Properties of Complex Numbers
Today I’ll collect a few basic properties of complex numbers.
First off, they form a vector space over the reals. We constructed them as an algebra — the quotient of the algebra of polynomials by a certain ideal — and every algebra is a vector space. So what can we say about them as a vector space? The easiest fact is that it’s two-dimensional, and it’s got a particularly useful basis.
To see this, remember that we have a basis for the algebra of polynomials, which is given by the powers of the variable. So here when we throw in the formal element 
![\mathbb{R}[i]](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BR%7D%5Bi%5D&bg=e6e6e6&fg=333333&s=0&c=20201002)
Given a polynomial 
We can peel off the constant and linear terms, and then pull out a factor of
Now this factor of 
Now the additive parts of the field structure are clear from the vector space structure here. We can write the sum of two complex numbers 
We can also write out products pretty simply, since we know the product of pairs of basis elements. The only one that doesn’t involve the unit of the algebra is 
Notice here that the field of real numbers sits inside that of complex numbers, using scalar multiples of the complex unit. This is characteristic of algebras, but it’s worth pointing out here. Any real number 
We also mentioned the symmetry between 
Studying this automorphism, and similar structures of other field extensions forms the core of what algebraists call “Galois theory”. I’m not going there now, but it’s a huge part of modern mathematics, and its study is ultimately the root of all of our abstract algebraic techniques. The first groups were automorphism groups shuffling around roots of polynomials.
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So the complex numbers form a vector space over the reals, but can we define a scalar multiplication to turn the reals into a vector space over the complex numbers?
sliderule, are you asking because you don’t already know the answer and want to know, or because you already know and are putting it out there as a poser? As a poser it’s semi-famous, and is mentioned for instance in Paul Halmos’s automathography.
Well, either way, here’s an answer. Defining such a scalar multiplication for a
-vector space structure on an additive abelian group 
amounts to giving a homomorphism
of rings with unit. We want to know if there exists such a ring homomorphism if A is the additive group of reals. But the additive group of reals is a vector space over the rationals (in a necessarily unique way) of dimension
, and therefore, as an abstract group, is also isomorphic to the additive group of complex numbers (as well as to 
for any 
no greater than 
). Using any one of these realizations, (1) amounts to giving a ring homomorphism
that is to say, a complex vector space structure on $\mathbb{C}^k$, which is trivial.
However, it is clear that this is highly nonconstructive: the abstract isomorphisms here implicitly involve the theorem that every vector space (here over
) has a basis, which relies on the axiom of choice.
(The
above means the complex vector space of dimension 
, not to be confused with the 
-fold cartesian product when 
is infinite.)
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To obtain the roots of the complex number a+ib, you find the arcotangent below 90 degrees of the complex ratio a/b which is then divided by the root required. This gives one of the roots of the complex ratio. For the other roots you add 360 degrees and then 720 degrees on to the arcotangent. In this way all the n nth roots of the complex number can be obtained.
Further to my comment of 21 May 2011, obtaining the root of a-ib is similar to obtaining the root of a+ib. The negative angles all suitably correspond to negative cotangents. On the other hand however the roots of -1 are not quite the same as the negatives of the roots of +1. They are for odd roots, but not for even roots which need a different type of calculation.
I can extend my comments of 16 July 2011 as follows. To convert the complex number a+ib into an imaginary number let a equal 0, and substitute the Cotes format for the arcotangent format so that 0+ib becomes cos90 +isin90 which is 0+i. Also cos180+isin180 becomes i squared which is -1, and cos369+isin360 becomes i to the power of 4 which is +1. This also applies to the division of 90 in that cos45+isin45 equals the square root of i. In this way the roots and powers of +1, -1, +i and -i can be obtained.
Further to the last sentence of my comments of 21 May 2011 which is ‘In this way all the n nth roots of the complex number can be obtained’, it seems that Hamilton’s quaternions equation i^2=j^2=k^2=ijk=-1 is incorrect in that -1 seems to have been given three square roots i, j,and k. It is k which seems to be wrong.
-1 cannot have more than two square roots, but it can have three cube roots which are cos60+isin60, cos180+isin180 which equals -1, and cos300+isin300.
Further to my comment of 5 April 2012, the roots of complex, and imaginary numbers have been irresponsibly ignored in most mathematical courses, as evidence for this, ask any advanced mathematician to describe how to compute the two square roots of the imaginary number i, he probably will not even be aware that there are two square roots.
See, now I know for sure that you’re either a crackpot or a troll. Just stop embarrassing yourself, Peter.
How about the following proof of Fermat’s Last Theorem.
If the equation is written as (p+q)^n-(p-q)^n, the binomial expansion will only have an integer nth root if p equals q for n greater than integer 2. If however n equals 2 then the Pythagorean Triples apply if both p and q have square roots, otherwise the rules for n greater than 2 apply even though Andrew Wiles does not seem to refer to this. These matters do not seem to have been discussed in John Armstrong’s recent PhD course.
The two square roots of the imaginary number i are cos45+isin45 and cos225+isin225.
Now, what are the two square roots of the imaginary number -i? Intuition could help with this.
In answer to the terribly difficult maths question I asked on 5 December 2012, the two square roots of the imaginary number -i are cos45-isin45 and cos225-isin225.
Further to my comment of 10 September 2012, Fermat’s Last Theorem can be proved in just over 400 words by recognising that it is a question of distinguishing rational from irrational numbers. If n is an integer greater than 1 then the nth root of 2 will always be irrational. But this irrationality will only be corrected in the binomial expansion if p equals q. Otherwise the whole expression is irrational which proves FLT.
Some doubts about the Riemann Hypothesis. Near the beginning of his 1859 paper Riemann incorrectly assumes that the complex variable s = (1/2) +ti is a zeta power. Riemann fails to recognise that an expresion containing an imaginary number such as (1/2) +ti cannot be a power unless the base is a log base such as e. The best known example of this is Cotes’s formula (not mentioned by Riemann) cosu + isinu equals e^(iu), where it is not possible for e to be meaningfully replaced by other values, also e^(1/2) X e^(iu) equals e^[(1/2) +iu]. This means that Riemann is badly wrong in applying as a power s =(1/2) + ti, and in failing to recognise t as an angle. It also means that practically all the arguments in his 1859 paper are fallacious.
A possible way of identifying primes is to apply Euler’s discovery and multiply the Harmonic series H by the product of denominator primes (1/2)(2/3)(4/5)(6/7)(10/11) ….=1. The harmonic series H is 1+(1/2)+(1/3) +(1/4) …..An investigation of the products below infinity should help to relate the primes to the Harmonic series below infinity, and hence to the log of the last term.
Further to my comment of 3 August 2013,
[ (1 + 1/r)^n – ( 1 – 1/r)^n]^1/n = 2^1/n [ (*n/1)^1/r + (*n/3)^1/r^3 ….]^1/n.
Let r = 1 so that p = q. This deliberately converts the 3 term assumption into 2 terms.
There can only be equality with 2 terms not the 3 terms assumed in Fermat’s Last Theorem. End
The complex number a+ib has the positive complex ratio a/b, whose arcotangent below 90 degrees can be divided by root n. That is [arcot(a/b)]/n. The cotangent of this result will produce one of the n roots of the complex ratio. This likewise applies to the complex number a-ib and the negative complex ratio -a/b whose negative arcotangent can also be divided by root n. The negative cotangent of this result will produce one of the roots of the negative complex ratio. The circular route of these negative complex ratios is opposite to the route of the positive complex ratios. End.