Modulus of z

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  • #1
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[tex]f(z) = |z|[/tex]

By the Cauchy-Riemann equations,

[tex]u_x = \frac{x}{\sqrt{x^2+y^2}}[/tex]

[tex]v_y = -v_x = 0[/tex]

[tex]u_y = \frac{y}{\sqrt{x^2+y^2}}[/tex]

Since the C.R. equations don't work at (0,0), how can show [itex]f(z)[/itex] is not holomorphic at (0,0)?
 

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  • #2
Dick
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[tex]f(z) = |z|[/tex]

By the Cauchy-Riemann equations,

[tex]u_x = \frac{x}{\sqrt{x^2+y^2}}[/tex]

[tex]v_y = -v_x = 0[/tex]

[tex]u_y = \frac{y}{\sqrt{x^2+y^2}}[/tex]

Since the C.R. equations don't work at (0,0), how can show [itex]f(z)[/itex] is not holomorphic at (0,0)?
You mean you are claiming the CR equations DO WORK at z=0, right? Actually, they don't. u isn't differentiable with respect to x at (0,0). u(x,0)=|x|.
 
  • #3
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You mean you are claiming the CR equations DO WORK at z=0, right? Actually, they don't. u isn't differentiable with respect to x at (0,0). u(x,0)=|x|.
I didn't claim they work at z = 0. I asked how to show they don't.
 
  • #4
A function is complex differentiable if their partial derivatives for u and v exist and they satisfy the C-R-eq. Since the p.d. for u do not exist, f(z) is not complex differentiable (in z=0). This means that f(z) is not holomorphic in z=0.
 
  • #5
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A function is complex differentiable if their partial derivatives for u and v exist and they satisfy the C-R-eq. Since the p.d. for u do not exist, f(z) is not complex differentiable (in z=0). This means that f(z) is not holomorphic in z=0.
So just take the limit of f(z) approaching from the x and y axis to show they limits are different. Thus, f(z) is not differentiable at z = 0?
 
  • #6
Dick
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So just take the limit of f(z) approaching from the x and y axis to show they limits are different. Thus, f(z) is not differentiable at z = 0?
That's the way to show f(z) is discontinuous at z=0. It's not. Write down a difference quotient for the x derivative of u(x,y) at z=0. (u(0+h,0)-u(0,0))/h. Show different ways of letting h->0 lead to different limits.
 
  • #7
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That's the way to show f(z) is discontinuous at z=0. It's not. Write down a difference quotient for the x derivative of u(x,y) at z=0. (u(0+h,0)-u(0,0))/h. Show different ways of letting h->0 lead to different limits.
This

[tex]\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}[/tex]

??
 
  • #8
Dick
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This

[tex]\lim_{h\to 0}\frac{\sqrt{(x+h)^2+y^2}-\sqrt{x^2+y^2}}{h}[/tex]

??
Yes. You were wondering what happens at z=0. So put x=0 and y=0.
 
  • #9
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Yes. You were wondering what happens at z=0. So put x=0 and y=0.
That is what I essentially did but my professor wrote, I need to consider (0,0) separately.
 
  • #10
Dick
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That is what I essentially did but my professor wrote, I need to consider (0,0) separately.
We ARE doing (0,0) separately. Away from (0,0) you can use your formulas for u_x and u_y. At (0,0) we are looking at the difference quotient. BTW what do you conclude from that?
 
  • #11
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We ARE doing (0,0) separately. Away from (0,0) you can use your formulas for u_x and u_y. At (0,0) we are looking at the difference quotient. BTW what do you conclude from that?
The limit is [itex]\pm 1[/itex].

What I was saying is on an assignment that is what my professor wrote even though I showed the [itex]\pm 1[/itex] too.
 
  • #12
Dick
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The limit is [itex]\pm 1[/itex].

What I was saying is on an assignment that is what my professor wrote even though I showed the [itex]\pm 1[/itex] too.
Since you can find two different limits that actually means the limit doesn't exist and the function isn't differentiable. Better have a talk with your professor about what is actually required here.
 

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