- #1
John O' Meara
- 330
- 0
Integrate [tex] \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)} \\[/tex] Therefore [tex] \frac{2}{1+\cos 2z} = \frac{2}{1 + \cos2x\cosh2y -\iota \sin2x\sinh2y}\\[/tex]
How do you split this fraction up to differentiate it w.r.t., [tex] u_x \mbox{,} v_y \\[/tex] i.e., the Cauchy - Riemann equations? It seems to me that [tex] sec^2 z [/tex] is not analytic.
Therefore integrate as follows:[tex] z(t) = t + (\frac{\pi}{4} -t)\iota \ \frac{dz}{dt} = 1 - \iota \\[/tex] Therefore the integral is [tex] \int_0^{\frac{\pi}{4}} \sec^2(t+(\frac{\pi}{4}-t)\iota)(1 - \iota ) dt\\[/tex]. Am I correct so far? Is z substituted in [tex] \sec^2z \ [/tex] done correctly. Thanks for helping.
How do you split this fraction up to differentiate it w.r.t., [tex] u_x \mbox{,} v_y \\[/tex] i.e., the Cauchy - Riemann equations? It seems to me that [tex] sec^2 z [/tex] is not analytic.
Therefore integrate as follows:[tex] z(t) = t + (\frac{\pi}{4} -t)\iota \ \frac{dz}{dt} = 1 - \iota \\[/tex] Therefore the integral is [tex] \int_0^{\frac{\pi}{4}} \sec^2(t+(\frac{\pi}{4}-t)\iota)(1 - \iota ) dt\\[/tex]. Am I correct so far? Is z substituted in [tex] \sec^2z \ [/tex] done correctly. Thanks for helping.