To determine whether $ \cos(z) = 0 $ implies that $ z $ must be real, we analyze the equation using the definition of cosine in the complex domain.
Key Idea: Using the Complex Definition of Cosine
The cosine function for a complex number $ z = x + iy $ (where $ x, y \in \mathbb{R} $) is defined as:
$$
\cos(z) = \frac{e^{iz} + e^{-iz}}{2}.
$$
Substituting $ z = x + iy $:
$$
\cos(z) = \frac{e^{i(x + iy)} + e^{-i(x + iy)}}{2} = \frac{e^{ix}e^{-y} + e^{-ix}e^y}{2}.
$$
Using Euler’s formula ($ e^{ix} = \cos(x) + i\sin(x) $), this expands to:
$$
\cos(z) = \frac{e^{-y}(\cos(x) + i\sin(x)) + e^y(\cos(x) – i\sin(x))}{2}.
$$
Simplify:
$$
\cos(z) = \frac{\cos(x)(e^{-y} + e^y) + i\sin(x)(e^{-y} – e^y)}{2}.
$$
Separate real and imaginary parts:
$$
\cos(z) = \cos(x)\cosh(y) + i\sin(x)\sinh(y).
$$
Analyzing $ \cos(z) = 0 $
For $ \cos(z) = 0 $, both the real and imaginary parts must be zero. That is:
- $ \cos(x)\cosh(y) = 0 $ (real part),
- $ \sin(x)\sinh(y) = 0 $ (imaginary part).
1. $ \cos(x)\cosh(y) = 0 $:
- $ \cosh(y) \neq 0 ) for all ( y \in \mathbb{R} ) because ( \cosh(y) = \frac{e^y + e^{-y}}{2} > 0 $.
- Therefore, $ \cos(x) = 0 $.
2. $ \sin(x)\sinh(y) = 0 $:
- $ \sinh(y) = 0 $ if and only if $ y = 0 $ because $ \sinh(y) = \frac{e^y – e^{-y}}{2} $.
- If $ y = 0 $, then $ \sin(x) $ can take any value, satisfying this equation.
Conclusion
- From the first equation, $ \cos(x) = 0 $, which implies $ x = \frac{\pi}{2} + n\pi $, where $ n \in \mathbb{Z} $.
- From the second equation, $ \sinh(y) = 0 $, which implies $ y = 0 $.
Thus, $ z = x + iy $ is real, specifically $ z = \frac{\pi}{2} + n\pi $, where $ n \in \mathbb{Z} $.
Therefore, $ \cos(z) = 0 $ implies $ z $ must be real.
