Complex Analysis - Differentiability

[NewtonianAlch]

Homework Statement

Show that the function f defined by $f(z) = 3\,{x}^{2}y+{y}^{3}-6\,{y}^{2}+i \left( 2\,{y}^{3}+6\,{y}^{2}+9\,x<br /> \right)$ is nowhere differentiable.

The Attempt at a Solution

Computing the C.R equations for this, I am left with

${y}^{2}+2\,y={\it xy}$

and

$x^2+(y-2)^2 = 1$

Now from the upper equation we can see that $x={\frac {{y}^{2}+2\,y}{y}}$

This means that there is a discontinuity at y = 0, and therefore this function is not differentiable?

I somehow don't think is right as it would mean it's only not differentiable at that point, not necessarily everywhere, how would I go about this?

 

[jackmell]

Homework Statement

Show that the function f defined by $f(z) = 3\,{x}^{2}y+{y}^{3}-6\,{y}^{2}+i \left( 2\,{y}^{3}+6\,{y}^{2}+9\,x<br /> \right)$ is nowhere differentiable.

The Attempt at a Solution

Computing the C.R equations for this, I am left with

${y}^{2}+2\,y={\it xy}$

and

$x^2+(y-2)^2 = 1$

Now from the upper equation we can see that $x={\frac {{y}^{2}+2\,y}{y}}$

This means that there is a discontinuity at y = 0, and therefore this function is not differentiable?

I somehow don't think is right as it would mean it's only not differentiable at that point, not necessarily everywhere, how would I go about this?

x and y are real numbers right? So, is there ever a real solution to those two equations? That is can you substitute x=f(y) of one into the other, make a polynomial out of it, then solve for the roots? If those roots are only complex, then the CR equations are never satisfied.

 

[Dustinsfl]

$$u_y = 3x^2 + 3y^2 - 12y = -9 = -v_x\Rightarrow x^2 + y(y -4) = -3$$

I think part of your CR has a mistake.

 

[NewtonianAlch]

$$u_y = 3x^2 + 3y^2 - 12y = -9 = -v_x\Rightarrow x^2 + y(y -4) = -3$$

I think part of your CR has a mistake.

I just completed the square, so y(y - 4) becomes (y-2)^2 and the addition of the 4 means the other side of equation becomes 1.

 

[NewtonianAlch]

x and y are real numbers right? So, is there ever a real solution to those two equations? That is can you substitute x=f(y) of one into the other, make a polynomial out of it, then solve for the roots? If those roots are only complex, then the CR equations are never satisfied.

Ah yes, I see now. Thanks for that!

I got two sets of complex roots because I solved for two values of y, and when I substituted those, both results were complex. If only one were complex would that mean it's still differentiable but at that point?

 

[jackmell]

It would be differentialbe where ever x and y are real and satisfy the CR equations.