- #1
Gh0stZA
- 25
- 0
Hello everyone,
The question:
My attempt:
I'll try my hand at the analytic part if I could get some clarification on this part first. :)
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. :)