2.2 Complex Logarithms

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We define the multivalued complex logarithm and discuss its branches and properties. We also define complex exponential functions.

1 The Multivalued Logarithm

Consider the function $w = f(z) = e^z = e^x \cis y$ . This function is $2\pi i$ -periodic, so it is not one-to-one. Hence, $e^z$ does not have a traditional inverse- the complex logarithm is multivalued. To define the complex log, consider a complex number $w_0$ in the image of $f(z) = e^z$ , i.e., let $z_0 = x_0 + iy_0$ and let $w_0 = e^{z_0}$ . Then $w_0 = e^{x_0} \cis (y_0)$ Due to the fact that $e^z$ is $2\pi i$ periodic, we can write $w_0 = e^{x_0} \cis (y_0+2k\pi )$ for any integer $k$ . See the figure below.

Since $w_0 = e^{x_0}\cis (y_0 + 2k\pi )$ , we can see that the modulus of $w_0$ is $|w_0| = e^{x_0}$ and an argument of $w_0$ is $y_0$ , so that $\arg (w_0) = \{y_0 + 2k\pi \}$

We wish to define the function $\log w_0$ so that the ouputs are the set of points $x_0 + i(y_0 + 2k\pi )$ . Thus we define $\log w_0 = \ln |w_0| +i \arg w_0$ . In the definition below, we use $z$ in the place of $w_0$ .

Multivalued Logarithm For $z \neq 0$ , we define the multivalued complex logarithm by $\log z = \ln |z| + i \arg z$

example 1 Find the values of $\log (1+i)$ .
We have $\log (1+i) = \ln |1+i| + i \arg (1+i) = \tfrac 12\ln 2 + i\left (\frac {\pi }{4} + 2k\pi \right )$ where $k \in \Z$ . Note that $\ln \sqrt 2 = \ln 2^{1/2} = \frac 12\ln 2$ .

(problem 1a) Find all values of the complex logarithm: $\; \log i$ .

$\frac {\pi }{2} + 2k\pi$
$1 + i\left (\frac {\pi }{2} + 2k\pi \right )$
$i\left (\frac {\pi }{2} + 2k\pi \right )$
$2k\pi i$

(problem 1b) Find all values of the complex logarithm: $\; \log (-3)$

$(2k+1)\pi + i\ln 3$
$\ln 3 + (2k+1)\pi i$
$3 + (2k+1)\pi i$
$(2k-1)\pi i$

(problem 1c) Find all values of the complex logarithm: $\; \log (i-1)$

$\left (\frac {3\pi }{4} + 2k \pi \right ) + i \ln \sqrt 2$
$\ln \sqrt 2 + i\left (\frac {3\pi }{4} + k \pi \right )$
$\ln 2 + i\left (\frac {3\pi }{4} + 2k \pi \right )$
$\tfrac 12 \ln 2 + i\left (\frac {3\pi }{4} + 2k \pi \right )$

Here is a video solution of problem 1c:

Some familiar properties of real logarithms carry over to the multivalued complex logarithm.

For $z_1, z_2 \neq 0$ , $\log (z_1\cdot z_2) = \log z_1 + \log z_2$ and for $z \neq 0$ $\log \left (\frac {1}{z}\right ) = - \log z$

The key to the proof of the first property is that the argument of a product is the sum of the arguments: $\arg (z_1 \cdot z_2) = \arg z_1 + \arg z_2$ As a result of this property of the multivalued argument, we have \begin{align*} \log (z_1\cdot z_2) &= \ln |z_1 z_2| + i\arg (z_1 z_2)\\ &= \ln |z_1| + \ln |z_2| + i\left [\arg z_1 + \arg z_2\right ]\\ &= \log z_1 + \log z_2 \end{align*}

The proof of the second property is similar (verify) and uses $\arg \left (\frac {1}{z}\right ) = -\arg z$

For $z_1, z_2 \neq 0$ $\log \left (\frac {z_1}{z_2}\right ) = \log z_1 - \log z_2$

2 Principal Log

Remarking that the complex exponential function, $e^z$ , is defined in terms of a radial part, $e^x=r$ , and an angular part $e^{iy}= \cis y$ , it is natural that the complex logarithm be defined in terms of the real natural logarithm of a modulus, and an argument function.

$\Log z$ For $z \neq 0$ , we define the principal branch of the complex logarithm by $\Log z = \ln |z| + i\Arg z$ where $\Arg z$ is the principal argument of $z$ .

First, we will show that $\Log z$ and $e^z$ are inverses.

$e^{\Log z} = e^{\ln |z| + i\Arg z} = e^{\ln |z|} \cis \Arg z = |z| \cis \Arg z =z$ in polar form (if $z=r\cis \theta$ then $r = |z|$ and $\theta \in \arg z$ ). Furthermore, $\Log e^z = \Log e^x\cis y = \ln e^x + i\Arg (\cis y) = x+iy$ if and only if $-\pi < y \leq \pi$ .

