Logarithm

Work in progress. This note is still being written and incomplete.

For a non-zero complex number zz, log⁡z\log z is the set of all w∈Cw \in \mathbb{C} with ew=ze^w = z.

Writing z=reiθz = re^{i\theta} with r=∣z∣r = |z| and w=u+ivw = u + iv, the condition ew=ze^w = z gives eu=re^u = r and v∈arg⁡zv \in \arg z. So

log⁡z=ln⁡∣z∣+iarg⁡z={ ln⁡∣z∣+i(Arg⁡z+2kπ):k∈Z },z≠0\log z = \ln|z| + i\arg z = \{\,\ln|z| + i(\operatorname{Arg} z + 2k\pi) : k \in \mathbb{Z}\,\}, \qquad z \ne 0

Here:

Arg⁡z\operatorname{Arg} z
principal argument of zz, the unique argument with −π<Arg⁡z≤π-\pi < \operatorname{Arg} z \le \pi
arg⁡z\arg z
{Arg⁡z+2kπ:k∈Z}\{\operatorname{Arg} z + 2k\pi : k \in \mathbb{Z}\}

log⁡z\log z is multi-valued.

Principal Value

The logarithm value deduced by setting k=0k=0.

Log⁡z=ln⁡∣z∣+iArg⁡z,z≠0\operatorname{Log} z = \ln|z| + i\operatorname{Arg} z, \qquad z \ne 0

Below you can change zz and see how Log⁡(z)\operatorname{Log}(z) changes.

Re Im 2.24 e 0.46i Log z = 0.80 + 0.46i

Analyticity

Log⁡z\operatorname{Log} z is analytic on C∖(−∞,0]\mathbb{C} \setminus (-\infty, 0] and

ddzLog⁡z=1z\frac{\text{d}}{\text{d}z}\operatorname{Log} z = \frac{1}{z}

Log⁡z\operatorname{Log} z is not analytic at any point of (−∞,0](-\infty, 0].

  • At z=0z = 0 it is undefined.
  • On the negative real axis Arg⁡z\operatorname{Arg} z jumps by 2π2\pi, so Log⁡z\operatorname{Log} z is not continuous there.

The negative real axis is the branch cut of Log⁡z\operatorname{Log} z. Any single-valued branch of log⁡z\log z is analytic on the plane minus a ray from 00.

00 is a branch point. No branch is analytic on a neighbourhood of 00 with only 00 removed, since going once around 00 changes arg⁡z\arg z by 2π2\pi.

Properties

Let z1z_1 and z2z_2 be non-zero.

  • log⁡(z1z2)=log⁡z1+log⁡z2\log(z_1 z_2) = \log z_1 + \log z_2
  • log⁡z1z2=log⁡z1−log⁡z2\log\dfrac{z_1}{z_2} = \log z_1 - \log z_2
  • log⁡1z=−log⁡z\log\dfrac{1}{z} = -\log z

Examples

log⁡(1+i)\log(1+i)

Solution Reveal Hide
  • ∣1+i∣=2|1+i| = \sqrt{2}
  • Arg⁡(1+i)=π4\operatorname{Arg}(1+i) = \dfrac{\pi}{4}
log⁡(1+i)=12ln⁡2+i(π4+2kπ),k∈Z\log(1+i) = \dfrac{1}{2}\ln 2 + i\left(\dfrac{\pi}{4} + 2k\pi\right), \qquad k \in \mathbb{Z}Log⁡(1+i)=12ln⁡2+iπ4\operatorname{Log}(1+i) = \dfrac{1}{2}\ln 2 + i\dfrac{\pi}{4}

log⁡i\log i

Solution Reveal Hide
  • ∣i∣=1|i| = 1
  • Arg⁡(i)=π2\operatorname{Arg}(i) = \dfrac{\pi}{2}
log⁡i=i(π2+2kπ),k∈Z\log i = i\left(\dfrac{\pi}{2} + 2k\pi\right), \qquad k \in \mathbb{Z}Log⁡i=iπ2\operatorname{Log} i = i\dfrac{\pi}{2}
Written by October 17, 2026 2 min read
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