Complex Logarithm Calculator

Math Algebra • Complex Numbers

Written by STEM Calculators Team Published December 20, 2025 Updated February 24, 2026

9. Complex Logarithm Calculator

Compute the principal complex logarithm \(\mathrm{Log}(z)=\ln|z|+i\,\mathrm{Arg}(z)\), and optionally list non-principal branches \(\log_k(z)=\ln|z|+i(\mathrm{Arg}(z)+2\pi k)\).

Inputs

Input mode

Angle input/display

Real part a

Imag part b

Modulus r

Angle θ

Branch cut / principal Arg range

In mathematics, \(\mathrm{Arg}(z)\) is in radians. If you choose Degrees, this tool shows both degrees and radians.

Custom cut angle α

α is the start of the Arg interval \((\alpha,\alpha+2\pi]\).

List non-principal branches

k range (−K … +K)

Verify \(e^{\mathrm{Log}(z)}\approx z\) Pick/drag \(z\) on graph

Ready

Complex plane

Drag to pan • wheel/pinch to zoom • drag the point to change \(z\) • dashed ray is the branch cut • purple arc is \(\mathrm{Arg}(z)\)

x: 0, y: 0, zoom: 1

Results

Click Calculate.

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Theory: 9. Complex Logarithm Calculator

The complex exponential \(e^{w}\) is periodic in the imaginary direction: \[ e^{w+2\pi i}=e^{w}. \] Because of this periodicity, the inverse function “\(\log\)” cannot be single-valued on all of \(\mathbb{C}\setminus\{0\}\). Instead, the complex logarithm is multi-valued.

1) Polar form and the multi-valued logarithm

For a nonzero complex number \(z\), write polar form \[ z = r(\cos\theta+i\sin\theta), \quad r=|z|>0. \] Any angle \(\theta\) representing \(z\) can be shifted by \(2\pi k\) without changing \(z\). Therefore the logarithm has infinitely many values:

\[ \log(z) = \ln r + i(\theta + 2\pi k), \qquad k\in\mathbb{Z}. \]

2) The principal logarithm \(\mathrm{Log}(z)\)

To make a single value, we choose an argument interval. The most common choice is \[ \mathrm{Arg}(z)\in(-\pi,\pi], \] which corresponds to a branch cut along the negative real axis. The principal logarithm is then

\[ \mathrm{Log}(z)=\ln|z|+i\,\mathrm{Arg}(z). \]

More generally, you can choose a start angle \(\alpha\) and force \[ \mathrm{Arg}(z)\in(\alpha,\alpha+2\pi], \] which moves the branch cut to the ray \(\{re^{i\alpha}: r\ge 0\}\). In a principal branch, the domain excludes both \(z=0\) and the cut ray.

3) Why \(e^{\mathrm{Log}(z)}=z\) still works

If \(\mathrm{Log}(z)=\ln r+i\Theta\) with \(\Theta=\mathrm{Arg}(z)\), then \[ e^{\mathrm{Log}(z)}=e^{\ln r}\,e^{i\Theta}=r(\cos\Theta+i\sin\Theta)=z. \] For any other branch \(\log_k(z)=\ln r+i(\Theta+2\pi k)\), the exponential still returns \(z\) because \(e^{i(\Theta+2\pi k)}=e^{i\Theta}\).

Solved example: \(\mathrm{Log}(-1)\)

Here \(z=-1\) has modulus \(|z|=1\) and (principal) argument \(\mathrm{Arg}(-1)=\pi\). So

\[ \mathrm{Log}(-1)=\ln(1)+i\pi=i\pi. \]

Note: the multi-valued logarithm also includes \(\log(-1)=i(\pi+2\pi k)\) for any integer \(k\).

Frequently Asked Questions

What is the principal complex logarithm Log(z)?

The principal complex logarithm is defined as Log(z) = ln|z| + i Arg(z), where Arg(z) is restricted to a chosen interval such as (−π, π]. This makes the logarithm single-valued on the complex plane excluding the branch cut and z = 0.

Why does the complex logarithm have multiple branches?

Angles that represent the same complex number differ by 2πk, so the logarithm values differ by i 2πk. This produces infinitely many values log_k(z) = ln|z| + i(Arg(z) + 2πk) for integers k.

What is a branch cut in the complex logarithm?

A branch cut is a ray removed from the domain so Arg(z) can be defined continuously on the remaining region. Changing the Arg interval, such as to (α, α+2π], moves the branch cut to the ray at angle α.

How do I choose between Arg(z) in (−π, π] and Arg(z) in (0, 2π]?

Both define a principal value but place the branch cut on different axes: (−π, π] typically cuts along the negative real axis, while (0, 2π] cuts along the positive real axis. The choice changes the reported Arg(z) and the imaginary part of Log(z) near the cut.

Why does e^{Log(z)} return z even though log(z) is multi-valued?

If Log(z) = ln|z| + i Arg(z), then e^{Log(z)} = e^{ln|z|} e^{i Arg(z)} = |z|(cos Arg(z) + i sin Arg(z)) = z. Adding i 2πk to a branch does not change the exponential because e^{i(Arg(z) + 2πk)} = e^{i Arg(z)}.

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Frequently Asked Questions

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