Complex numbers Watch

Lunu
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#1
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Does Re(z)Im(z) + iIm(z) = xy +iy

If this is true then why can't I prove it's differentiable?
u(x,y) = xy and v(x,y) = y

∂u/∂x = y
∂u/∂y = x
∂v/∂x = 0
∂v/∂y = 1

I must have done something wrong because the Cauchy- Riemann equations won't hold and it says it's differentiable in the question.
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Lunu
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Can anyone answer if Re(z)Im(z) + iIm(z) = xy + iy?
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Zacken
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(Original post by Lunu)
...
Apologies for butting my unknowledgeable head in but I don't know much, if any, complex analysis so will be useless. Here's my attempt anyway (until somebody who knows anything pops by): Are you sure your thing is differentiable? Because I know that \Re (z) isn't differentiable and it's making me doubt the differentiability of your expression, so I was wondering if you could upload a picture or such of the entire question.
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rayquaza17
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Yeah I don't think that sounds right. Unless I've gotten my numbers wrong, take this for example:

z=3+2i
x=3,y=2
re(z)=3,im(z)=2
re(z)im(z)+iim(z)=3*2+2i=6+2i

If the CR relations don't hold, then it's not complex differentiable everywhere. It sounds like your question is wrong if it says it is differentiable everywhere!

EDIT: Math12345 is correct, I have also added in the word "everywhere" to my post.
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poorform
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(Original post by Lunu)
Does Re(z)Im(z) + iIm(z) = xy +iy

If this is true then why can't I prove it's differentiable?
u(x,y) = xy and v(x,y) = y

∂u/∂x = y
∂u/∂y = x
∂v/∂x = 0
∂v/∂y = 1

I must have done something wrong because the Cauchy- Riemann equations won't hold and it says it's differentiable in the question.
Post photo of question please.
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Lunu
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#6
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Show thatf(z) = Re(z)Im(z) + iIm(z)is differentiable at only one point in C, and find this point. I think this point must be z=0?
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Math12345
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It is complex differentiable where the Cauchy-Riemann equations hold. y=1 and x=0. So the point is (0,1).
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Lunu
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(Original post by Math12345)
It is differentiable where the Cauchy-Riemann equations hold. y=1 and x=0. So the point is (0,1).
Thank you!!
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Zacken
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(Original post by Lunu)
Show thatf(z) = Re(z)Im(z) + iIm(z)is differentiable at only one point in C, and find this point. I think this point must be z=0?
Seriously? This is very different to the question you originally asked. Urgh.
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Lunu
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#10
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(Original post by Zacken)
Seriously? This is very different to the question you originally asked. Urgh.
I'm sorry I didn't understand the question at all so I got confused.
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Math12345
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From the working out in the op, it was obvious the question was at what points is the function complex differentiable at.
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Math12345
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#12
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(Original post by Lunu)
Actually isn't it (0, -1)?
Look at the Cauchy- Riemann equations
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rayquaza17
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(Original post by Lunu)
I'm sorry I didn't understand the question at all so I got confused.
Don't worry about being confused and making a mistake, we're here to help you. The only thing that matters is you understand the question and what to do now!

Also the CR relations are: du/dx=dv/dy and du/dy=-dv/dx. So I think (0,1) is correct.
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Lunu
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(Original post by rayquaza17)
Don't worry about being confused and making a mistake, we're here to help you. The only thing that matters is you understand the question and what to do now!

Also the CR relations are: du/dx=dv/dy and du/dy=-dv/dx. So I think (0,1) is correct.
Thanks
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poorform
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#15
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#15
(Original post by Lunu)
Thanks
Keep in mind Cauchy-Riemann must hold for f to be complex differentiable at a point but it is not sufficient to show differentiability at a point. (That is there are functions that satisfy C-R at a point but are not differentiable at said point).

To ensure complex differentiability you need to make sure C-R is satisfied and the partial derivatives are also continuous.
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Implication
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#16
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(Original post by Lunu)
Does Re(z)Im(z) + iIm(z) = xy +iy
since the rest of your post (the hard part) has been answered I'm going to go with a tentative 'yes, if z=x+iy'
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Gregorius
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#17
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(Original post by poorform)
Keep in mind Cauchy-Riemann must hold for f to be complex differentiable at a point but it is not sufficient to show differentiability at a point. (That is there are functions that satisfy C-R at a point but are not differentiable at said point).

To ensure complex differentiability you need to make sure C-R is satisfied and the partial derivatives are also continuous.
Just to emphasize poorform 's remark and to offer a slight clarification: When one is considering complex differentiability at a point, the CR equations offer a necessary but not sufficient condition for differentiability. So in this problem, use the CR equations to find the only candidate possibilities for pointwise differentiability - and then prove from first principles whether differentiability holds at that point. That is, look at the limit as h \rightarrow 0 of

 \displaystyle \frac{f(z+h) - f(z)}{h}

where h = \alpha + i \beta is a general complex number.

So here, you are looking at the limit as  \alpha, \beta \rightarrow 0 of

 \displaystyle \frac{f(0 + i + \alpha + i \beta) - f(0 + i)}{\alpha + i \beta}

An analytic (or holomorphic) function is one that is complex differentiable in an open set; this is where one starts to look at the continuity of the partial derivatives in the CR equations.
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