Euler's formula defined $e^{i\theta}$ for real $\theta$. Combining it with the ordinary $e^x$ gives the exponential of any complex number. The result is entire, is its own derivative, and has one feature the real exponential lacks: it repeats.
🔑 Key idea
For $z = x + iy$, $\;e^z = e^x(\cos y + i\sin y)$. Then $|e^z| = e^x$, $y$ is an argument of $e^z$, $e^z \ne 0$ for every $z$, $e^{z + w} = e^ze^w$, $(e^z)' = e^z$, and
$$e^{z + 2\pi i} = e^z,$$
so $e^z$ is periodic with period $2\pi i$. The equation $e^z = w$ (with $w \ne 0$) has infinitely many solutions $z = \ln|w| + i(\theta + 2\pi k)$, where $\theta$ is any argument of $w$.
💡 Why it works
The definition is the only one that keeps $e^{x + iy} = e^xe^{iy}$. With $u = e^x\cos y$ and $v = e^x\sin y$: $u_x = e^x\cos y = v_y$ and $u_y = -e^x\sin y = -v_x$, so the Cauchy–Riemann equations hold everywhere and $(e^z)' = u_x + iv_x = e^z$. Since $\cos$ and $\sin$ have period $2\pi$ in $y$, so does $e^z$ in the imaginary direction. To solve $e^z = w$, match moduli ($e^x = |w|$, so $x = \ln|w|$) and angles ($y$ equals an argument of $w$, which is only fixed up to $2\pi k$).
$e^z$ sends the horizontal strip $0 < \operatorname{Im} z < \pi$ onto the upper half-plane: each horizontal line goes to a ray from $0$, each vertical segment to a half circle. Shifting the strip up by $2\pi$ would land on the same picture again.
$z = \ln 2 + (2k + 1)\pi i$ for every integer $k$: for example $\ln 2 + \pi i$ and $\ln 2 - \pi i$.
Answer: $z = \ln 2 + (2k + 1)\pi i$
✏️ Worked example: Modulus and argument
Find $|e^{3 - 4i}|$ and an argument of $e^{3 - 4i}$.
$|e^{x + iy}| = e^x = e^3$.
An argument is $y = -4$ (radians; it is not in $(-\pi, \pi]$, but $-4 + 2\pi$ is, if the principal value is needed).
Answer: $e^3$; $-4$
The picture is worth remembering. A vertical line $\operatorname{Re} z = c$ goes to the circle $|w| = e^c$, traced once for every vertical step of $2\pi$; a horizontal line $\operatorname{Im} z = d$ goes to the ray from $0$ at angle $d$. So every nonzero $w$ is hit, and hit infinitely often, once in each horizontal strip of height $2\pi$. Only $w = 0$ is missed.
⚠️ Common mistake
Expecting $e^z = w$ to have one solution. In real calculus $e^x = 5$ has exactly one solution. In the complex plane $e^z$ repeats every $2\pi i$, so if $z_0$ is a solution, so are $z_0 \pm 2\pi i$, $z_0 \pm 4\pi i$, and so on. Check your list: all solutions share one real part and differ by multiples of $2\pi i$. (And $e^z = 0$ has no solution at all, since $|e^z| = e^x > 0$.)
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.