Chaitanya's Random Pages

February 9, 2014

The validity of the index laws for complex numbers

Filed under: mathematics — ckrao @ 11:47 am

In an earlier post we saw that some of the index laws fail when the base is negative or zero. Now we shall see what occurs when the allowed values of base and exponent are extended to be complex numbers.

For z_1 \in \mathbb{C}\backslash \{ 0\} we can proceed analogously to the real case and define

\displaystyle z_1^{z_2} = \exp(z_2 \log (z_1)).\quad \quad(1)

However what are the exponential and logarithm of a complex number? Last time we defined the logarithm before the exponential and we can do the same here. For complex z we can define \log(z) as \int_1^z \frac{1}{t}\ \text{d}t as for the real case, but note that this time it is a contour integral along any path from 1 to z not including the origin. By allowing the path to wind around the origin any number of times, we find the function is no longer single-valued. For example if we choose the contour to be the unit circle from +1 to itself in an anticlockwise direction, then the contour may be parametrised by t = \cos \theta + i \sin \theta for \theta \in [0, 2\pi] (\text{d}t = (-\sin \theta + i \cos \theta)\text{d}\theta) and we have

\begin{aligned} \log(1) &= \int_1^1 \frac{1}{t}\ \text{d}t\\ &= \int_0^{2\pi} (\cos \theta - i \sin \theta)(-\sin \theta + i \cos \theta)\text{d}\theta\\ &= \int_0^{2\pi} i( \cos^2 \theta + \sin^2 \theta )\ \text{d}\theta \\ &= 2\pi i. \end{aligned}

More generally if w is a logarithm of z, so is w + 2\pi i n where n is an integer. Defining \exp(w) for complex w to be z where w is a logarithm of z, we have

z = \exp(w + 2\pi i n) = \exp(w),\quad n \in \mathbb{Z}.

(Note that \displaystyle \exp(\log (z)) = z while \displaystyle \log(\exp (z)) = z + 2\pi i k.)

Based on these definitions we can show that \log(z_1z_2) = \log(z_1) + \log(z_2) and \exp(z_1 + z_2) = \exp(z_1)\exp(z_2) in an analogous way to the real case, except that the first equation is to be interpreted as an equality of sets of values rather than individual values. Note that for this reason we have to be careful when adding or subtracting logarithms: for example for complex numbers,

\log(z) - \log(z) = \log(1) = 2\pi i n \neq 0

and

\begin{aligned} \log(z^2) &= \log (z) + \log (z)\\ &= (\log (z) + 2\pi i m) + (\log (z) + 2\pi i n)\\ &= 2\log (z) + 2\pi i k, \quad k \in \mathbb{Z}\\ \text{while }\quad 2 \log (z) &= 2 (\log (z) + 2\pi i n)\\ &= 2\log (z) + 4 \pi i n, \quad n \in \mathbb{Z}. \end{aligned}

Hence we cannot write a\log (z) + b \log (z) = (a+b)\log (z) without paying special attention to the values of a,b,z. If we want to know when \log (z^c) = c \log (z), we can verify that \log (z^c) = c \log (z) + 2\pi i k, k \in \mathbb{Z} which is only equal to c \log (z) = c (\log (z) + 2\pi i m) when cm covers all integers, i.e. if c = 1/n for some non-zero integer n:

\displaystyle \log\left(z^{1/n}\right)= \frac{1}{n} \log(z), \quad\quad n \in \mathbb{Z}.

Now in general,

\begin{aligned} z_1^{z_2} &= \exp(z_2 \log (z_1))\\ &= \exp(z_2 (\log (z_1) + 2\pi i n) )\\ &= \exp(z_2 \log (z_1))\exp(2\pi i n z_2).\quad\quad (2)\end{aligned}

This will be multi-valued if nz_2 takes on non-integer values, as n varies over the integers. It will only be single-valued if z_2 is an integer. For example treated as a complex power, 2^{1/2} will have two values: \sqrt{2} and -\sqrt{2} while 2^{1/3} will take three values. The number 2^{\sqrt 2} will have infinitely many complex values \exp(\sqrt{2} \log (2))\exp(2\pi i n \sqrt{2}), n \in \mathbb{Z} although only one of them is real-valued. Note that through (2) we can work out quantities such as:

  • (-1)^{\sqrt{2}} = \exp(i \pi \sqrt{2})\exp(2\sqrt{2} \pi i n), n \in \mathbb{Z} (infinitely many non-real values!)
  • i^i = \exp(-\pi/2)\exp(-2\pi n), n\in \mathbb{Z} (infinitely many real values!).

Also note from (2) that

\log(z_1^{z_2}) = z_2 \log (z_1) + 2\pi i k = z_2 \textrm{Log} (z_1) + z_2 2\pi i k_1 + 2\pi i k_2.

One can define the principal value of the logarithm \textrm{Log}(z) to be that with imaginary part in the interval (-\pi, \pi]. Similarly one can define the principal value of the power function as

\displaystyle z_1^{z_2} := \exp(z_2 \textrm{Log} (z_1)).\quad \quad (3)

This gives single-valued results but they may not be as expected. For example, since \textrm{Log}(-1) = i \pi, (-1)^{1/3} = \exp(i \pi/3) rather than the real-valued root -1. However we can now say a \textrm{Log}(z) + b \textrm{Log}(z) = (a+b)\textrm{Log}(z), being single-valued.

