Is Cosh(4z) Integrable Along Any Path Between -πi/8 and πi/8?

[John O' Meara]

Integrate cosh(4z) w.r.t., z, for any path from $$\frac{-\pi i}{8} \inbox{to } \frac{\pi i}{8}$$. If the function is analytic, i.e., obeys Cauchy - Riemann equations we can integrate as in standard calculus.
$$\frac{{\partial \cosh(4(x+iy)}}{{\partial x}} = a\sinh(4x + 4iy) \\ \frac{{\partial \cosh(4x +4iy)}}{{\partial y }} = 4i\sinh(4x + 4iy)\\$$. Therefore $$u_x$$ and $$v_y$$ are not equal, therefore the cosh(4z) is not analytic. So we integrate as follows: C (the path of integration) can be represented by z(t)= 0 + it, $$\frac{-\pi}{8} \leq t \leq \frac{\pi}{8}\\$$ Hence $$\dot z(t) = \iota \\$$ and f(z(t)) = cosh(it), therefore $$\int_C \cosh(z) dz = \int_{\frac{-\pi}{8}}^\frac{\pi}{8} \cosh(it)\iota dt \\$$ = sinh(it) evulated at the respective limits = $$2\sinh(\frac{\iota\pi}{8}) \$$
I don't think that this is correct, I think my Z(t) = 0 +it is wrong.

 

[Dick]

cosh(4z) IS analytic. You have to split the function into real and imaginary parts before applying CR. cosh(z)=(e^z+e^-z)/2. Now e.g. e^z=e^(x+iy)=e^x*e^(iy)=e^x*(cos(y)+i*sin(y)). So the split for e^z is u(x,y)=e^x*cos(y) and v(x,y)=e^x*sin(y) and it IS analytic. Similarly for cosh(z).

 

[John O' Meara]

Thanks for the reply Dick. As a matter of interest do you know how "Cauchy" is pronounced? E.g., Ca-u-chy?

 

[Dick]

Thanks for the reply Dick. As a matter of interest do you know how "Cauchy" is pronounced? E.g., Ca-u-chy?

Co-she, 'co' as in coordinate and 'she' as in she. You're welcome.

 

[Kummer]

You can use the fundamental theorem of calculus here, it is easy to find the primitive of cosh(4z)