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2. Week 2: More on \(\exp\) and \(\log\)

≪ 1. Week 1: Complex Numbers, Alone and in Packs | Table of Contents | 3. Week 3: Complex Derivatives ≫

One way to define the logarithm of a real number \(x\) is as follows: if \(y\) is a real number such that \(e^y=x\), then \(\log x=y\). In words, the logarithm is the inverse map to the exponential function, formalised by the hopefully familiar identities \[e ^{\log x} = \log(e^x) = x\] for all real numbers \(x > 0\). The restriction that \(x>0\) is important when working with real numbers: there exists no real number \(y\) such that \(e^y = -1\), for instance.

This suddenly changes when we enter the land of complex numbers. We have for real numbers \(x\) and \(y\) that \[e ^{x+iy} = e^x e ^{iy}.\] Interpreted in polar form, this is the point a distance of \(e^x\) away from the origin and making an angle \(y\) with the real axis.

Thus, writing \(z = r e ^{i \theta }\) for any nonzero complex number, this tells us that \[z = r e ^{i \theta } = e ^{\log r} e ^{i \theta } = e ^{\log r + i \theta }.\] This works for any nonzero complex number, including \(z=-1\)! We therefore define the complex logarithm, denoted \(\operatorname{Log}\), as follows: \[\operatorname{Log} z = \left\lbrace \log r + i \theta + 2 \pi i n : n \in \mathbb{Z} \right\rbrace,\] where \(z=re ^{i \theta }\) is the polar form of \(z\).

Note that this is a whole set of numbers: adding multiples of \(2 \pi\) to the argument of \(z\) does not change what \(z\) is, but it does change the complex logarithm of \(z\)!

Example 1.

  1. Compute \(\operatorname{Log}(-1)\).
  2. Compute \(\operatorname{Log}(9+12i)\)
Solution to part 1
First, we write write \(-1\) in polar form. It is a distance of \(r=1\) from the origin, and it makes an angle of \(\theta = \pi \) with the positive real axis. Thus, \(-1 = r e ^{i \theta } = 1\cdot e ^{i \pi }\). Using the above, with \(\log r = 0\) and \(\theta = \pi \), \[\operatorname{Log}(-1) = \left\lbrace i \pi + 2 \pi i n : n \in \mathbb{Z} \right\rbrace.\] You can verify that \(e ^{3 \pi i} = -1\), showing that both \(3 \pi i\) and \(\pi i\) are equally valid yet distinct choices for the complex logarithm of \(-1\).

Let’s think about what the exponential map does geometrically, beginning first with analysing the image of the horizontal strip \[\left\lbrace x+iy : x\in \mathbb{R}, y \in \left[ -\pi , \pi \right] \right\rbrace\] under \(\exp\). One can imagine rolling up this strip into a cylinder, then staring through the cylinder from the right side (from “\(+\infty\)”). The cylinder will fill up your entire field of view, minus a hole at the origin! Vertical lines in the strip got wrapped around into circles around the cylinder, which appear as circles centred at the origin. Horizontal lines stayed straight, and they now appear to radiate out from the origin.

The global picture for the exponential map is slightly different. Repeating the above over and over and covering the complex plane with such horizontal strips, the exponential map is coiling the complex plane \(\mathbb{C}\) into a helical spiral rather than a many-layered cylinder, then squashing it down onto \(\mathbb{C} \setminus \left\lbrace 0 \right\rbrace\). Each strip of height \(2 \pi \) corresponds to one coil of the spiral.

I’m sorry I couldn’t draw a better spiral. I was having a lot of trouble getting the perspective right, and I gave up on drawing the blue circles coiling up the spiral. But I hope this gets the image across. The point is that the exponential map performs this coiling and squashing action, taking infinitely “layers” of the spiral and compressing them down into one punctured complex plane. Note how points that are multiples of \(2 \pi i\) apart are first coiled onto vertically overlapping points, then squashed down into a single point. This reflects how multiples of \(2 \pi \) in the argument of a complex number doesn’t change the number itself.

Homework Problem 2.

Sketch \(L_1\cup L_2\cup L_3\) and its image under the map \(f(z) = e ^{z}\), where \[\begin{align*} L_1 &= \left\lbrace x+iy : -\infty < x \leq \ln 3, y = 0 \right\rbrace, \\ L_2 &= \left\lbrace x+iy : -\infty < x \leq \ln 3, y = \frac{\pi }{2} \right\rbrace, \textrm{and}\\ L_3 &= \left\lbrace x+iy : x = \ln 3, 0 \leq y \leq \frac{\pi }{2} \right\rbrace. \end{align*} \]

The job of the complex logarithm is thus to undo this action of the exponential map: it needs to “lift” \(\mathbb{C}\setminus \left\lbrace 0 \right\rbrace\) up onto the spiral somehow.

We thus arrive at a grave issue. The real-valued logarithm is continuous (even smooth) on its entire domain of the positive real numbers. However, we have a geometric obstruction to defining a continuous complex logarithm on its entire domain \(\mathbb{C}\setminus \left\lbrace 0 \right\rbrace\): when we lift the punctured plane back up to the helical spiral, there ends up being a point of discontinuity. We have to “pull apart” \(\mathbb{C}\setminus \left\lbrace 0 \right\rbrace\) along some line in order to go back up to the spiral. Moreover, it’s hard to decide which branch of the spiral to pick. Who’s to say one branch is “better” or “more appropriate” than the other?

In the above diagram, the pink line had to be separated onto two adjacent “levels” of the spiral. Two nearby points (marked in red) on either side of of the pink line thus get pulled far apart, making the action of the complex logarithm discontinuous.

The fix is to make what’s called a “branch cut”. By excising the pink link from the domain of \(\operatorname{Log}\), suddenly we can find a continuous way to “lift” the slit plane back up to the spiral — we no longer have to separate one line into two.

Of course, this is not the only way to make a branch cut, but it’s one of the simplest ways. You can cut out any ray extending from the origin, and you can pick any branch of the spiral that you want. In fact, you can even cut out a curve starting from the origin and extending to \(\infty\) without self-intersections, then define the complex logarithm from there! Why you would do this is beyond me.

There is a standard choice of the complex logarithm, where one excises the negative real axis from the punctured complex plane and picks the branch of the spiral that contains the real axis. This way, \(\operatorname{Log}(1) = 0\) (the way it should be)! This is called the “principal branch” of the complex logarithm.

We should remark that the complex logarithm is often expressed as \[\operatorname{Log} z = \log \left\lvert z \right\rvert + \operatorname{Arg} z.\] The discontinuity of \(\operatorname{Log}\) on its full domain reflects the impossibility of defining a continuous \(\operatorname{Arg}\) function on \(\mathbb{C}\setminus \left\lbrace 0 \right\rbrace\)! The principal branch of the logarithm also corresponds to the standard choice of the argument.

Problem 3.

Define the set \[A = \left\lbrace re ^{i \theta } : 1 < r < 2, 0 < \theta < \frac{\pi }{4} \right\rbrace.\]

  1. Sketch \(A\) as a region of the complex plane.
  2. Sketch the inverse image of \(A\) under \(e^z\), i.e. sketch the set \[\left\lbrace z\in \mathbb{C} : e^z\in A \right\rbrace.\]
  3. Sketch the image of \(A\) under the principal branch of the complex logarithm \(\operatorname{Log}\), i.e. sketch the set \[\left\lbrace \operatorname{Log}(z) : z\in A \right\rbrace.\]
  4. What’s the difference between 2 and 3?