We would like to know which of the index laws hold. In the remainder of the post we verify the identities summarised in the following table. The real number case was already treated in this post.

Real numbers a,b Complex numbers z,z_1,z_2,a,b
Positive real x,y

Multiple-valued power

z_1^{z_2} = \exp(z_2 \log z_1)

Single-valued power

z_1^{z_2} = \exp(z_2 \textrm{Log} z_1)

(1)  x^a x^b = x^{a+b}  z^{a+b} a subset of  z^a z^b  z^a z^b = z^{a+b}
(2)  (x^a)^b = x^{ab}  z^{ab} a subset of (z^a)^b  (z^a)^b = z^{ab}\exp(2\pi i b n_0)
(3)  (xy)^a = x^a y^a  (z_1 z_2)^a = z_1^a z_2^a  (z_1 z_2)^a = z_1^a z_2^a \exp(2 \pi i a n_{+})
 (4)  x^0 = 1  z^0 = 1   z^0 = 1
 (5)  1^a = 1  1^a = \exp(a 2\pi i k)   1^a = 1
 (6)  x^{-a} = 1/x^a  z^{-a} = 1/z^a but z^{-a}z^a = \exp(2\pi i k)  z^{-a} = 1/z^a and  z^{-a}z^a = 1
 (7)  x^a / x^b = x^{a-b}  z^{a-b} a subset of z^{a}/z^{b}   z^a / z^b = z^{a-b}
 (8)  (x/y)^a = x^a/y^a  (z_1/z_2)^a = z_1^a/z_2^a  (z_1/z_2)^a = \frac{z_1^a}{z_2^a}\exp(2\pi i a n_{-})

Note that in the table, n_0, n_{+},n_{-} are particular integers chosen to enable equality.

For verifying identity (1) in the multi-valued power case we have

\begin{aligned} z^{a+b} &= \exp((a+b)\log (z))\\ &= \exp\left((a+b)(\textrm{Log} (z) + 2\pi i k)\right)\\ &= \exp\left((a+b)\textrm{Log} (z) \right) \exp\left( 2\pi i k (a+b)\right) \end{aligned}

while

\begin{aligned}z^a z^b &= \exp(a \log(z)) \exp(b \log(z))\\&= \exp\left(a (\textrm{Log} z + 2\pi ik) \right)\exp\left(b (\textrm{Log} z + 2\pi in) \right)\\ &= \exp\left((a+b)\textrm{Log} (z) \right) \exp\left(2\pi i (ka + nb) \right).\end{aligned}

This shows that the set of values of z^{a+b} is a subset of the set of values of z^a z^b. In the single-valued case,

\begin{aligned} z^az^b &= \exp(a \textrm{Log} (z)) \exp(b \textrm{Log} (z))\\ &= \exp(a \textrm{Log} (z) + b \textrm{Log} (z))\\&= \exp((a+b) \textrm{Log} (z))\\ &= z^{a+b}.\end{aligned}

For identity (2) in the multi-valued power case we have

\begin{aligned} (z^a)^b &=(\exp(a \log (z)))^b\\ &= \exp(b \log(\exp(a \log( z))))\\ &= \exp(b(a \log (z)+2 \pi ik))\\ &= \exp(ba \log (z)) \exp(2 \pi ibk)\\ &= z^{ab}\exp(2\pi ibk). \end{aligned}

This shows that the set of values of z^{ab} is a subset of the values of (z^a)^b. We have the equality (z^a)^b = z^{ab} if \exp(2\pi ibk) =1, or bk \in \mathbb{Z} for all k, which is true if b \in \mathbb{Z}. In the single-valued case, the integer k is chosen so that a \textrm{Log}(z)+2 \pi ik has imaginary part in the interval (-\pi, \pi].

For identity (3) in the multi-valued power case we have

\begin{aligned} (z_1 z_2)^a &= \exp(a \log(z_1 z_2))\\ &= \exp(a(\log (z_1)+\log (z_2)))\\ &= \exp(a \log (z_1))\exp(a \log (z_2))\\ &= z_1^a z_2^a.\end{aligned}

In the single-valued power case,

\begin{aligned} (z_1 z_2)^a &= \exp(a \textrm{Log}(z_1 z_2))\\ &= \exp(a(\textrm{Log} (z_1)+\textrm{Log} (z_2) + 2\pi i a n_{+}))\\ &= \exp(a \textrm{Log} (z_1))\exp(a \textrm{Log} (z_2))\\ &= z_1^a z_2^a \exp(2 \pi i an_{+}). \end{aligned}

Here the value n_{+} is chosen so that \textrm{Log} (z_1)+\textrm{Log} (z_2) + 2\pi i a n_{+} has imaginary part in the interval (-\pi, \pi].

Identity (4) comes from setting b=0 in identity (1). Identity (5) results from setting y or z_2 to 1 in identity (3). Identity (6) results from setting b = -a in identity (1) and using identity (4). Finally identities (7) and (8) follow from identities (1) and (3).

The moral of all this is that care is to be taken when applying the index laws to complex numbers (or indeed even when adding logarithms) by virtue of the multi-valued nature of the complex logarithm.

Reference

H. Haber, The complex logarithm, exponential and power functions, UC Santa Cruz Physics 116A notes (2011) available at scipp.ucsc.edu/~haber/ph116A/clog_11.pdf

Advertisements

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: