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The exponential turns a real part into a modulus and an imaginary part into an argument, which makes it periodic with period $2\pi i$; the logarithm inverts it and therefore has infinitely many values, so it needs a branch and a cut before it is a function at all.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the complex exponential from the two parts of its argument, use its periodicity, solve exponential equations by matching modulus and argument, compute the principal logarithm and say what the other logarithms are, define a complex power and count how many values it has, and explain what a branch cut is and why one is unavoidable.
Polar form, and the roots of unity: the first place where the argument being fixed only up to whole turns produced $n$ answers instead of one. The logarithm is the same phenomenon with the division left out, so instead of $n$ answers there are infinitely many.
The complex exponential is $e^{z} = e^{x}(\cos y + i \sin y)$. A function is periodic with period $p$ when $f(z + p) = f(z)$ for all $z$. A branch of a multivalued expression is a single-valued choice made continuously on some region; a branch cut is the curve along which that choice is broken. $\operatorname{Log} z$ with a capital letter is the principal logarithm, the one whose imaginary part is the principal argument.
The exponential. For $z = x + iy$,
$$e^{z} = e^{x}\left(\cos y + i\sin y\right),$$
which is exactly the polar form of a complex number with modulus $e^{x}$ and argument $y$. So the exponential converts the real part into a modulus and the imaginary part into an argument. It obeys $e^{z + w} = e^{z}e^{w}$ as usual, it is never zero, and its image is the whole plane except the origin.
And it is periodic. $e^{z + 2\pi i} = e^{z}$ for every $z$: adding $2\pi i$ adds a whole turn to the argument, which is no change at all. The real exponential is one-to-one; the complex one repeats every horizontal strip of height $2\pi$. That single difference is responsible for everything that follows.
The logarithm. Solving $e^{w} = z$ for $z \ne 0$ gives
$$\log z = \ln|z| + i\arg z,$$
and since $\arg z$ is only fixed up to whole turns, $\log z$ is fixed only up to adding $2\pi i k$. It is not a function until a choice is made. The principal logarithm $\operatorname{Log} z = \ln|z| + i\operatorname{Arg} z$ makes the choice by taking the principal argument, and it is continuous everywhere except on the negative real axis and at the origin — the branch cut, where the argument jumps from just below half a turn to just above minus half a turn.
Powers follow. $z^{c} = e^{c\log z}$, and it inherits the ambiguity: one value when $c$ is a whole number, $n$ values when $c = 1/n$, infinitely many otherwise. $i^{\,i} = e^{-\pi/2}$ is the famous case.
And so do the trigonometric functions. $\cos z = \tfrac12(e^{iz} + e^{-iz})$ and $\sin z = \tfrac{1}{2i}(e^{iz} - e^{-iz})$ extend the real ones, keep every identity, and are unbounded: $\cos(iy) = \cosh y$ grows without limit. The familiar bound between $-1$ and $1$ was a fact about the real axis and not about the functions.
Another way: picture
The exponential takes the horizontal line $y = c$ to the ray of argument $c$ from the origin, and the vertical segment $x = c$, $0 \le y < 2\pi$ to the whole circle of radius $e^{c}$. So a horizontal strip of height $2\pi$ is wrapped exactly once round the punctured plane, like a sheet rolled into a cylinder, and the next strip up is wrapped over it again. Choosing a branch of the logarithm is choosing which strip to unroll into.
Another way: steps
A branch cut looks like an administrative decision and is in fact a statement about the shape of the plane.
Walk once anticlockwise round a circle enclosing the origin, keeping track of the argument continuously. You return to the point you started from with an argument that has grown by $2\pi$ — and therefore with a logarithm that has grown by $2\pi i$. Nothing went wrong; the argument simply cannot be defined continuously on a loop that encircles the origin.
So a single-valued logarithm needs a region in which no loop encircles the origin, and cutting the plane along a curve from the origin to infinity produces one. The negative real axis is the usual choice; any curve would do, and the function it produces differs from the usual one only in where it jumps.