Example 2 Find $\Log z$ for $z = -5, i, 1+i, 1-i$ and $-3-i\sqrt 3$ .
\begin{align*} & \Log (-5) = \ln |-5|+ i \Arg (-5) = \ln 5 + i\pi \\ & \Log (i) = \ln |i|+ i \Arg (i) = \ln 1 + i\pi /2 = i\pi /2\\ & \Log (1+i) = \ln |1+i|+ i \Arg (1+i) = \ln \sqrt 2 + i\pi /4\\ & \Log (1-i) = \ln |1-i|+ i \Arg (1-i) = \ln \sqrt 2 - i\pi /4\\ & \Log (-3-i\sqrt 3) = \ln |-3-i\sqrt 3|+ i \Arg (-3-i\sqrt 3) = \ln 2\sqrt 3 + i7\pi /6 \end{align*}

(problem 2) Find $\Log z$ for $z = -2, -i, -2-2i, -4+4i$ and $1+i\sqrt 3$ .
\begin{align*} & \Log (-2) = \answer{\ln 2 + i\pi }\\ & \Log (-i) = \answer{-i\pi /2}\\ & \Log (-2-2i) = \answer{\frac 32 \ln 2 - i3\pi /4}\\ & \Log (-4+4i) = \answer{\frac 52 \ln 2 +i3\pi /4}\\ & \Log (1+i\sqrt 3) = \answer{\ln 2+ i\pi /3} \end{align*}

Here is a video solution of one part of problem 2:

The properties of the multivalued complex logarithm do not carry over to the principal branch without modification.

For $z_1, z_2 \neq 0$ $\Log (z_1\cdot z_2) = \Log z_1 + \Log z_2$ if and only if $-\pi < \Arg z_1 + \Arg z_2 \leq \pi$

and for $z \neq 0$ $\Log \left (\frac {1}{z}\right ) = - \Log z$ if and only if $z$ is not a negative real number.

Under what conditions can we expect $\Log \left (\frac {z_1}{z_2}\right ) = \Log z_1 - \Log z_2?$

3 Other Branches

Let $\alpha \in \R$ and consider the interval $(\alpha , \alpha + 2\pi ]$ . For any non-zero $z \in \C$ , there is an element of the set $\arg z$ which belongs to this interval. Hence, we can define a branch of the complex logarithm as follows.

Let $\alpha \in \R$ and let $\theta _\alpha$ be an argument of $z$ in the interval $(\alpha , \alpha + 2\pi ]$ . Then we define $\log _\alpha z = \ln |z| +i\theta _\alpha$

Example 3 Find the indicated logarithm. The expression $\arg _\alpha (z)$ is used below to indicate the element of the set $\arg z$ which lies in the interval $(\alpha , \alpha + 2\pi ]$ .
\begin{align*} i) \, & \log _0(5) = \ln |5|+ i \arg _0(5) = \ln 5 + i2\pi \\ ii) \, & \log _\pi (i) = \ln |i|+ i \arg _\pi (i) = \ln 1 + i5\pi /2 = i5\pi /2\\ iii) \, & \log _{2\pi }(-2) = \ln |-2|+ i \arg _{2\pi }(-2) = \ln 2 + i3\pi \\ iv) \, & \log _0(-1-i)= \ln |-1-i|+ i \arg _0(-1-i) = \ln \sqrt 2 + i5\pi /4\\ v) \, & \log _{4\pi }(\sqrt 3 + i) = \ln |\sqrt 3 + i|+ i \arg _{4\pi }(\sqrt 3+i) = \ln 2 + i25\pi /6 \end{align*}

(problem 3) Find the following complex logarithms:
\begin{align*} i) \, & \log _0(3) = \answer{\ln 3 + i2\pi }\\ ii) \, & \log _\pi (-3) = \answer{\ln 3 + i3\pi }\\ iii) \, & \log _{3\pi }(-i) = \answer{ i7\pi /2}\\ iv) \, & \log _{2\pi }(-1+i) = \answer{\ln \sqrt 2 +i11\pi /4}\\ v) \, & \log _0(1- i\sqrt 3) = \answer{\ln 2+ i5\pi /3} \end{align*}

Here is a video solution of problem 3, part ii:

4 Complex Powers

We define power functions with complex powers. To do so, recall that the real functions $e^x$ and $\ln x$ are inverses, so that for $x>0$ , we can write $x = e^{\ln x}$ Furthermore, a property of logarithms allows us to write $\ln (x^n) = n \ln x$ Combining these gives $x^n = e^{n\ln x}$

For $c \in \C$ and $z \neq 0$ , we define $z^c = e^{c \log z}$ and the principal branch of the complex power function is given by $z^c = e^{c \Log z}$

example 4 Find all values of $i^i$ and highlight the principal value.
Using the definition of a complex exponent, we have $i^i = e^{i\log i} = e^{i(\pi /2 + 2k\pi )i}$ $= e^{-(\pi /2 + 2k\pi )}$ The principal value occurs when $k=0$ . It is $i^i = e^{-\pi /2} \approx 0.20788...$ Note that the values of $i^i$ are real!

(problem 4) Find all values of each of the exponential expressions and highlight their principal values.
\begin{align*} i) &\; 4^i; \;\; \mbox{Principal value:}\;\; \answer{e^{i\ln 4}}\\ ii) & \;(1+i)^{\pi i}; \;\;\mbox{Principal value:}\;\; \answer{e^{-\pi ^2/4 + \pi i \ln (\sqrt 2)}}\\ iii) & \;(-1)^{\sqrt 2}; \;\;\mbox{Principal value:}\;\; \answer{e^{i\pi \sqrt 2}} \end{align*}

Here is a video solution of problem 4, part iii:

2024-09-27 14:05:53