John O' Meara
- 325
- 0
Integrate \int_C \sec^2 z dz \ \mbox{any path from } \ \frac{\pi}{4} \mbox{ to } \frac{\pi\iota}{4} \\ \sec^2 z = \frac{1}{\cos^2 z} \ \mbox{ which is equal to } \ \frac{1}{2(1+\cos 2z)} \\ Therefore \frac{2}{1+\cos 2z} = \frac{2}{1 + \cos2x\cosh2y -\iota \sin2x\sinh2y}\\
How do you split this fraction up to differentiate it w.r.t., u_x \mbox{,} v_y \\ i.e., the Cauchy - Riemann equations? It seems to me that sec^2 z is not analytic.
Therefore integrate as follows:z(t) = t + (\frac{\pi}{4} -t)\iota \ \frac{dz}{dt} = 1 - \iota \\ Therefore the integral is \int_0^{\frac{\pi}{4}} \sec^2(t+(\frac{\pi}{4}-t)\iota)(1 - \iota ) dt\\. Am I correct so far? Is z substituted in \sec^2z \ done correctly. Thanks for helping.
How do you split this fraction up to differentiate it w.r.t., u_x \mbox{,} v_y \\ i.e., the Cauchy - Riemann equations? It seems to me that sec^2 z is not analytic.
Therefore integrate as follows:z(t) = t + (\frac{\pi}{4} -t)\iota \ \frac{dz}{dt} = 1 - \iota \\ Therefore the integral is \int_0^{\frac{\pi}{4}} \sec^2(t+(\frac{\pi}{4}-t)\iota)(1 - \iota ) dt\\. Am I correct so far? Is z substituted in \sec^2z \ done correctly. Thanks for helping.