Cauchy-Riemann Equations - Complex Analysis

[Gh0stZA]

Hello everyone,

The question:

Find all the points where $$f(z) = (x^2 + y^2 -2y) + i(2x-2xy)$$ is differentiable, and compute the derivative at those points.

Is the function above analytic at any point? Justify your answer clearly.

My attempt:

$$u (x,y) = x^2 + y^2 - 2y$$
$$v (x,y) = 2x - 2xy$$

$$u_x = 2x$$
$$v_y = -2x$$

$$u_y = 2y - 2$$
$$v_x = 2 - 2y$$

However Cauchy-Riemann states that $$u_x = v_y$$ so my reasoning is $$v_y = -v_y$$ and that is only true where $$v_y = 0$$. That is to say: $$-2x = 0 \rightarrow x = 0$$.

But if $$x=0$$ then $$v(x,y) = 0$$ and $$u(x,y) = y^2 - y$$

We then continue: By Cauchy-Riemann:
$$u_y = -v_x$$

But if $$v(x,y) = 0$$ then $$v_x = 0$$
And as such: $$2y - 2 = 0$$
$$y = 1$$


Does this mean the function is only differentiable at (0,1) ?

The derivative of the function:
$$f'(z_0) = u_x + iv_x = 2x + i(2-2y)$$

At the point (0,1):
$$f'(z_0) = 0 + i (2-2) = 0$$

I'll try my hand at the analytic part if I could get some clarification on this part first. :)

 

[HallsofIvy]

Everything is correct except your statement

$f'(z_0)= 2x+ i(2- 2y)$

If f is differentiable only at (0, 1), that makes no sense except for z= i.

As for analytic- a function is analytic at a point if and only if it is differentiable in some neighborhood of that point.

 

[snipez90]

HallsofIvy was a bit too generous in saying everything is correct.

You cannot evaluate the function at x = 0 and then compute the second set of Cauchy-Riemann equations as you did. This amounts to evaluating a real function f(x,y) at x = 0, computing the partial derivative with respect to y, and then claiming that the result is actually $\frac{\partial f}{\partial y}$.

You must compute the Cauchy-Riemann equations first, then look at the set of (x,y) that satisfy the equations. Then you can determine where the function is analytic (by HallsofIvy's given definition), if anywhere.

 

[Gh0stZA]

You must compute the Cauchy-Riemann equations first, then look at the set of (x,y) that satisfy the equations. Then you can determine where the function is analytic (by HallsofIvy's given definition), if anywhere.

So basically you're saying the part about figuring that $$x = 0$$ should just shift down a bit?