Lesson 3: Power Series & Elementary Functions

$e^{i\pi}=-1$ is a familiar fact, usually justified via the polar-multiplication picture from Lesson 1: rotate by $\pi$, land on $-1$. This lesson gets there a different way: from the power series definition of $e^z$ itself, which turns out to be the more fundamental route.

Convergence

The complex exponential is defined the same way as its real counterpart, as a power series: $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$ Applying the ratio test, the ratio of consecutive terms $\to 0$ as $n\to\infty$ for any $z$, so the radius of convergence is $R=\infty$: $e^z$ is entire (holomorphic on all of $\mathbb{C}$). Contrast with $\sum n!\,z^n$, whose ratio $\to\infty$: radius of convergence $R=0$, converging only at $z=0$.

But what does a complex series converging actually look like? The simplest testbed is the geometric series $\sum_{n=0}^\infty z^n$, radius of convergence exactly $1$, and its partial sums can be watched directly:

Each partial sum $S_N = 1 + z + \dots + z^N$ is a point in the plane, and consecutive sums differ by one term $z^N$, a step that rotates by $\arg z$ and rescales by $|z|$ every time. Inside the unit circle the steps shrink geometrically and the path spirals into $\frac{1}{1-z}$; outside, each step is longer than the last and the path escapes; on the circle itself the steps never shrink and the sums circle forever. That threshold (shrinking terms inside, growing terms outside) is what "radius of convergence" means, and the $\varepsilon$-disk overlay turns the definition of convergence into something you can find with a slider: for every $\varepsilon$ there's an $N$ past which the path never leaves the disk. (That definition gets its full treatment in Lesson 5a.)

Euler's formula, derived rather than assumed

Substitute $z=iy$ into the series and write out the first several terms of $e^{iy}$. Separating them by whether $i$'s power is even or odd — using $i^2=-1$, $i^3=-i$, $i^4=1,\dots$ — the even-indexed terms reassemble into $\cos y$'s Taylor series, and the odd-indexed terms (each carrying one factor of $i$) reassemble into $i\sin y$'s. That is, $$e^{iy} = \cos y + i\sin y.$$ Euler's formula isn't a separate assumed identity: it's a consequence of the power series definition of $e^z$, which is exactly what resolves the gap the polar-multiplication picture leaves open.

Cauchy-Riemann check for $e^z$

Writing $e^z=e^{x+iy}=e^x\cos y + ie^x\sin y$, so $u=e^x\cos y$, $v=e^x\sin y$: $$u_x = e^x\cos y = v_y, \qquad u_y=-e^x\sin y=-v_x$$ Cauchy-Riemann holds everywhere, confirming directly that $e^z$ is entire (consistent with the $R=\infty$ result above, now via Lesson 2's machinery).

Key takeaways: $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$, radius of convergence $R=\infty$ (entire). Euler's formula, derived from the series: $e^{iy}=\cos y + i \sin y$; $e^{i\pi}=-1$.

Complex trig functions

Once $e^z$ is defined, $\cos z$ and $\sin z$ follow by inverting Euler's formula: $$\cos z = \frac{e^{iz}+e^{-iz}}{2}, \qquad \sin z = \frac{e^{iz}-e^{-iz}}{2i}$$ These are genuinely different from their real counterparts in one striking way: they're unbounded. Take $z=i$: $$\cos(i) = \frac{e^{i\cdot i}+e^{-i\cdot i}}{2} = \frac{e^{-1}+e^{1}}{2} \approx 1.543$$ $\cos(i)>1$: impossible for real cosine, perfectly fine for the complex version, since the usual $[-1,1]$ bound was a fact about restricting to the real line, not a fact about the function itself.

Key takeaway: $\cos z = \frac{e^{iz}+e^{-iz}}{2}$, $\sin z = \frac{e^{iz}-e^{-iz}}{2i}$, unbounded on $\mathbb{C}$, e.g. $\cos(i)=\frac{e^{-1}+e}{2}>1$.

The complex logarithm and branch cuts

Defining $\log z$ means solving $e^w=z$ for $w$. Writing $z=re^{i\theta}$ and using the same "match in polar form" reasoning as the roots-of-unity derivation in Lesson 1: $e^w=z$ needs $w=\ln r + i\theta$, but since $\theta$ is only defined mod $2\pi$, so is $w$: there are infinitely many valid values of $w$, differing by $2\pi ik$: $$\log z = \ln|z| + i(\theta+2\pi k), \qquad k\in\mathbb{Z}$$ $\log z$ is genuinely multivalued. Picking the principal argument gives the principal branch, $$\operatorname{Log} z = \ln|z| + i\,\operatorname{Arg}(z), \qquad \operatorname{Arg}(z)\in(-\pi,\pi].$$

This forces a discontinuity. As $z$ crosses the negative real axis, $\operatorname{Arg}(z)$ jumps from just under $\pi$ to just above $-\pi$, a jump of nearly $2\pi$ in an otherwise continuous function. The ray where this happens (the negative real axis, for the principal branch) is called a branch cut: removing it from the domain is exactly what makes $\operatorname{Log}$ single-valued and continuous everywhere else.

The four views below make each piece of that story visible. As you scroll, the pinned demo switches views to follow the argument, and every view stays fully interactive (drag the probe, spin the surface) the whole way.

Every full loop around the origin shifts $\operatorname{Arg}(z)$ by a full $2\pi$ (the multivaluedness of $\log z$ made visible), and the branch cut is precisely the ray where that walk gets interrupted to force a single, consistent choice.

Key takeaways: $\log z$: if $e^w=z$, $w=\ln|z|+i(\theta+2\pi k)$, $k\in\mathbb{Z}$, multivalued. Principal branch: $\operatorname{Log}z=\ln|z|+i\operatorname{Arg}(z)$, $\operatorname{Arg}(z)\in(-\pi,\pi]$. Branch cut: the ray (negative real axis, for the principal branch) where $\operatorname{Arg}(z)$ (and hence $\operatorname{Log}z$) is discontinuous.