On a linear algebra exam, I was asked something like, “If A is a real n\times n matrix and A^{2} is invertible, prove that A is also invertible.”

Apparently, a number of students attempted the problem this way:

A^{2} * (A^{2})^{-1} = A * (A * A^{-1}) * A^{-1} = A * A^{-1} = 1

and concluded that A had an inverse.  What’s wrong with this argument?  The astute reader will note that we’re actually assuming that A^{-1} exists to prove its existence!  That’s a little circular.  Instead, we’re going to prove something slightly more general.

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This is going to be a quick post, and a wrap-up to the other post I made about this.  I realized that, besides the Cauchy-Riemann equations being satisfied, very little else needs to happen such that a function is complex differentiable at a point z_{0}.  What else, specifically?  Let’s make this a nice little theorem.

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A small tangent, because I enjoyed this so much in class.

Let’s call a polynomial of n variables a homogeneous polynomial if every term has the same degree.  Recall that we compute the degree of a term by summing the powers of each variable in the term.

For example 3x_{1}^{2}x_{2}^{4}x_{3}^{4} has degree 10.  For another example, 2x_{1}^{3}x_{2}^{3} has degree 6.  So, an example of a homogenous polynomial in three variables is

f(x_{1}, x_{2}, x_{3}) = 2x_{1}^{3}x_{2}^{1} + 5x_{1}^{2}x_{3}^{2} + x_{3}^{4}

and this polynomial is of degree 4.

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When we do differential calculus, we work in the real numbers more times than not — we’re used to them, and they’re easy to manipulate.  A good number of applications of calculus also can naturally be worked on using the real numbers.

But there are some obvious drawbacks to using the real numbers — the most obvious being that not every polynomial actually has a root in the real numbers.  In fact, some polynomials will have all of their roots in the complexes.  This is kind of crappy if we’d like to know everything there is to know about a certain function, and so we often will extend our field to become the complex numbers.

I’m not going to run through what complex numbers are, or what the field topology looks like, or anything like that.  I’m going to assume that you know this.  The first point that we’re going to go through is just what it means for a function to be continuous — and, really, if a function is not continuous, what chance does it have of being differentiable?

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