Exercise 5.5

     Penrose is giving us the following definition for e^z, where z is a complex number:

And says that we must use (1) to show the following:

e^a+b=e^ae^b

(2)

      In order to do this, we have to apply (1) to both sides of (2) and show that they give equivalent quantities.  Let's first consider e^a+b, which according to (1) expands as follows:

    This can be expanded as follows for the first few terms:

Which suggests a general pattern for the expansion up to the nth term, and continuing onwards (using Penrose's hint for the coefficient of the nth term of the expansion of (a+b)ⁿ:

Which in turn can be written more concisely as:

Now we look at the right hand side of (2).  Using (1):

Which expands as:

Comparing (6) and (7) we see that the overall equality boils down to whether:

The easiest way I've found to show (8) is to re-examine the products on the left hand side in a different way:

This appears to be exactly the operation described by the right hand side of (8), which we derived from the expression for e^(a+b).  In this way we have demonstrated (2).  

Here Goes...

This is as far as I got in Road when I was initially working through it in March and April of last year.  I was stumped and remain stumped by the second part of the challenge Penrose sets in Exercise 5.4, which is to show without detailed calculation or trigonometry that multiplication of complex number z by another complex number w corresponds to two geometric transformations of the representation of z on the complex plane: (a) a uniform expansion/contraction and (b) a non-reflective rotation.  Penrose hints that this is due to two properties of complex numbers z and w:

w(z₁+z₂) = wz₁+wz₂

(1)

w(iz) = i(wz)

(2)

    Penrose also tells us that (1) implies the preservation of linearity in the multiplication of complex numbers; and that (2) implies the preservation of right angles in the multiplication of complex numbers.  Which leaves me confused as to whether we are allowed to use the fact that multiplication of a complex number by i corresponds to a rotation of z through a right angle in the complex plane, or whether we still have to prove that.  My guess is that this is simply "guilty knowledge" that will have to be shown logically, but that ends up sucking for me because while (a) is fairly easy to show in a way that I think is consistent, (b) has proven very difficult to show without making use of trigonometry and/or "detailed calculation," whatever the hell "detailed calculation" actually means. Anyhoo, I'll talk more about this when I get around to talking about possible approaches to showing (b). 

     In general, Penrose seems to want for us to be able to think about what's happening with these complex number in terms of geometry, not algebra, and to use "mostly" geometric arguments to demonstrate these properties of complex number multiplication.  Since I've always been a more rote-calculation person it's a good exercise for my brain to try the harder, or at least different, geometric approach.

Setup

      Just to make sure I know what I'm talking about throughout, here are some basic definitions.  For the rest of the time we'll be talking about two arbitrary complex numbers z and w defined as follows:

z = a+bi

w = c+di

  

      Where a, b, c, and d are all real numbers.  Also we will be talking about representing z and w as vectors on the "complex plane" which is defined by a horizontal "real" axis and a vertical "imaginary" axis, as shown in the following diagram:

    We are also going to need to assume some givens about how these vectors can be "added" geometrically.  For example, z can be represented in the complex plane as the "resultant" vector z of vectors a and b, whose magnitudes correspond to the real numbers a and b respectively.  Finding the resultant vector from component vectors on the complex plane works according to the same parallelogram rule as works for vectors on the real plane:

Uniform Expansion/Contraction

     Show that multiplying z by w involves a uniform expansion/contraction.  Not sure what Penrose means by "uniform, " but I assume that he means an expansion/contraction by a constant factor.  In any case he seems to be asking about how multiplication affects the geometric length of the vector z in the complex plane  Let's represent this length as |z|.  Penrose is asking us to look at |wz|, i.e. the magnitude of the length of the geometric representation of the complex product wz, and see whether it differs from |z| by a constant factor.

      We begin by looking at a bit of algebra (which hopefully doesn't count as the sort of "detailed calculation" which Penrose is disallowing):

|wz| = |z(c+di)| = |cz| + z(di)| = c|z| + id|z|

(3)

     This assumes some things about two complex numbers z and w that seem to me to be true:

|zw| = |z||w|

|z+w| = |z| + |w|

|i| = i

     The first of these facts is the most important for us.  In particular, we need to show in a simpler case that |bz| = b|z| where b is a real number.  

     We can show this quite easily by interpreting bz as z added to itself b times, which can be represented in the complex plane as a simple geometric addition of the vector z to itself b times (b is a scalar value):

     Inspecting this we can see that |bz| = b|z|, which makes me feel better about saying |bz|=b|z|.

   

     Now according to (3) |zw| can be rewritten as c|z| + id|z|.  Since |z| is a scalar value, we can interpret this sum as a third complex number whose geometric interpretation is::

     Which means that the magnitude |z(c+di)| depends on the magnitude |z| multiplied by constant factors c and d.  Given (3), this means that |wz| also depends on these constant factors.  Which means that the multiplication of z by w results in a third complex number, whose magnitude is related to the magnitude of z by a constant factor.  Which is (I think) what Penrose wanted us to prove.

