Solve for z in the equation sin(z) = 2

Note that |sin(x)| <= 1 for all real x, so our solution must be complex if one exists at all
Euler's formula gives that exp(ix) = cos(x) + isin(x), and hence exp(-ix) = cos(x) - isin(x) since cos is even (so cos(-x) = cos(x)), and sin is odd (and so sin(-x) = -sin(x)).
Thus combining the two equations yields the identity sin(z) = (1/(2i))(exp(iz) - exp(-iz))
Remember that we are solving for sin(z) = 2, so proceed by writing w for exp(iz) and we have that
w - (1/w) = 4i noting that exp(-iz) = 1/(exp(iz)) = 1/w
Multiplying by w yields the quadratic in w: w^2 - 4wi - 1 = 0 and solving gives w = i(2 ± sqrt(3))
Writing z = x + iy where x, y are real, we see that w = exp(iz) = exp(i(x + iy)) = exp(-y + ix) = exp(-y)(cos(x) + isin(x))
Hence equating real and imaginary parts with the known values of w, we see that exp(-y)cos(x) = 0, and since exp is non-negative for any real argument, it must be that cos(x) = 0 and so x = (pi/2) + n(pi) where n is any integer
Now we deal with y, which is slightly more tricky. we have that exp(-y)sin(x) = 2 ± sqrt(3)
Note that when cos(x) = 0, sin(x) = 1 or -1, in fact sin(x) = 1 when x = (pi/2) + 2n(pi), sin(x) = -1 when x = (pi/2) + (2n+1)(pi)When sin(x) is negative, exp(-y) is negative and hence has no solution since y is real.
Thus we find the x = (pi/2) + 2n(pi), and y = log(2 ± sqrt(3))
Hence the final solution is z = (pi/2) + 2n(pi) + i(log(2 ± sqrt(3))