Logarithm of a Complex Number

Follows from z = re^(iθ) by taking ln of both sides: ln z = ln r + iθ. Since arg(z) is multi-valued, ln z is also multi-valued — the general value is ln|z| + i(Arg(z) + 2nπ). The principal value uses Arg(z) ∈ (−π, π].

The logarithm of a complex number is the natural extension of $\ln$ to $\mathbb{C}$. It is multi-valued — and understanding why is as important as knowing the formula.

Derivation

Write $z$ in Euler's form:

$$z = |z|\, e^{i\theta}, \qquad \theta = \arg(z)$$

Express the modulus as an exponential:

$$z = e^{\ln|z|} \cdot e^{i\theta} = e^{\ln|z| + i\theta}$$

So $z = e^w$ where $w = \ln|z| + i\theta$. Taking the natural logarithm of both sides:

$$\ln z = \ln|z| + i\arg(z)$$

Why It Is Multi-Valued

The argument $\arg(z)$ is not unique — it is defined up to multiples of $2\pi$. Any value $\theta + 2k\pi$ ($k \in \mathbb{Z}$) is a valid argument, and each gives a different value of $\ln z$:

$$\ln z = \ln|z| + i(\operatorname{Arg}(z) + 2k\pi), \qquad k \in \mathbb{Z}$$

This is the general value of $\ln z$. There are infinitely many values, one for each integer $k$.

The principal value uses $k = 0$, i.e., $\operatorname{Arg}(z) \in (-\pi, \pi]$:

$$\operatorname{Log}\, z = \ln|z| + i\operatorname{Arg}(z)$$

Consequences

$\ln$ of a negative real number: For $z = -a$ ($a > 0$), $|z| = a$ and $\operatorname{Arg}(z) = \pi$:

$$\operatorname{Log}(-a) = \ln a + i\pi$$

This is why $\ln(-1) = i\pi$ — Euler's identity $e^{i\pi} = -1$ read backwards.

$\ln$ of $i$: $|i| = 1$, $\operatorname{Arg}(i) = \pi/2$:

$$\operatorname{Log}(i) = \ln 1 + i\frac{\pi}{2} = \frac{i\pi}{2}$$

Check: $e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2) = i$ ✓

Where This Appears in JEE

Problems involving $i^i$, $(-1)^{\sqrt{2}}$, or general complex exponents $z^w = e^{w \ln z}$ all use this formula. The multi-valuedness means such expressions have infinitely many values — JEE problems typically ask for the principal value.