Properties of the Complex Exponential Function

Recall from The Complex Exponential Function page that if $z = x + yi \in \mathbb{C}$ then we defined the complex exponential function by:

(1)

\begin{align} \quad e^z = e^{x + yi} = e^x e^{yi} = e^x (\cos y + i \sin y) \end{align}

We will now look at some basic properties regarding the complex exponential function - many of which are analogous to the real exponential function but we must indeed prove these for the complex case!

Proposition 1: If $z, w \in \mathbb{C}$ then: a) $e^{z} \neq 0$. b) $e^{\pi i} = -1$ (Euler's Formula). c) $e^z = 1$ if and only if $z = 2k\pi i$ for some $k \in \mathbb{Z}$. d) $e^{z + w} = e^z \cdot e^w$. e) $\mid e^z \mid = e^x$ (where $z = x + yi$. f) $e^0 = 1$, $e^{\frac{\pi}{2}i} = i$, $e^{\pi i} = -1$, and $e^{\frac{3\pi}{2}i} = -i$.

• Proof of a) Let $z = x + yi \in \mathbb{C}$. Then $x, y \in \mathbb{R}$, and $e^z = e^x (\cos y + i \sin y )$. Notice that $e^x > 0$ for all $x \in \mathbb{R}$, so we will look at the other factor, $\cos y + i \sin y$. This factor equals zero if and only if $\cos y = 0$ and $\sin y = 0$. But no such $y \in \mathbb{R}$ exists. So $\cos x + i \sin y \neq 0$ for all $y \in \mathbb{R}$. Thus $e^z \neq 0$. $\blacksquare$

• Proof of b) Substituting $x = \pi i$ into the complex exponential function yields:

(2)

\begin{align} \quad e^{\pi i} = e^{0 + \pi i} = e^0 (\cos \pi + i \sin \pi) = 1(-1 + 0i) = -1 \end{align}

• Therefore $e^{\pi i} = -1$. $\blacksquare$

• Proof of c) $\Rightarrow$ Let $z = x + yi$ and suppose that $e^z = 1$. Then $e^x (\cos y + i \sin y) = 1$. So $\sin y = 0$, i.e., $y = k\pi$ for some $k \in \mathbb{Z}$. Since $e^x \cos y = 1$ and $e^x > 0$ we must have that $\cos y > 0$ and so $y = 2k\pi $ for some [[$ k \in \mathbb{Z}$ in which case we have that $e^x = 0$, i.e., $x = 0$. So $z = 2k\pi i$ for some $k \in \mathbb{Z}$.

• $\Leftarrow$ Suppose that $z = 2k \pi i$ for some $k \in \mathbb{Z}$. Then:

(3)

\begin{align} \quad e^{z} = e^{2k \pi i} = e^0 (\cos (2k \pi) + i \sin (2k \pi)) = 1(1 + 0i) = 1 \quad \blacksquare \end{align}

• Proof of d) Let $z = x_1 + y_1i, w = x_2 + y_2i \in \mathbb{C}$. Then:

(4)

\begin{align} \quad e^{z} \cdot e^{w} &= e^{x_1 + y_1i} \cdot e^{x_2 + y_2i} \\ & = [e^{x_1} (\cos y_1 + i \sin y_1)] [ e^{x_2} (\cos y_2 + i \sin y_2)] \\ &= e^{x_1}e^{x_2}(\cos y_1 + i \sin y_1)(\cos y_2 + i\sin y_2) \\ &= e^{x_1}e^{x_2}[(\cos y_1 \cos y_2 - \sin y_1 \sin y_2) + i(\cos y_1 \sin y_2 + \sin y_1 \cos y_2)] \\ &= e^{x_1}e^{x_2} [\cos (y_1 + y_2) + \sin (y_1 + y_2)] \\ &= e^{x_1 + x_2} [\cos (y_1 + y_2) + \sin (y_1 + y_2)] \\ &= e^{z + w} \quad \blacksquare \end{align}

• Proof of e) Let $z = x + yi \in \mathbb{C}$. Then:

(5)

\begin{align} \quad \mid e^z \mid = \mid e^{x + yi} \mid = \mid e^x(\cos y + i \sin y) \mid = \mid e^x \mid \mid \cos y + i \sin y \mid = \mid e^x \mid \cdot \sqrt{\cos^2 y + \sin^2 y} = \mid e^x \mid \cdot 1 = \mid e^x \mid \end{align}

• We know that $x \in \mathbb{R}$ and $e^x > 0$ so $\mid e^z \mid = e^x$. $\blacksquare$

• Proof of f) We have that:

(6)

\begin{align} \quad e^0 = \cos 0 + i \sin 0 = 1 + i \cdot 0 = 1 \end{align}

(7)

\begin{align} \quad e^{\frac{\pi}{2}i} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = 0 + i \cdot 1 = i \end{align}

(8)

\begin{align} \quad e^{\pi i} = \cos \pi + i \sin \pi = -1 + i \cdot 0 = -1 \end{align}

(9)

\begin{align} \quad e^{\frac{3\pi}{2}i} = \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} = 0 + i \cdot -1 = -i \quad \blacksquare \end{align}