1. ## 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?

2. 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 $\displaystyle \cos z=\frac{e^{iz}+e^{-iz}}{2}\,,\,\,\sin z=\frac{e^{iz}-e^{-iz}}{2i}\Longrightarrow\,0=\cos z-i\sin z=e^{-iz}$

Putting $\displaystyle z=x+iy\,,\,\,x\,,\,y\,\in\mathbb{R}$, we get $\displaystyle 0=e^{-iz}=e^{-ix+y}=e^y(\cos x-i\sin x)\Longrightarrow\,\cos x=\sin x= 0$, and since these two real functions never vanish at the same point we get that there are no roots.

Tonio