The Logarithmic Function

Consider $z$ any nonzero complex number. We would like to solve for $w,$ the equation $\begin{array}{r} (1) & e^{w} = z . \end{array}$ If $Θ = Arg (z)$ with $- π < Θ \leq π,$ then $z$ and $w$ can be written as follows $\begin{array}{r} z = r e^{i Θ} and w = u + i v . \end{array}$ Then equation ( $1$ ) becomes $e^{u} e^{i v} = r e^{i Θ} .$ Thus, we have $\begin{array}{r} e^{u} = r and v = Θ + 2 n π \end{array}$ where $n \in Z .$ Since $e^{u} = r$ is the same as $u = \ln r,$ it follows that equation ( $1$ ) is satisfied if and only if $w$ has one of the values

\begin{array}{r} w = \ln r + i (Θ + 2 n π) (n \in Z) . \end{array}

Therefore, the (multiple-valued) logarithmic function of a nonzero complex variable $z = r e^{i Θ}$ is defined by the formula

\begin{array}{r} (2) & \log z = \ln r + i (Θ + 2 n π) (n \in Z) . \end{array}

Example 1: Calculate $\log z$ for $z = - 1 - \sqrt{3} i .$

Solution: If $z = - 1 - \sqrt{3} i,$ then $r = 2$ and $Θ = - \frac{2 π}{3} .$ Hence

\log (- 1 - \sqrt{3} i) = \ln 2 + i (- \frac{2 π}{3} + 2 n π) = \ln 2 + 2 (n - \frac{1}{3}) π i

with

n \in Z .

The principal value of $\log z$ is the value obtained from equation ( $2$ ) when $n = 0$ and is denoted by $Log z .$ Thus $Log z = \ln r + i Θ .$ The function $Log z$ is well defined and single-valued when $z \neq 0$ and that $\log z = Log z + 2 n π i (n \in Z)$ This is reduced to the usual logarithm in calculus when $z$ is a positive real number.

Example 2: Calculate $\log (1)$ and $\log (- 1) .$

Solution: From expression ( $2$ ) $\log (1) = \ln 1 + i (0 + 2 n π) = 2 n π i (n \in Z)$ and $\log (- 1) = \ln 1 + i (π + 2 n π) = (2 n + 1) π i (n \in Z)$ Notice that $Log (1) = 0$ and $Log (- 1) = π i .$

Expression ( $2$ ) is also equivalent to the following: $\begin{array}{rcl} \log z & = & \ln | z | + i arg (z) \\ = & \ln | z | + i Arg (z) + 2 n i π (n \in Z) \end{array}$

Some basic properties of the function $\log z$ are the following:

$\log (z_{1} z_{2}) = \log z_{1} + \log z_{2}$
$\log (\frac{z_{1}}{z_{2}}) = \log z_{1} - \log z_{2}$
There may hold $Log (z_{1} z_{2}) \neq Log z_{1} + Log z_{2}$

Branches of logarithms

From definition ( $2$ ) let $θ = Θ + 2 n π$ ( $n \in Z$ ), so we can write $\begin{array}{r} (3) & \log z = \ln r + i θ . \end{array}$

Now, let $α$ be any real number. If we restrict the value of $θ$ so that $α < θ < α + 2 n π$ , then the function

\begin{array}{r} (4) & \log z = \ln r + i θ (r > 0, α < θ < α + 2 π), \end{array}

with components

\begin{array}{r} u (r, θ) = \ln r, v (r, θ) = θ, \end{array}

is a single-value and continuous function in the stated domain.

A branch of a multiple-valued function $f$ is any single-valued function $F$ that is analytic in some domain at each point $z$ of which the value $F (z)$ is one of the values of $f .$ The requirement of analyticity, of course, prevents $F$ from taking on a random selection of the values of $f .$ Observe that for each fixed $α,$ the single-valued function ( $4$ ) is a branch of the multiple-valued function ( $3$ ). The function

\begin{array}{r} (5) & Log z = \ln r + i Θ (r > 0, - π < θ < π), \end{array}

is called the principal branch.

A branch cut is a portion of a line or curve that is introduced in order to define a branch $F$ of a multiple-valued function $f .$ Points on the branch cut for $F$ are singular points of $F,$ and any point that is common to all branch cuts of $f$ is called a branch point. The origin and the ray $θ = α$ make up the branch cut for the branch ( $4$ ) of the logarithmic function. The branch cut for the principal branch ( $5$ ) consists of the origin and the ray $= π .$ The origin is evidently a branch point for branches of the multiple-valued logarithmic function.

We can visualize the multiple-valued nature of $\log z$ by using Riemann surfaces. The following interactive images show the real and imaginary components of $\log (z) .$

Real component of $\log z$

Imaginary component of $\log z$ : Each branch is identified with a different color.

Special care must be taken in using branches of the logarithmic function, especially since expected identities involving logarithms do not always carry over from calculus.

Final remark

Notice that for $z \neq 0,$ we have

\begin{array}{r} (6) & e^{\log z} = z and \log (e^{z}) = z + 2 n π i \end{array}

with

n \in Z .

Example 3: Calculate $e^{\log z},$ and $\log (e^{z})$ for $z = 4 i .$

Solution: If $z = 4 i,$ then $e^{z} = e^{4 i} .$ Hence $\log (e^{4 i}) = 4 i + 2 n π i$ with $n \in Z .$ On the other hand, we have $e^{\log (4 i)} = 4 i .$