Just began the study of complex analysis. Let $$ f(x,y) = x^2 - y^2 + 2 i xy - x - iy. $$ I need to determine if this function is analytic. This means I have to show the partials satisfy the Cauchy-Riemann equations, and that the partials are continuous. So in this case we have $u(x,y) = x^2 - y^2 - x$ and $v(x,y) = 2xy - y$. Now \begin{align*} \frac{\partial v}{\partial y} = 2x - 1 = \frac{\partial u}{\partial x} \end{align*} and \begin{align*} - \frac{\partial u}{\partial y} = - (-2y) = 2 y = \frac{\partial v}{\partial x} \end{align*} Hence the Cauchy-Riemann equations are satisfied, which is a necessary condition for being analytic, but not sufficient. Now I have to show the partials are continuous? How do I do that?
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$\begingroup$Let $g$ define by $g(z)=z^2-z$. $g$ is clearly analytic on $\mathbb C$. You have that $f(x,y)=g(z(x,y))$ which is a composition of two analytic function.
$\endgroup$ $\begingroup$We have that $u = u(x,y)$ and $v=v(x,y)$ are polynomial functions of $x$ and $y$. If you know that polynomials are continuous functions and that derivatives of polynomials are again polinomials, you're done.
$\endgroup$ $\begingroup$Another way is to transform your function into a function of $\overline{z}$, where $\overline{z}$ is the conjugate of $z$, i.e.,
$$ z= x + iy ,\quad \overline{z} = x -iy.$$
In terms of $z$ and $\overline{z}$, we can write $x$ and $y$ as
$$x= \frac {z+\overline{z}}{2} $$$$y= \frac { z - \overline{z}}{2i},$$
The given function is written in terms of $z$ and $\overline{z}$. An analytic function should be a function of $z$ exclusively.
The simplest example likely is$$ \begin{align} f(z) &= x+iy \\ &= \frac {z+\overline{z}}{2}+ i\frac{z-\overline{z}}{2i} \\ &= \frac {z+\overline{z}}{2}+ \frac{z-\overline{z}}{2} \\ &= z, \end{align} $$which means $f$ is analytic.
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