Home » Blog » Prove that the complex exponential function is never zero

Prove that the complex exponential function is never zero

by RoRi
  1. Prove that e^z \neq 0 for all z \in \mathbb{C}.
  2. Find all z \in \mathbb{C} such that e^z = 1.
    1. Proof. Let z = a + bi for real numbers a,b. Then from the definition of the complex exponential, and the fact that e^{x+y} = e^x e^y for complex numbers x and y from Theorem 9.3 (page 367 of Apostol) we have,

          \[ e^z = e^{a+bi} = e^a (\cos b + i \sin b). \]

      But from the properties of the real exponential function we know e^a \neq 0 for any a \in \mathbb{R}. Furthermore, we know \cos b + i \sin b \neq 0 for all b \in \mathbb{R} since

          \[ \cos b + i \sin b = 0 \quad \implies \quad \cos b = \sin b = 0 \quad \implies \quad \sin^2 b + \cos^2 b = 0, \]

      which contradicts the Pythagorean identity (\cos^2 + \sin^2 = 1). Hence, e^z \neq 0 for any z \in \mathbb{C}. \qquad \blacksquare

    2. We compute

          \[ e^z = 1 \quad \implies \quad e^a \cos b + i (e^a \sin b) &= 1. \]

      Then, setting real and imaginary parts equal this implies

          \[ e^a \cos b = 1 \quad \text{and} \quad e^A \sin b = 0. \]

      This implies a = 0 and b = 2n \pi. Thus,

          \[ z = 2n \pi i. \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):