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Complex numbers

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.
Reply 1
Avatar for Lunu
Lunu
OP
1
Can anyone answer if Re(z)Im(z) + iIm(z) = xy + iy? :frown:
Reply 2
Avatar for Zacken
Zacken
22
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 ℜ(z)\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.
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.
(edited 8 years ago)
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.
Reply 5
Avatar for Lunu
Lunu
OP
1
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?
It is complex differentiable where the Cauchy-Riemann equations hold. y=1 and x=0. So the point is (0,1).
(edited 8 years ago)
Reply 7
Avatar for Lunu
Lunu
OP
1
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!!
Reply 8
Avatar for Zacken
Zacken
22
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.
Reply 9
Avatar for Lunu
Lunu
OP
1
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.
From the working out in the op, it was obvious the question was at what points is the function complex differentiable at.
Original post by Lunu
Actually isn't it (0, -1)?


Look at the Cauchy- Riemann equations
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! :smile:

Also the CR relations are: du/dx=dv/dy and du/dy=-dv/dx. So I think (0,1) is correct.
(edited 8 years ago)
Reply 13
Avatar for Lunu
Lunu
OP
1
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! :smile:

Also the CR relations are: du/dx=dv/dy and du/dy=-dv/dx. So I think (0,1) is correct.


Thanks :smile:
Original post by Lunu
Thanks :smile:


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.
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'
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→0h \rightarrow 0 of

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

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

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

f(0+i+α+iβ)−f(0+i)α+iβ \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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