Solve the equation sinz=2 for z by:
a) equating real parts and then imaginary parts in that equation
b) using sin^-1(z)=-ilog[iz+(1-z^2)^1/2]© BrainMass Inc. brainmass.com September 21, 2018, 9:41 am ad1c9bdddf - https://brainmass.com/math/complex-analysis/solve-the-equation-sin-z-2-347449
In this problem you have to use the definition of sin(z) for complex z. Let's first discuss analtic continuation of real function to complex arguments before discussing the specific problem. The following theorem is called the analytic continuation theorem and it is very useful:
If two analytic functions f and g defined on some subset V of C, are equal to each other on some set of points z_n, such that the z_n has a limit point in V and the z_n are all different, then f(z) = g(z) everywhere on V.
This theorem implies that given some real function, you can only find one unique analytic function on the complex plane that reduces to that real function when restricted to the real line. To see this, suppose that one real function like sin(x) has two continuations to the complex plane, f(z) and g(z). But then f(z) and g(z) are equal to each oter on the real line and on the real line there is plenty of room to find a converging sequence of different points, e.g. x_n = 1/n. So, f(z) must be equal to g(z), at least if we demand that both functions are analytic. Note that for complex functions, complex differentiability implies that the function is analytic (and vice versa, of course).
What is also useful about this theorem is that it implies that the familiar relations/identities valid for real arguments generalize for complex aguments. You don't have to do tedious computations using the definition for complex arguments to verify these. Let's consider two examples. The definition
exp(i x) = cos(x) + i sin(x) (1)
together with the property exp(a + b) = exp(a) exp(b), implies that:
exp(z) = exp(x+iy) = exp(x) [cos(y) + i sin(y)] (2)
So, this defines the exponential function on the complex ...
We explain the way complex analytic functions like sin(z) are defined from the definition of their real counterparts. The solution is then explained in detail.