This is worth holding on to, because it is the same phenomenon as the one that makes $\oint dz/z = 2\pi i$ rather than $0$ in unit 3. A loop round the origin picks up $2\pi i$, whether you are following an argument or integrating $1/z$, and those two statements are eventually the same statement. An argument is fixed only up to whole turns, and every strangeness of the logarithm, the power and the root comes from that one sentence. Choosing a branch is choosing which turn to call the right one, and the cut is where the choice comes apart. A calculation that treats an argument as a single number is a calculation that will contradict itself the first time it crosses the cut.
Every identity of the real logarithm has a complex version that holds up to a whole turn, and dropping that qualification produces confident nonsense.
$\operatorname{Log}(zw) = \operatorname{Log} z + \operatorname{Log} w$ is false in general. Take $z = w = -1$: the left side is $\operatorname{Log} 1 = 0$ and the right side is $i\pi + i\pi = 2\pi i$. The equation holds for $\log$, the multivalued object, as an equality of sets — not for the principal branch.
$\log(e^{z}) = z$ is false in general for the same reason: it recovers $z$ only up to a whole multiple of $2\pi i$, because the exponential threw that information away. The other way round, $e^{\log z} = z$ is always true, on any branch.
$\cos z$ is not bounded by $1$. On the imaginary axis it is the hyperbolic cosine and grows without limit, and that fact is doing real work later: it is why $\cos$ is not a counterexample to Liouville's theorem and why the equation $\cos z = 5$ has solutions.
$e^{2 + i\pi}$: the real part $2$ gives the modulus $e^{2}$, and the imaginary part $\pi$ gives the argument, half a turn.
Real part to modulus, imaginary part to argument.
Half a turn out at distance $e^{2}$ is the negative real number $-e^{2}$.
The exponential of a complex number can be negative.
$\operatorname{Log}(-1 + i)$: the modulus is $\sqrt{2}$ and the principal argument is three quarters of half a turn, $3\pi/4$.
Modulus and argument first.
So $\operatorname{Log}(-1 + i) = \tfrac12\ln 2 + 3\pi i/4$, and the other logarithms are this plus any whole multiple of $2\pi i$.
One principal value, infinitely many logarithms.
The modulus of $-1$ is $1$, so the real part of $z$ is $\ln 1 = 0$.
Match the modulus first.
An argument of $-1$ is $\pi$ plus any whole number of turns, so $z = i\pi + 2\pi i k$ for every whole number $k$.
Let $z = 3 + i\theta$, where $\theta$ is the angle of $26$ degrees written in radians. Fill in the table.
| Value | |
|---|---|
| The power of e giving the modulus of the exponential of z | |
| The argument in degrees of the exponential of z | |
| The power of e giving the modulus of the exponential of twice z | |
| The argument in degrees of the exponential of twice z | |
| The power of e giving the modulus of the exponential of z plus two pi i | |
| The argument in degrees of the exponential of z plus two pi i |
Does this explain why the complex logarithm takes more than one value: the exponential is never zero?
The equation $e^{z} = 8$ has infinitely many solutions. How many of them have imaginary part strictly between $-5 \times 2\pi$ and $5 \times 2\pi$?
Answer:
For a non-zero $z$, match each expression to how many values it has.
| Exactly one value | Exactly one value, because the power is a whole number | Exactly $4$ values | Infinitely many values | |
|---|---|---|---|---|
| $e^{z}$ | ||||
| $z^{4}$ | ||||
| $z^{1/4}$ | ||||
| $\log z$ |
Put the steps of computing the principal logarithm of $1 + i$ in order.
Number the steps in order (write the number in the box):
Using principal values, $\left(i^{\,i}\right)^{3} = e^{k\pi/2}$ for one real number $k$. What is $k$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $z = 6 + i\theta$, where $\theta$ is the angle of $22$ degrees written in radians. Fill in the table.
| Value | |
|---|---|
| The power of e giving the modulus of the exponential of z | |
| The argument in degrees of the exponential of z | |
| The power of e giving the modulus of the exponential of twice z | |
| The argument in degrees of the exponential of twice z | |
| The power of e giving the modulus of the exponential of z plus two pi i | |
| The argument in degrees of the exponential of z plus two pi i |
You can compute with the complex exponential and with the logarithm, and you can say why the logarithm needs a branch before it is a function. Say in your own words what goes wrong on a loop round the origin. Next: limits and continuity, where the plane starts to demand more than the line ever did.
9. Your turn: every solution of $e^{z} = -1$, step 2