cot z= -3i what is the value of z?
Some relations we can use.
cos(z) = (1/2)(e^(iz) + e^(-iz)), sin(z) = (1/2)(e^(iz) – e^(-iz))
Let u = e^(iz) in the expression cot(z) = -3i, then
cot(z) = cos(z)/sin(z) = (e^(iz) + e^(-iz)) / (e^(iz) – e^(-iz)) = (u + u^(-1)) / (u – u^(-1))
(u^2 + 1) / (u^2– 1) = -3i
u^2 + 1= -3i.(u^2 – 1)
u^2(1 + 3i) = -1 + 3i
u^2 = -(1 - 3i) / (1 + 3i) = -(1 – 3i)^2 / {(1 + 3i)(1 – 3i) = -(1 – 3i)^2 / ( 1 + 9)
u^2 = -(1 – 6i – 9) / 10
u^2 = (8 + 6i)/10
u = √((4 + 3i)/5)
Then, using e^(iz) = u,
e^(iz) = √((4 + 3i)/5)
iz = ln{√((4 + 3i)/5)}
z = -i. ln{√((4 + 3i)/5)}