Final answer:
The solution to z^6 = -1 can be found using the polar form representation of -1 and finding the sixth roots of unity. The equation sin(z) = 50 has infinite solutions derived from the complex arcsine of 50 and the periodicity of the sine function. Both sets of solutions can be represented on the complex plane.
Step-by-step explanation:
The question presents two separate mathematical challenges involving complex numbers and trigonometric functions, respectively. To find all solutions for the equation z6 = -1, we must recognize that -1 can be represented as cos( π ) + i*sin( π ) in polar form. Since the roots of unity in the complex plane are evenly spaced, each solution is 60 degrees (π/3 radians) apart. Therefore, the solutions can be expressed as z = exp((2kπ + π)/6) for k = 0, 1, 2, 3, 4, 5. For the equation sin(z) = 50, recall that the sine function in the complex plane is not bounded as it is in the real domain, so it can achieve values greater than 1. Hence, there will be an infinite number of solutions in the form z = (-1)nasin(50) + nπ, where asin(50) represents the complex arcsine of 50, and n is an integer.