# The Unapologetic Mathematician

## 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 $i$, its powers form a basis of the ring $\mathbb{R}[i]$. But we have a relation, and that cuts things down a bit. Specifically, the element $i^2$ is the same as the element $-1$.

Given a polynomial $p$ in the “variable” $i$, we can write it as $c_ni^n+...+c_2i^2+c_1i+c_0$

We can peel off the constant and linear terms, and then pull out a factor of $i^2$: $(c_ni^{n-2}+...+c_2)i^2+(c_1i+c_0)$

Now this factor of $i^2$ can be replaced by $-1$, which drops the overall degree. We can continue like this, eventually rewriting any term involving higher powers of $i$ using only constant and linear terms. That is, any complex number can be written as $c_0+c_1i$, where $c_0$ and $c_1$ are real constants. Further, this representation is unique. This establishes the set $\{1,i\}$ as a basis for $\mathbb{C}$ as a vector space over $\mathbb{R}$.

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 $a_1+b_1i$ and $a_2+b_2i$ simply by adding the components: $(a_1+a_2)+(b_1+b_2)i$. We get the negative of a complex number by taking the negatives of the components.

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 $i\otimes i\mapsto-1$. So in terms of components we can write out the product of the complex numbers above as $(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i$.

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 $a$ can be considered as the complex number $a+0i$. This preserves all the field structures, but it ignores the order on the real numbers. A small price to pay, but an important one in certain ways.

We also mentioned the symmetry between $i$ and $-i$. Either one is just as valid as a square root of $-1$ as the other is, so if we go through consistently replacing $i$ with $-i$, and $-i$ with $i$, we can’t tell the difference. This leads to an automorphism of fields called “complex conjugation”. It sends the complex number $a+bi$ to its “conjugate” $a-bi$. This preserves all the field structure — additive and multiplicative — and it fixes the real numbers sitting inside the complex numbers.

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.

August 8, 2008 - Posted by | Fundamentals, Numbers

## 25 Comments »

1. […] start by noting that since the real numbers sit inside the complex numbers, we can consider any real polynomial as a complex polynomial. If the polynomial has a real root , […]

Pingback by Factoring Real Polynomials « The Unapologetic Mathematician | August 14, 2008 | Reply

2. […] else can we use? The complex numbers form a two-dimensional vector space over the reals, which means that as a vector space we have the […]

Pingback by Some Topological Fields « The Unapologetic Mathematician | August 26, 2008 | Reply

3. […] course, the second of these equations is just the complex conjugate of the first, and so it’s unsurprising. The first, however, is called “Euler’s […]

Pingback by Sine and Cosine « The Unapologetic Mathematician | October 13, 2008 | Reply

4. […] and nondegenerate. But there’s no way for such a form to be positive-definite. Indeed, we saw that there isn’t even a notion of “order” on the field of complex numbers. They […]

Pingback by Complex Inner Products « The Unapologetic Mathematician | April 22, 2009 | Reply

5. 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? Comment by sliderule | May 11, 2009 | Reply

6. 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. Comment by Todd Trimble | May 11, 2009 | Reply

7. Well, either way, here’s an answer. Defining such a scalar multiplication for a $\mathbb{C}$-vector space structure on an additive abelian group $A$ amounts to giving a homomorphism $\mathbb{C} \to \hom(A, A) \qquad (1)$

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 $c = 2^{\aleph_0}$, and therefore, as an abstract group, is also isomorphic to the additive group of complex numbers (as well as to $\mathbb{C}^2, \mathbb{C}^3, \ldots, \mathbb{C}^k$ for any $k$ no greater than $c$). Using any one of these realizations, (1) amounts to giving a ring homomorphism $\mathbb{C} \to \hom(\mathbb{C}^k, \mathbb{C}^k)$

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 $\mathbb{Q}$) has a basis, which relies on the axiom of choice. Comment by Todd Trimble | May 12, 2009 | Reply

8. (The $\mathbb{C}^k$ above means the complex vector space of dimension $k$, not to be confused with the $k$-fold cartesian product when $k$ is infinite.) Comment by Todd Trimble | May 12, 2009 | Reply

9. […] that we have a natural basis for the complex numbers as a vector space over the reals: . If we ask that this natural basis be […]

Pingback by Complex Numbers and the Unit Circle « The Unapologetic Mathematician | May 26, 2009 | Reply

10. 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. Comment by Peter L. Griffiths | May 21, 2011 | Reply

11. 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. Comment by Peter L. Griffiths | July 16, 2011 | Reply

12. 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. Comment by Peter L. Griffiths | August 19, 2011 | Reply

13. 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. Comment by Peter L.Griffiths | March 2, 2012 | Reply

14. -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. Comment by Peter L. Griffiths | April 5, 2012 | Reply

15. 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. Comment by Peter L. Griffiths | August 1, 2012 | Reply

16. See, now I know for sure that you’re either a crackpot or a troll. Just stop embarrassing yourself, Peter. Comment by John Armstrong | August 1, 2012 | Reply

17. 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. Comment by Peter L. Griffiths | September 10, 2012 | Reply

18. The two square roots of the imaginary number i are cos45+isin45 and cos225+isin225. Comment by Peter L. Griffiths | December 3, 2012 | Reply

19. Now, what are the two square roots of the imaginary number -i? Intuition could help with this. Comment by Peter L. Griffiths | December 5, 2012 | Reply

20. 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. Comment by Peter L. Griffiths | March 24, 2013 | Reply

21. 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. Comment by Peter L.Griffiths | August 3, 2013 | Reply

22. 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. Comment by Peter L. Griffiths | November 30, 2013 | Reply

23. 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. Comment by Peter L. Griffiths | August 24, 2014 | Reply

24. 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 Comment by Peter L. Griffiths | July 7, 2016 | Reply

25. 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. Comment by Peter L. Griffiths | July 14, 2016 | Reply