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→Unique square root of -1 necessary for complex analysis: new section
::(5) "but elementary proofs are not always preferred ..." by whom? One doesn't allow a fox to guard a henhouse, and neither mathematicians nor politicians should be setting their own rules as they go, or else before long all sorts of nonsensical "proofs" might be asserted by some academics at [[liberal]] institutions.--[[User:Aschlafly|Andy Schlafly]] 00:46, 8 April 2012 (EDT)
::(5) "but elementary proofs are not always preferred ..." by whom? One doesn't allow a fox to guard a henhouse, and neither mathematicians nor politicians should be setting their own rules as they go, or else before long all sorts of nonsensical "proofs" might be asserted by some academics at [[liberal]] institutions.--[[User:Aschlafly|Andy Schlafly]] 00:46, 8 April 2012 (EDT)
== Unique square root of -1 necessary for complex analysis ==
Dear Mr. Schlafly,
Where are you getting your information that complex analysis requires there to be a unique square root of −1? I am currently taking a complex analysis course, and I understand that there are two square roots of −1, namely ''i'' and −''i''.<sup>1</sup> I don't see how this undermines complex analysis. It may help to examine how complex numbers are derived in the first place. (The interested reader may consult ''Complex Function Theory'', 2nd ed., by Sarason, sections I.1 and I.2.)
#We define a set <math>\mathbb{C}</math> to consist of all ordered pairs of real numbers.
#We define addition of elements of <math>\mathbb{C}</math> as follows:
#:<center><math>(a, b) + (c, d) := (a + c, b + d)</math></center>
#We define multiplication of elements of <math>\mathbb{C}</math> as follows:
#:<center><math>(a, b) \cdot (c, d) := (ac - bd, ad + bc)</math></center>
#We verify the following about <math>+</math>:
##It is commutative
##It is assoiciatve
##There exists an element of <math>\mathbb{C}</math>, namely <math>0 := (0, 0)</math>, such that for all <math>a \in \mathbb{C}</math>,
<math>a + 0 = a = 0 + a</math>
##For every element <math>a \in \mathbb{C}</math> there exists an additive inverse <math>-a</math> such that
<math>a + (-a) = 0 = (-a) + a</math>
##Therefore, <math>\mathbb{C}</math> is an [[abelian group]] under addition.
#We verify the following about <math>\cdot</math>:
##It is commutative
##It is associative
##It has the distributive property over addition
##There exists an element of <math>\mathbb{C}</math>, namely <math>1 := (1, )</math>, such that for all <math>a \in \mathbb{C}</math>,
<math>a \cdot 1 = a = 1 \cdot a</math>
##For every nonzero element <math>a \in \mathbb{C}</math> there exists a multiplicative inverse (reciprocal) <math>a^{-1}</math> such
that <math>a \cdot a^{-1} = 1 = a^{-1} \cdot a</math>
##Therefore, <math>\mathbb{C}</math> is a [[field]].
#We consider the subset <math>S = \{(a, 0)|a \in \mathbb{R}\} \subseteq \mathbb{C}</math> We note that there is a natural [[isomorphism]]
from <math>S</math> to <math>\mathbb{R}</math>. Thus, we can associate elements of <math>\mathbb{C}</math> with zero second coordinate with real numbers.
#As notation, we write the ordered pair <math>(a, b)</math> as <math>a + bi</math>. This is just notation. We note that it does not constitute an [[abuse of notation]] since <math>(a, b) = (a, 0) + (0, b)</math>, so the complex number notated by <math>a + bi</math> is really the sum
of the complex numbers <math>a + 0i</math> and <math>0 + bi</math>.
#We can also use natural shorthand for the <math>a + bi</math> notation. In particular, we can write <math>i</math> to represent the element
<math>(0, 1) \in \mathbb{C}</math>. Note that <math>i^2 = (0, 1)\cdot(0, 1) = (-1, 0) = -1</math>.
This is how complex numbers are typically derived. (One could of course, define them to be the cosets in the quotient field <math>\mathbb{R}[x]/(x^2 + 1)</math>; this is clearly equivalent to the derivation above.) Please let me know where you see a fault in this derivation.
<sup>1</sup>Incidentally, [[Galois theory]] tells us that there cannot be a unique nonreal solution to a polynomial with real coefficients. Considering <math>\mathbb{C}</math> as a field extension of <math>\mathbb{R}</math>, we note that the [[Galois group]] of <math>\mathbb{C}/\mathbb{R}</math>, which consists of all automorphisms of <math>\mathbb{C}</math> that fix elements of <math>\mathbb{R}</math>, has precisely two members: the identity function and the function taking complex numbers to their complex conjugate (which we will denote <math>\phi</math>. Suppose that
<math>f(x) = a_0 + a_1 x + \cdots + a_n x^n</math> is a polynomial with real coefficients that has a root <math>r</math>. Then,
<math>0 = a_0 + a_1 r + \cdots + a_n x^n</math>. We apply the aforementioned <math>\phi</math> to both sides; the application of this function leaves <math>0</math> and all the coefficients <math>a_0, \ldots, a_n</math> unchanged. What we get is
<math>0 = a_0 + a_1 \bar{r} + \cdots + a_n \bar{r}^n</math>. This means that <math>\bar{r}</math>, the complex conjugate of <math>r</math>,
is also a root of <math>f</math>. Thus, any polynomial in real coefficients with a nonreal root <math>r</math> also has
<math>\bar{r}</math> as a root. In particular, complex square roots of <math>-1</math> are roots of the polynomial
<math>x^2 + 1</math>, and since we know this polynomial has no real solutions, the two roots guaranteed by the [[fundamental theorem of algebra]] must be a complex conjugate pair. [[User:GregG|GregG]] 01:13, 8 April 2012 (EDT)
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