I'm working through a practice exam and can only do part of this problem.
Let $f(z) = x^3 + iy^3$. Find all points where the Cauchy-Riemann equations are satisfied, and give a brief explanation of why $f(z)$ is differentiable but not analytic at these points.
I can do the first part:
$u(x,y)=x^3$ and $v(x,y)=v^3$
$u_x=3x^2$, $v_y=3y^2$
$u_y=0, v_x=0$
So $u_y=-v_x$, but the Cauchy-Riemann equations are only satisfied, that is $u_x=v_y$ or $3x^2=3y^2$when $x=y$.
How do I address being differentiable and not analytic?
Best Answer
By definition a complex function is analytic at a point if it is differentiable not only at the point itself, but in its neighborhood. If you want formally, a function is analytic in $z_0$ if there exists $\epsilon>0$ such that the function is differentiable at every point $z$ which satisfies $|z-z_0|<\epsilon$. So being differentiable at a point and being analytic at a point are two different things.
For example, let's look at your function. You found it is differentiable only at the points of the set $\{x+iy: x^2=y^2\}$. Easy to see that this set has no interior points. If you want formally, let's assume $x^2=y^2$ for some number $z=x+iy$, and take any $\epsilon>0$. Then look at the point $w=(x+\frac{\epsilon}{2}+iy)$. The distance of that point from $z$ is smaller than $\epsilon$ but $(x+\frac{\epsilon}{2})^2 \ne y^2$, so $f$ is not differentiable at $w$. Hence there is no neighborhood of $z$ in which $f$ is differentiable at every point, so $f$ is not analytic at $z$.