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Showing g(z) is constant in C

Given that a function g=u+ivg=u+iv is differentiable on C\mathbb{C}, then given that v=u3v=u^3 I need to show g(z)g(z) is constant in C\mathbb{C}.

I know that I need to use that on a given disc D(0;R)D(0;R) then g′(z)=0g'(z) = 0 and g differentiable   ⟹  g(z)=c\implies g(z)=c for some constant cc on D(0;R)D(0;R), although I'm pretty much stuck at this point...

I initially tried using Cauchy Riemann equations out of desperation but I don't get anything useful... can anyone help me get the ball rolling? :s-smilie:
Reply 1
Avatar for Bananas01
Bananas01
What level is this?
Original post by Bananas01
What level is this?


Uni, although I think it'll probably be something straightforward that I just haven't spotted (as is the usual case).
Reply 3
Avatar for Bananas01
Bananas01
Original post by wanderlust.xx
Uni, although I think it'll probably be something straightforward that I just haven't spotted (as is the usual case).


Just wondering because I'm doing uni maths next year.
Reply 4
Avatar for SimonM
SimonM
18
It comes out quite quickly using Cauchy-Riemann equations...
Original post by SimonM
It comes out quite quickly using Cauchy-Riemann equations...


So g(z)=u(x,y)+iv(x,y)=u(x,y)+iu3(x,y)g(z) = u(x,y) + iv(x,y) = u(x,y) + iu^3(x,y) and g diff   ⟹  \implies ∂u∂x=∂(u3)∂y\dfrac{\partial u}{\partial x} = \dfrac{\partial (u^3)}{\partial y} and ∂u∂y=−∂(u3)∂x\dfrac{\partial u}{\partial y} = - \dfrac{\partial (u^3)}{\partial x}.

Then what can I do? :s-smilie: Or am I taking the wrong approach here?
Reply 6
Avatar for DFranklin
DFranklin
18
du^3/dy = 3u^2 du/dy (all d's partial).
Original post by DFranklin
du^3/dy = 3u^2 du/dy (all d's partial).


:facepalm: dear god im being slow

So basically I get (3u4+1)∂u∂x=0   ⟹  ∂u∂x=0  ⟹  g′(z)=0  ⟹  g(z)=c(3u^4 +1) \dfrac{\partial u}{\partial x} = 0 \ \implies \dfrac{\partial u}{\partial x} = 0 \implies g'(z) = 0 \implies g(z) = c

Buuuuut as a final question, I don't understand what happens if u4=−13u^4 = -\dfrac{1}{3}? Or does this just mean g(z)=c  ⟺  u4≠−13g(z) = c \iff u^4 \not= -\dfrac{1}{3}?
(edited 13 years ago)
Reply 8
Avatar for DFranklin
DFranklin
18
Isn't in 9u^4+1? Although it doesn't really matter.

Moving on, from context, u is real, so u^4 >=0. (If u is complex valued, f doesn't have to be constant. e.g. just take u = z).
Original post by DFranklin
Isn't in 9u^4+1? Although it doesn't really matter.

Moving on, from context, u is real, so u^4 >=0. (If u is complex valued, f doesn't have to be constant. e.g. just take u = z).


yeah i missed writing out a 3 somewhere, it is in fact 9. I understand, cheers! :smile:

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