Let $S$ denote the set $\{z \in \mathbb{C} : Im(z) > 0\}$. Suppose that $f: S \to \mathbb{C}$ is analytic and continuous on the interval $(0,1)$. It is given that $f(x) = x^4 - 2x^2$ on this interval.
My question is this:
I know I can expand $f$ to be analytic on $S \cup \{z \in \mathbb{C} : Im(z) < 0\} \cup (0,1)$. After doing so, if $u$, $v$ denote the real and imaginary parts of $f$ we have that on $(0,1)$ $v = 0$ and so $\frac{dv}{dx}(x,0) = -\frac{du}{dy}(x,0) = 0 = \frac{dv}{dy}(x,0) = \frac{du}{dx}(x,0)$, so this would imply that $u$ is contant in $(0,1)$ and is a contradiction to $f(x) = x^4 - 2x^2$ on the interval.
What am I doing wrong?
Your $v_y=0$ assumption is the error. We know $v$ is constant as we change $x$ but not as we change $y$. Take the function $f(x,y)=x+iy$, it has $v=0$ on $(0,1)$, but clearly $v_y=1$.