You are Here: Home >< Maths

# Complex numbers

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
1. 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.
2. Can anyone answer if Re(z)Im(z) + iIm(z) = xy + iy?
3. (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 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.
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 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.
5. (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.
6. 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?
7. It is complex differentiable where the Cauchy-Riemann equations hold. y=1 and x=0. So the point is (0,1).
8. (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!!
9. (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.
10. (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.
11. From the working out in the op, it was obvious the question was at what points is the function complex differentiable at.
12. (Original post by Lunu)
Actually isn't it (0, -1)?
Look at the Cauchy- Riemann equations
13. (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.
14. (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
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.
16. (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'
17. (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 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.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 15, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams