Abstract
This article will build on the hints about treating the complex numbers as a branched surface, briefly described and pictured in section 4.2 of Using a particular set of conventions, all the problems described in can be removed, and the rules described there as applying only to reals generalized to complex numbers. A better approach will be given to the problem of log being multi-valued. While there is obviously nothing new here, a particular choice of conventions will be argued to provide clean packaging of complex arithmetic and elementary functions.
Definitions
It is assumed that the reader has read the two links above, and has a general understanding of complex numbers. Some notations are introduced which are non-standard, but make this presentation easier.
It will be crucial to distinguish complex numbers considered as a branched surface from complex numbers considered as a plane of ##a+b\mathbf {i}## values. Functions and operations will be explicitly defined as operating on one or the other of these. I will call the branched surface Z, versus C for the simple complex plane. I define the projection operator pr(z) mapping points of Z to points of C. This function is obviously single valued but not bijective. I will exclusively use ##(r,\theta)## to represent a point of Z, shorthand for ## r e^{\mathbf {i} \theta}##. For C, I will always use ##a+b\mathbf {i}## notation. Then ##pr(r,\theta) = r \cos(\theta)+ r \sin(\theta)\mathbf {i}##. Equality in Z in does not imply any actions of the projection operator, so ##(r,\theta)## and## (r,\theta+2\pi)## are not equal. Within Z, the relation between these two points would be expressed as ##pr((r,\theta)) = pr((r,\theta+2\pi))##. As is commonly done, ##Re(z##) returns real part of z. When applied to Z, it implicitly means ##Re(pr(z)##). Similarly ##Im(z)## for the imaginary part.These functions are ‘naturally’ functions from C to R. For values of Z, ##Arg(z)=\theta##. Within Z, zero is a special value which will be represented simply as 0. ##Arg(0)## is undefined. In places where many authors distinguish the principle branch of Z, I will speak of C instead. Except in a few passing references, no use will be made of the notion of a principle branch.
Arithmetic
Addition is naturally an operation on C: ##a+b\mathbf{i}~~+~~ c+d\mathbf{i} = (a+c)+(b+d)\mathbf{i}##. But we may say ##z_1+z_2## for values of Z, implicitly performing ##pr(z_1)+pr(z_2)##. In this usage, addition maps values of Z to C. Similarly for subtraction. No definition is given for addition or subtraction operating Z to Z.
Multiplication in Z and C are different. ##(r_1,\theta_1) * (r_2,\theta_2) = (r_1 r_2,~\theta_1+\theta_2)##. In C: ##(a+b\mathbf{i})*(c+d\mathbf{i}) = (ac-bd)+\mathbf{i}(bc+ad)##. Obviously, however, ##pr(z_1*z_2) = pr(z_1)*pr(z_2)##. For division, ##(r_1/r_2,~\theta_1-\theta_2) ##and ##[(ac+bd)+\mathbf{i}(bc-ad)]/(c^2+d^2)##. Again, ##pr(z_1/z_2) = pr(z_1)/pr(z_2)## holds true. Thus we distinguish multiplication and division operating Z to Z from these operations operating C to C.
Exponentiation
Exponentiation will be defined as operating from C to Z. It is, in fact, a bijection between C and Z-0. We define:
##\exp(a+b\mathbf{i})= (e^a,b)##
The shorthand ##\exp(z)## implicitly means ##\exp(pr(z))##.
We now generalize to any base in Z. This is defined as a mapping from Z and C to Z, that is, Z^C in Z. It is strictly single valued. It happens that for some exponents, applied to the set of all z with the same projection, the set of results has only a finite number of projected values, while for others it is infinite (e.g. rational real exponent versus irrational real exponent). But thus has nothing to do with anything being multivalued.
Given ##z=(r,\theta)## not 0, ## c=a+b\mathbf{i}##, we define ##z^c = (r^a e^{-b\theta}, a \theta + b \log r)##. This is motivated by writing ##(r,\theta)## as ##e^{(\log r + \mathbf{i} \theta)}## and applying the power ##a+b\mathbf{i}## using normal rules. However, this is taken as definition. For 0 , the only powers allowed are ##a+0\mathbf{i}##, with ##a >0##, and the result is 0. Anything else is undefined for 0. As usual, writing ##z_1^{z_2}## is a shorthand for ##z_1^{pr(z_2)}##. It is crucial to keep this in mind.
It is easy to see that given this definition, ##exp(c)## is the same as ##(e,0)^c##. Note that, e.g. ##(e,2\pi)^c## produces a completely different value in Z.
Note that for a given exponent, and the infinite set of values of Z with the same projection, the set of distinct projections of the results will normally be infinite. However, for a rational real exponent, this latter set will be finite. For an integer exponent, all the infinite distinct results in Z will have the same projection. However, in all cases, the infinite set of z with the same projection, maps (with a given exponent in C), to an infinite set of distinct results in Z.
Laws for exponents
We can now state precise generalizations of laws that hold for reals. All 3 rules presented in have natural generalizations. We will also apply these results to the 4 specific problems presented in this reference, showing how each is resolved without problem with our conventions.
Rule 1
For ##z_1## not 0, $${\large (z_1^{z_2})^{z_3} = z_1^{z_2 z_3} = z_1 ^ {pr(z_2) pr(z_3)}}$$
The result is in Z. Note, this actually holds for 0 if ##z_2## and ##z_3## are positive real numbers. Or, you can say, whenever the expressions are defined, it is true. It is easy to see this follows from the definitions above.
Rule 2
For ##z_1## and ##z_2## not zero, $$\large{(z_1 z_2)^{z_3} = z_1^{z_3} z_2^{z_3} = (z_1 z_2)^{pr(z_3)} = z_1^{pr(z_3)} z_2^{pr(z_3)}}$$
The result is in Z. As above, you can include zero if you say “whenever the expressions are defined”. Again, it is straightforward to demonstrate this from the definitions.
Rule 3
For z not 0 or (1,0), $$\large{z^{c_1} = z^{c_2} \Leftrightarrow c_1 = c_2}$$ Obviously also, $$\large{z^{z_1} = z^{z_2} \Leftrightarrow pr(z_1) = pr(z_2)}$$
For the right handed implication, this requires some work to show from the definitions, but it is true and related to the fact that over C: ##c_1 c_2 = c_1 c_3 \Rightarrow c_2 = c_3##, as long as c1 is not zero.
Application to Examples
Example A
The first problem example from can be presented as the following manipulations, whose last step could never happen over positive reals (but seems reasonable in the presented example).
$$x = y^2 = \sqrt{y^2}\sqrt{y^2} = \sqrt{y^2 y^2} = \sqrt{y^4} = -x$$
The problem is all in the last step. The laws presented above state there should be no problem if the operations are done correctly on their domains. Specifically, maintaining ##arg(z)## information resolves the apparent problem. For ##pr(z)=-1##, the possible elements of Z are ##(1, (2k+1)\pi)##. There are only two cases distinguishable via projection, exemplified by ##(1,\pi)## and ##(1,3\pi)##. Choosing ##(1,\pi)## gives:
$$(1,\pi) = (1,\pi/2)^2 = \sqrt{(1,\pi)}\sqrt{(1,\pi)} = \sqrt{(1,\pi)(1,\pi)} = \sqrt{(1,2\pi)}= (1,\pi)$$
or, giving projections of each step:
$$-1=\mathbf{i}^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=-1$$
In short, maintaining arg information, relying on exponentiation being a bijection, we are forced to pick the correct answer in the last step, leading to no anomaly.
Following through with ##(1,3\pi)## gives:
$$(1,3\pi) = (1,3\pi/2)^2 = \sqrt{(1,3\pi)}\sqrt{(1,3\pi)} = \sqrt{(1,3\pi)(1,3\pi)} = \sqrt{(1,6\pi)}= (1,3\pi)$$
or, in projections (adding an extra step for clarity):
$$-1 = (-\mathbf{i})^2 = (-\sqrt{-1})(-\sqrt{-1}) =\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = -1$$
again, totally consistent once you are forced to pick the correct square root result in the last step.
Example B
For this example, any choice in Z for i works out the same, projectively, so just choose ##(1,\pi/2)##. Again, the whole problem is in the last step, which needs arg information to avoid ambiguous wrong choice:
$$(1,\pi/2) = (1,\pi)^{1/2} = (1/\pi)^{2/4} = ((1,\pi)^2)^{1/4} = (1,2\pi)^{1/4} = (1,\pi/2)$$
or projectively:
$$\mathbf{i} = (-1)^{1/2} = (-1)^{2/4} = ((-1)^2)^{1/4} = 1 ^ {1/4} = \mathbf{i}$$
Obviously, any of ##\{1,-1,\mathbf{i},-\mathbf{i}\}## raised to the 4th power produces 1. Carrying arg information forces the correct choice in the last step.
Example C
Ignoring log, which I’ll discuss later, the argument of this example is ##exp(2\pi\mathbf{i})=1##,## exp(0) = 1##, therefore ##2\pi\mathbf{i}=0##. To apply this rule correctly, we note that exp() maps C to Z, so ##exp(2\pi\mathbf{i})=(1,2\pi)## While ##exp(0) = (1,0)##. But these are different in Z, so the rule 3 stated above simply does not apply.
Example D
Here we have ##exp(1+2\pi\mathbf{i})=(e,2\pi)##. Then, we call this just e, and expect that raising to ##(1+2\pi\mathbf{i})## should again be e. It is claimed in the discussion that this shows that ##(x^a)^b = x^{(ab)}## is flawed for complex numbers. However, general exponentiation maps from Z and C to Z, and exp() is equivalent to exponentiation of (e,0). So, the correct check on this law is:
##exp(1+2\pi\mathbf{i}) = (e,2\pi)##, then ##(e,2\pi)^{(1+2\pi\mathbf{i})}= (e^{(1-4\pi^2)}, 4\pi )##
compared to ## exp(1 – 4\pi^2+4\pi\mathbf{i}) = (e^{(1-4\pi^2)}, 4\pi)##
which match. The real flaw in this claimed problem is the erroneous expectation that if ##pr(z_1) = pr(z_2)##, then ##(z_1)^a = (z_2)^a##. This is just wrong.
A recent youtube example
A youtube video gave an incomplete (but mostly correct) solution to the following initially strange problem:
Find x such that ##1^x = 4##.
Obviously, over the reals, it is nonsense. Further, if 1 is taken to be (1,0) it also has no solutions: ##(1,0)^{(a+bi)} = 1##, period. However, if you allow for ##(1,2k\pi)##, k not 0, then there are solutions. There are even more solutions if 4 is treated as ##(4,2m\pi)##, here no need to disallow m=0. Thus, the most general statement would be:
##(z_1)^x = z_2##, such that ##pr(z_1)=1##, and ##pr(z_2)=4##.
We have:
$$(1,2k\pi)^{a+b\mathbf{i}} = (e^{-2bk\pi},2ak\pi) = (4,2m\pi)$$
giving ##a = m/k##, ##b = -log(2)/k\pi##
so the solutions are: ##(m/k) – (log(2)/k\pi)\mathbf{i}##
again verifying that k=0 is no good. If you want to require that 4 be treated as (4,0), then the real part above becomes zero. You can describe this by saying the solution maps 1 on the kth branch of Z to 4 on the mth branch of Z.
Logarithms
Exponentiation was treated as single valued, with no loss of generality and better behavior (e.g. allowing generalizations of all the common laws of exponents) by considering as operating from Z and C to Z. This is in contrast to treating it as operating on C, whence you are led to the multi-value complications.
A similar approach is taken for logarithms. Note, we have stated that exp(c) is a bijection from C to Z-0. Thus, it has a bijective inverse from Z-0 to C. This is the logarithm. Specifically:
$$log(r,\theta) = log(r) + \theta\mathbf{i}\tag{1}$$
What happened to the other values? They simply come from solving a projective equation:
$$pr(e^w) = pr(z)$$
which implies solutions of form:
$$e^w = (r,\theta+2k\pi) \tag{2}$$
from which, with our log as a simple function, you get:
$$w = log(r) + (\theta+2k\pi)\mathbf{i}\tag{3}$$
For each k, this is the value of w in C which maps (e,0) to the kth branch of ##(r,\theta)## with ##\theta \in (-\pi,\pi]##. As will be seen later, this is not really the complete solution set for the equation ##e^w=z##. Also, see the last of the examples in the prior section.
Note, a mechanical trick for writing a projective equation as a series of equations on Z is to replace each projection with multiplication by ##(1,2k\pi)##: ##e^w (1,2k\pi) = z (1,2m\pi)##. Then, division by ##(1,2k\pi)## shows that (2) above is sufficiently general.
There is no need to introduce a multi-valued log to solve such equations – just as we kept exponentiation single valued in all cases. However, it is worth noting that a better notion than a multi-valued log, if you must, is treat each choice of k in (3) above as defining a different bijective function from Z-0 to C. Thus a family of functions rather than a multi-valued function. This allows us to see why the standard principal value approach loses most nice properties of logs. Consider, that for each function in this family, each branch of Z is mapped to horizontal strip of C. Further, for each k, a given branch of Z is mapped to a different strip in C. The notion of making log single valued by choosing k so that a given z is mapped to an Im() value in ##(-\pi,\pi]##, then applies a different function to each branch of Z. So instead of a family of well behaved functions on Z, we create a function on Z with a different mapping for each branch. Critically, this new function, while single valued, is not bijective. This is what breaks all nice properties of log. I consider all this pointless, and the simple log defined in (1) suffices for all purposes.
A little more insight into the principle value log arises from treating it as function from C to C (for this purpose, I will call it logc):
$$logc(a+bi) = .5 log(a^2+b^2) + ang(a,b)\mathbf{i}$$
where ang returns the angle ##\theta## such that ##\sqrt{a^2+b^2}\cos(\theta)=a## and ##\sqrt{a^2+b^2}\sin(\theta)=b## such that ##\theta## in ##(-\pi,\pi]##. Obviously, it is Arctan(b/a) with some quadrant adjustments and edge cases handled. Obviously, ang(0) is undefined. Thus defined, logc is a bijective function from C to a horizontal strip of C. With our conventions, logc(z) would implicitly be logc(pr(z)). However, even the most basic properties of log fail for logc:
$$logc (c_1 c_2) \neq logc (c_1) + logc(c_2)$$
Specifically, it is true that ##Re(logc (c_1 c_2) ) = Re(logc (c_1)) + Re(logc(c_2))##. However, this fails for the imaginary parts, precisely due to forcing the imaginary part to ##(-\pi,\pi]##.
In contrast, with our definition of log (from Z to C, inverse of exp), we have;
$$log (z_1 z_2) = log (z_1) +log(z_2)$$
$$log(z_1^{z_2}) = pr(z_2)log(z_1)$$
$$exp(log(z)) = z$$
$$log(exp(z)) = pr(z)$$
We can further define:
$${\large log_{z_1} (z_2) = log(z_2)/log(z_1)}$$
as a well behaved single valued function from Z and Z to C. It has the expected properties with respect to log of products and log of powers. Also:
$${\large log_{z_1} (z_1^{z_2}) = pr(z_2)}$$
$${\large z_1^{log_{z_1}(z_2)} = z_2}$$
Addendum: More General Logarithmic equation
Though equation (3), at the beginning of this section, is often presented as the general solution to: ##e^w=z## (see ) , this is not actually correct. Consider, in common multivalue parlance, that one might say ##e^.5=\pm\sqrt{e}##. One might then expect that for ##z=-\sqrt{e}##, equation (3) would include .5 as a solution. Noting that the principle branch Z value for ##-\sqrt{e}## is ##(\sqrt{e},\pi)##, equation(3) only gives solutions of the form ##.5+(2k+1)\pi\mathbf{i}##, which does not include .5. The problem is we have allowed free branch choice on z but not on e. The more general family of solutions is given by:
$$(e,2m\pi)^w=(r,\theta+2k\pi)$$
Again, using our simple, single value log, this parameterized equation is readily solved:
$$w=\frac {\log(r)+(\theta+2k\pi)\mathbf{i}} {1+2m\pi\mathbf{i}} $$
Then, for k=0, m=1, ##(r,\theta)=(\sqrt{e},\pi)##, you, indeed get .5 as a possible value for w. If you are really want all solutions of ##e^w=z## to be your ‘multivalued log’, this is the formula that should be used. For each choice (m,k) this is the value of ##w\in C## which maps e on branch m to branch k of a given principle branch ##(r,\theta)##.
Similarly, the general solution for w for:
$$(z_1)^w=z_2$$
allowing all branch choices for ##z_1## and ##z_2## is:
$$w=\frac{log(r_2)+(\theta_2+2k\pi)\mathbf{i}}{log(r_1)+(\theta_1+2m\pi)\mathbf{i}}$$
This would be the maximally multi-valued definition of ##\log_{z_1}(z_2)##.
Hyperbolic and Trig functions
Given the definition of exp(z), the hyperbolic functions (sinh, cosh, etc.) immediately generalize to complex numbers. However, the addition operations in their definitions, combined with the action of exp (it projects its argument), means that the hyperbolic functions are single valued functions mapping C to C. Writing sinh(z) is shorthand for sinh(pr(z)). An equation
$$\sinh(z) =
must be taken as ##sinh(z) = pr (
Given that exp over complex numbers has its usual properties, the standard hyperbolic trig identities remain true over complex numbers.
There are different approaches to generalizing trig functions (sin, cos, etc.) to complex numbers. I will simply define them from the hyperbolic functions, as follows (justifications can be found in any text on complex numbers):
$$\sin(z) = -\mathbf{i}\sinh(\mathbf{i}z)$$
$$\cos(z) = \cosh(\mathbf{i}z)$$
This immediately implies trig functions are also mappings from C to C. An explicit definition useful for direct computation (that can be derived from the definitions above) is:
$$\sin(a+b\mathbf{i}) = \cosh(b)\sin(a) + \mathbf{i}\sinh(b)\cos(a)$$
$$\cos(a+b\mathbf{i}) = \cosh(b)\cos(a) – \mathbf{i}\sinh(b)\sin(a)$$
From the definition in terms of hyperbolic functions, most standard trig identities follow from those for hyperbolic functions. For example:
$$\sin^2(z) + \cos^2(z) = -sinh^2(\mathbf{i}z) + cosh^2(\mathbf{i}z) = 1$$
Other identities (e.g. periodicity in terms of real part of argument) follow from the explicit formulas in terms of a+bi.
Conclusion
While the definitions and conventions presented here are a bit nonstandard, it is hoped that the reader can see the advantages gained from treating all functions as single valued, on either the branched complex surface or the ‘projected’ simple complex plane. Essentially all laws for reals generalize, and many apparent anomalies are resolved.
Appendix
I thought I would go over another problem making the rounds. This one has little to do with exponentiation or logs, but it does illustrate branch choice considerations. Consider solving the following two equations:
$$\sqrt{z}+\sqrt{-z}=36$$
$$\sqrt{z}-\sqrt{-z}=36$$
One assumption would be to allow z and -z to be from any branches, i.e. interpret the equations as:
##\sqrt{z_1}+\sqrt{z_2}=36## such that ##pr(z_1)=-pr(z_2)##
##\sqrt{z_1}-\sqrt{z_2}=36## such that ##pr(z_1)=-pr(z_2)##
(In ‘mutlivalue’ language, this means you can pick ‘any’ square root for z or -z, independently).
Another assumption would be to require that the square roots are principle square roots. We will solve the more general case first. The answer for the restrictive assumption then trivially follows.
We have ##z_1=(r,\theta)## and ##z_2=(r,\theta+(2k+1)\pi)##. The square roots are ##(\sqrt{r},\theta/2)## and ##(\sqrt{r},\theta/2+\pi/2+k\pi)##. The imaginary parts must cancel. This means, for both equations, the following must be true (because the multiples of ##\pi## don’t change absolute value of imaginary part):
$$|\sin(\theta/2)|=|\sin(\theta/2+\pi/2)|$$
The only way for this to be true is for ##\theta/2=\pi/4+m\pi/2##. However, we also want the real part to be positive (for either the + problem or the – problem). This means we must have ##\theta/2=2m\pi\pm\pi/4##. For the addition problem, the imaginary parts must be opposite signs, real parts same sign. This requires (after some thought), for ##\sqrt{z_1}## and ##\sqrt{z_2}##:
$$(\sqrt{r},2m\pi+\pi/4),(\sqrt{r},2k\pi-\pi/4)$$
or the reverse (i.e. the first is ##\sqrt{z_2},##, second ##\sqrt{z_1}##).
For the subtraction problem, the imaginary parts must be same sign, real parts opposite sign. This leads to either of the following possibilities:
##(\sqrt{r},2m\pi+\pi/4),(\sqrt{r},2k\pi+3\pi/4)## or ##(\sqrt{r},2m\pi-\pi/4),(\sqrt{r},2k\pi-3\pi/4)##
Ignoring r for the moment, the projections of the square roots for the addition problem are:
##\sqrt{2}/2+\mathbf{i}\sqrt{2}/2## and ##\sqrt{2}/2-\mathbf{i}\sqrt{2}/2##
For the subtraction problem, there are two distinct cases, projectively:
##\sqrt{2}/2+\mathbf{i}\sqrt{2}/2## and ##-\sqrt{2}/2+\mathbf{i}\sqrt{2}/2##, or
##\sqrt{2}/2-\mathbf{i}\sqrt{2}/2## and ##-\sqrt{2}/2-\mathbf{i}\sqrt{2}/2##.
For r, we must (from above) require that ##\sqrt{2r}=36##, thus ##r=648##. So, finally, for the addition problem we have:
##a=(648,4m\pi+\pi/2)##, ##-a=(648,4k\pi-\pi/2)##, or the reverse. In projective terms, ##\pm648\mathbf{i}##.
For the subtraction problem we have two cases:
##a=(648,4m\pi+\pi/2)##,##-a=(648,4k\pi+3\pi/2)## and ##a=(648,4m\pi-\pi/2)##,##-a=(648,4k\pi-3\pi/2)##.
Projectively, this is the same as for the addition problem, but we must make different branch choices (ignoring r) for i or -i, for the two different problems.
For the subtraction problem, we see that it is impossible for both a and -a to be on the principle branch, while this is possible for the addition problem.Thus, only the addition problem has a solution over principle square roots.
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