Thread: Roots of equation

Problem statement is to find all roots of cos(z) - isin(z) = 0

I know that e^(-iz) = cos(z) - isin(z)

If I were to express e^(-iz) = e^y*e^(-ix) = 0
(since z = x + iy)
either e^y = 0 or e^(-ix) = 0 or both.
e^(-ix) = cos(x) - isin(x). This is never 0.
e^y as a real function of y is also never 0.

Thus, how do I go about finding the roots to this equation?

Originally Posted by amoeba

Problem statement is to find all roots of cos(z) - isin(z) = 0

I know that e^(-iz) = cos(z) - isin(z)

If I were to express e^(-iz) = e^y*e^(-ix) = 0
(since z = x + iy)
either e^y = 0 or e^(-ix) = 0 or both.
e^(-ix) = cos(x) - isin(x). This is never 0.
e^y as a real function of y is also never 0.

Thus, how do I go about finding the roots to this equation?

As

Putting , we get , and since these two real functions never vanish at the same point we get that there are no roots.

Tonio