963 - Math - Complex Analysis

Use the Maclaurin series of $f(z) = \cos^2 z - \sin^2 z$ to compute integral on Curve for $\int_{|z| = 2} \frac{\cos^2 z - \sin^2 z}{z^{53}} dz$

Solution.

Let's consider $f(z) = \cos^2 z - \sin^2 z = \cos 2z$ then the Maclaurin series for it $f(z) = \cos 2z = \sum_{k=0}^{\infty} (-1)^k \frac{(2z)^{2k}}{(2k)!}$ .

Thus, $\int_{|z| = 2}\frac{\cos^2z - \sin^2z}{z^{53}} dz = \int_{|z| = 2}\frac{\cos 2z}{z^{53}} dz = 2\pi i$ re $\frac{\cos 2z}{z^{53}} = 2\pi i(-1)^{26}\frac{2^{52}}{(52)!} = \pi i\frac{2^{53}}{(52)!}$

Answer: $\int_{|z| = 2}\frac{\cos^2z - \sin^2z}{z^{53}} dz = \pi i\frac{2^{53}}{(52)!}$

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