Solution to these problems, where z is a complex number

In a nutshell, the overlap between what my complex analysis professor teaches and what he assigns homework over is the null set. Anyway, here's the problems:
1) Solve: sin(z) = (3/4) + i(1/4)
2) Solve: sin(z) = 4

For 1), I've tried writing sin(z) in the form [e^iz - e^-iz]/2i. Then I multiplied through by an e^iz and obtained the following:

e^2iz + [1/2 - (3/2)i]*(e^iz) - 1 = 0

Used the quadratic formula and found the roots of e^iz. However, this leaves me with a really nasty equation, which I have to take the natural log of and divide by i to solve for z. Surely there's an easier way, right?

For 2, I'm tempted to say that there are no solutions for z which solve the equation (since sin(x) can never be greater than 1 for a real number x). However, I'm afraid this may be an oversimplification.

Any help would be appreciated.

(1) There isn't an easier way to do this than what you did, but it is not as hard as you think.

The equation you have so far, e^(2iz) + [1/2 - (3/2)i]e^(iz) - 1 = 0, is correct. I'm going to multiply both sides of this by 2 to make the formatting easier, and it will probably make the numbers easier to work with as well, giving:
2e^(2iz) + (1 - 3i)e^(iz) - 2 = 0.

Then, by the Quadratic Formula:
e^(iz) = {-(1 - 3i) ± √[(1 - 3i)^2 - 4(2)(-2)]}/[2(2)]
= [-1 + 3i ± √(1 - 6i + 9i^2 + 8)]/4
= [-1 + 3i ± √(1 - 6i - 9 + 8)]/4, since i^2 = -1
= [-1 + 3i ± √(-6i)]/4.

We now need to compute the square root of √(-6i).

The square root (or any root for that matter) of a complex number is another complex number, so suppose that √(-6i) has a square root of a + bi where a and b are real numbers. Then:
a + bi = √(-6i).

Squaring both sides of this gives:
(a + bi)^2 = -6i ==> (a^2 - b^2) + 2abi = -6i.

Comparing real and imaginary parts yields the system:
a^2 - b^2 = 0 and ab = -3.