- #1

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The question:

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

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

My attempt:

[tex]u (x,y) = x^2 + y^2 - 2y[/tex]

[tex]v (x,y) = 2x - 2xy [/tex]

[tex]u_x = 2x[/tex]

[tex]v_y = -2x[/tex]

[tex]u_y = 2y - 2[/tex]

[tex]v_x = 2 - 2y[/tex]

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

But if [tex]x=0[/tex] then [tex]v(x,y) = 0[/tex] and [tex]u(x,y) = y^2 - y[/tex]

We then continue: By Cauchy-Riemann:

[tex]u_y = -v_x[/tex]

But if [tex]v(x,y) = 0[/tex] then [tex]v_x = 0[/tex]

And as such: [tex]2y - 2 = 0[/tex]

[tex]y = 1[/tex]

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

The derivative of the function:

[tex]f'(z_0) = u_x + iv_x = 2x + i(2-2y)[/tex]

At the point (0,1):

[tex]f'(z_0) = 0 + i (2-2) = 0[/tex]

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