I have just started studying complex analysis and i am stuck with one question.
My book says, the derivative of an analytic polynomial with respect to $z$ is equal to the partial derivative of that polynomial with respect to $x$, $i.e.$ $P'(z) = P_x$.
For example,
$P(x+iy)$ = $x^3-3xy^2-x+i(3x^2y-y^3-y)$
then its derivative should be equal to $3x^3-3y^2-1+6ixy$.
I don't quite understand this. It would be nice if someone could explain this correlation between the polynomial being analytic and its derivative being given by, $P'(z) = P_x$, to me.
Best Answer
The question is a bit confusing, because the book seems to be trying to teach you about general complex functions when it says "polynomial". It is true in general that if you have a function $f:\mathbb R^2\to \mathbb C$, then the function $x+iy \mapsto f(x,y)$ is complex differentiable (on an open set) exactly if $$ \frac{\partial}{\partial y}f(x,y) = i\frac{\partial}{\partial x}f(x,y) \tag{*}$$ (in that open set). This is just a (slightly unusual) way to write the Cauchy-Riemann equations.
Assuimg that you have defined the complex derivative as something like $$ f'(z) = \lim_{h\to 0\in\mathbb C}\frac{f(z+h)-f(z)}{h} $$ Once you know that $(*)$ holds, there's a theorem guaranteeing you that this limit exists, so if you want to find it you can choose to look at only real $h$s -- because if the quotient gets close to some number for all small enough complex $h$, then in particular small enough real $h$ will allow you to get close to it. And then we have $$ f'(z) = \lim_{h\to 0\in\mathbb R}\frac{f(z+h)-f(z)}{h} $$ but that is exactly the definition of $\frac{\partial}{\partial x}f(x+iy)$.
Actually, whenever you have a complex polynomial -- that is, something like $f(z)=z^3-z$ that can be written as a polynomial in the complex variable rather than the real and imaginary part separately -- then all of this will automatically be true, because the rules for differentiating symbolically that we know and love from the real case continue to hold for $\mathbb C$.
In fact it happens that $f(x+iy)=x^3-3xy^2-x+i(3x^2y-y^3-y)$ and $f(z)=z^3-z$ is the same function. (Try writing it out!). So to differentiate it, you can just differentiate $f(z)=z^3-z$ symbolically to get $3z^2-1$, which is much faster than messing around with $x$s and $y$s.
It also happens to hold that if $P$ is a polynomial in two variables with complex coefficients, then the function $x+iy\mapsto P(x,y)$ is analytic exactly if it can be written as a polynomial in one complex variable. And if that happens, the coefficients naturally have the be the same as the coefficients of $P(x,0)$ (because otherwise the functions wouldn't match on the real line). So an alternative way to check if a polynomial in two variables corresponds to an analytic function is to check whether $P(x+iy,0)=P(x,y)$.