Rotation

      Now we have to show that multiplying z by w also involves a "non-reflective" rotation.  Since (as will be clear shortly) multiplying z by the factor i² = -1 is analogous to a geometric reflection of z through the origin, I guess that Penrose did not mean for us to consider "degenerate" cases such as when w equals i, -1, 1, or 0.  Even so, we will need to start out by considering what happens, geometrically in the complex plane, when you multiply z by the "special" value of i.  Algebraic multiplication by i of z = a+bi results in a new number z' = -b+ai.  A relationship between z and z' becomes evident upon examinations of their analagous vectors z and z' in the complex plane:

      We are most interested in the angle between z and z'.  If you imagine the two vectors defining triangles with one side along the real and imaginary axes respectively you will see that the triangles are similar.  Therefore the angle γ between z and the real axis is equal to the angle between z' and the imaginary axis.  Define β as the angle between z and the imaginary axis.  We know that γ + β equals 90 degrees.  Then note that angle between z and z' is equal to β plus the angle between z' and the imaginary axis.  But this is equal to β+γ as well, therefore the angle between z and z' is equal to 90 degrees.  And so multiplication of the complex number z by i is analogous to a counterclockwise 90-degree rotation of  z in the complex plane.  

      Now consider multiplication of z  by another arbitrary complex number w.  We can represent w by the corresponding vector w on the complex plane.  Suppose that w has components w_R and w_I:

     Examining the above, and using (1), we see that:

zw = z(w_R+w_I) = zw_R+zw_I

     Or algebraically:

zw = zw_R+zw_I

(4)

     However again by examination we know that w_I is nothing more than i multiplied by some real constant C, while w_R is nothing more than a real constant D.  Therefore (4) can be rewritten as:

zw = z(iC)+zD

     Using (2) we slightly rewrite this as:

zw = iCz+Dz

zw = iCz+Dz

(5)

     We have shown that multiplication of a vector such as z by a constant such as C or D results in  a dilation or contraction.  We have also shown that a multiplication of a vector in the complex plane by i results in a 90-degree counterclockwise rotation.  Therefore, the vector product zw is equivalent to combining a counterclockwise 90-degree rotation and dilation/contraction of z with another, not necessarily identical contraction/dilation of z:

     One component of the resultant vector equal to zw will be a stretched out/compressed copy of z rotated ninety degrees, and the other will be a non-rotated copy of z stretched out/compressed by a different amount.  Assuming that w is not one of various "degenerate" cases, the product of the two will therefore be a contraction/expansion combined with a rotation which is what we were supposed to show.

Exercise 4.6.  Given a general complex polynomial equation P = 0, where polynomial P is equal to

and z and a₀...a_n are all complex expressions, show that (1) must be factorable into the following form:

Where b₁...b_n are also complex numbers.

     I guess one way to start is by showing that if b_x is a solution to (1), then any arbitrary (z-b_x) divides evenly into (1).  First substitute b_x into (1) and solve for a₀:

     Now divide P by z-b_x to arrive at the following:

     Substitute a₀ by the result in (3) to obtain:

     Which can be rearranged into:

     Or:

      Which shows that any z-b_x will divide evenly into all terms of the quotient in (6).  Now, using (3), 

      So z-b_x also divides into P.  Since this is true for any b_x, we can say that

      Q is the remainder which, when multiplied with the divisor D in (8), will provide a fully factored expression for P, i.e. QD = P.  To find Q, note that a sequence of products such as found in D can be expanded as follows:

      Where all A_x incorporate products and sums of the solutions b_x.  It's worth noting that all A_x as well as the final standalone constant (the product of all the b_x) correspond to the various A_x found in (1).  However (1) also has a factor a_n that is a coefficient for z_n.  a_n is therefore the missing factor Q needed to ensure that P perfectly factors Q. (DUBIOUS!)  Upon multiplying (9) by a_n, each coeffficient and the final standalone constant will equal each coefficient andd standalone constant in (1).  Therefore the final factorization for (1) is:

Shit's Messed Up

      Everything about this solution seemed to go smoothly until, unfortunately, the key moment when I had to figure out wtf happened to a_n.  I am not very happy that a_n did not arise naturally but instead had to be artificially multiplied in to (9) based on examination of (1): it seems too much like I am using my guilty knowledge of what the answer is supposed to be and then jerry-rigging the solution accordingly.  On the other hand, since z_n has a coefficient of 1 in (9), I suppose that you could always factor out the term a_n with the result that that coefficient becomes 1/a_n.  It still seems shitty and I'd be grateful if anyone knows how to go through these steps in a way that a_n pops up naturally as a coefficient for z_n.  Another potentially much more serious problem with the whole thing is that it doesn't use any of the properties of complex numbers.  Which is troubling.

     I apologize for the tininess of most of the inline math, it's ridiculously hard to read. Blogspot's image upload system is pretty crappy (as is their too-small text window, grah!).  But the only alternative I have is to basically buy  another box and start up my own, fully customizable blogging engine which would just take too much work.  Oh, if there was just a way to upload these things as inline-viewable PDFs!