# Complex numbers

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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.

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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#3

(Original post by

...

**Lunu**)...

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#4

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

EDIT: Math12345 is correct, I have also added in the word "everywhere" to my post.

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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#5

(Original post by

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.

**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.

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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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#7

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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(Original post by

It is differentiable where the Cauchy-Riemann equations hold. y=1 and x=0. So the point is (0,1).

**Math12345**)It is differentiable where the Cauchy-Riemann equations hold. y=1 and x=0. So the point is (0,1).

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#9

(Original post by

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?

**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?

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(Original post by

Seriously? This is very different to the question you originally asked. Urgh.

**Zacken**)Seriously? This is very different to the question you originally asked. Urgh.

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#11

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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#12

(Original post by

Actually isn't it (0, -1)?

**Lunu**)Actually isn't it (0, -1)?

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#13

(Original post by

I'm sorry I didn't understand the question at all so I got confused.

**Lunu**)I'm sorry I didn't understand the question at all so I got confused.

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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(Original post by

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.

**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.

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#15

(Original post by

Thanks

**Lunu**)Thanks

To ensure complex differentiability you need to make sure C-R is satisfied and the partial derivatives are also continuous.

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#16

(Original post by

Does Re(z)Im(z) + iIm(z) = xy +iy

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

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#17

(Original post by

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.

**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.

**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 of

where is a general complex number.

So here, you are looking at the limit as of

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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