Cauchy Riemann complex function real and imaginary parts

Redwaves
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Homework Statement
Find real and imaginary parts of a complex function
Relevant Equations
##f(z)= \frac{1}{1+z}##
Hi,
I have to find the real and imaginary parts and then using Cauchy Riemann calculate ##\frac{df}{dz}##
First, ##\frac{df}{dz} = \frac{1}{(1+z)^2}##

Then, ##f(z)= \frac{1}{1+z} = \frac{1}{1+ x +iy} => \frac{1+x}{(1+x)^2 +y^2} - \frac{-iy}{(1+x^2) + y^2}##
thus, ##\frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{-(1+x)^2 + y^2}{((1+x)^2 + y^2)^2} + \frac{i2y(1+x)}{((1+x)^2 + y^2)^2}##

However, it doesn't match with ##\frac{1}{(1+z)^2}##,where is my error.
 
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Redwaves said:
Hi,
I have to find the real and imaginary parts and then using Cauchy Riemann calculate ##\frac{df}{dz}##
First, ##\frac{df}{dz} = \frac{1}{(1+z)^2}##

Then, ##f(z)= \frac{1}{1+z} = \frac{1}{1+ x +iy} => \frac{1+x}{(1+x)^2 +y^2} - \frac{-iy}{(1+x^2) + y^2}##
thus, ##\frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{-(1+x)^2 + y^2}{((1+x)^2 + y^2)^2} + \frac{i2y(1+x)}{((1+x)^2 + y^2)^2}##

However, it doesn't match with ##\frac{1}{(1+z)^2}##,where is my error.
Does if match ##\dfrac{-1}{(1+z)^2}## ?
 
SammyS said:
Does if match ##\dfrac{-1}{(1+z)^2}## ?
As far as I can tell, I don't get the same answer.
 
Redwaves said:
As far as I can tell, I don't get the same answer.
You have that ## \dfrac{1}{1+ x +iy} = \dfrac{1+x}{(1+x)^2 +y^2} - \dfrac{-iy}{(1+x^2) + y^2} ## .

Check that.

I get ##\dfrac{1+x}{(1+x)^2 +y^2} - \dfrac{iy}{(1+x^2) + y^2} ##
 
SammyS said:
You have that ## \dfrac{1}{1+ x +iy} = \dfrac{1+x}{(1+x)^2 +y^2} - \dfrac{-iy}{(1+x^2) + y^2} ## .

Check that.

I get ##\dfrac{1+x}{(1+x)^2 +y^2} - \dfrac{iy}{(1+x^2) + y^2} ##
I have the same on paper, typo here, sorry.
 
Redwaves said:
I have the same on paper, typo here, sorry.
Does that solve your problem ?
 
SammyS said:
Does that solve your problem ?
I still don't get the same answer.
 
Maybe I was too subtle in Post #2 .

I asked that question because ##\dfrac{df}{dz} = \dfrac{-1}{(1+z)^2} ##

You have the wrong sign.
 
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The thing is I don't get 1 at numerator.
##\frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{-(1+x)^2 + y^2}{((1+x)^2 + y^2)^2} + \frac{i2y(1+x)}{((1+x)^2 + y^2)^2}##
 
  • #10
You shouldn't expect 1 in the numerator. Here's a nudge. You have
$$-\frac{1}{(1+z)^2} = -\frac{1}{[(1+x)+iy]^2}.$$ Note that the denominator isn't real.
 
  • #11
Follow @vela 's suggestion. Otherwise, if you really want to get a 1 in the numerator, rationalize the numerator for ##\frac{df}{dz} =
\dfrac{-(1+x)^2 + y^2 + i2y(1+x)}{((1+x)^2 + y^2)^2} ##

That might turn out to be much more work.
 
  • #12
SammyS said:
Follow @vela 's suggestion. Otherwise, if you really want to get a 1 in the numerator, rationalize the numerator for ##\frac{df}{dz} =
\dfrac{-(1+x)^2 + y^2 + i2y(1+x)}{((1+x)^2 + y^2)^2} ##

That might turn out to be much more work.
If I don't get ##-\frac{1}{(1+z)^2}## How should I know I have the right answer?
##\frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{-(1+x)^2 + y^2}{((1+x)^2 + y^2)^2} + \frac{i2y(1+x)}{((1+x)^2 + y^2)^2} = -\frac{1}{(1+z)^2}##
I don't see how this is right.
 
  • #13
Did you try what @SammyS suggested?
 
  • #14
vela said:
Did you try what @SammyS suggested?
##\dfrac{-(1+x)^2 + y^2 + i2y(1+x)}{((1+x)^2 + y^2)^2}##

Should the denominator have a i like, ##\dfrac{-(1+x)^2 + y^2 + i2y(1+x)}{((1+x)^2 + iy^2)^2}## to rationalize?
 
  • #15
No, you can't just arbitrarily insert a factor of ##i## into the denominator. The suggestion was to rationalize the numerator, which is already complex.
 
  • #16
Redwaves said:
If I don't get ##-\frac{1}{(1+z)^2}## How should I know I have the right answer?
##\frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{-(1+x)^2 + y^2}{((1+x)^2 + y^2)^2} + \frac{i2y(1+x)}{((1+x)^2 + y^2)^2} = -\frac{1}{(1+z)^2}##
I don't see how this is right.
Why do you think it's wrong? Just because two expressions don't look the same doesn't mean they aren't equivalent. For example, ##\cos^2 x + \sin^2 x = 1##.

The obvious thing to do is express the RHS as a function of ##x, y##.
 
  • #17
vela said:
Did you try what @SammyS suggested?
The algebra can be a bit tricky and/or very messy.

It requires you to multiply by the complex conjugate of the numerator and then divide by the same thing. Leave the denominators alone until after multiplying and simplifying the numerators.

When working with the numerator(s) it may be useful to substitute a single variable such as a for (1+x) .

I.e. Write ## y^2 - (1+x)^2 + i2y(1+x)## as ## y^2 - a^2 + i2ya## ,

then multiply ##\left( (y^2 - a^2) + i2ya \right) \left( (y^2 - a^2) - i2ya \right) ## and simplify.

(Edited per @PeroK 's post, below.)
 
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  • #18
SammyS said:
When working with the numerator(s) it may be useful to substitute a single variable such as u for (1+x) .
Note that ##u## is already defined. The algebra is easy enough with ##1+x##, IMO.
 
  • #19
PeroK said:
Note that ##u## is already defined. The algebra is easy enough with ##1+x##, IMO.
Thanks. I have changed from ##u## to ##a##.

I agree that the algebra isn't too difficult. Just trying to avoid having OP multiply out the ##(1+x)^2## worse yet, multiplying out ##(1+x)^4## .
 
  • #20
SammyS said:
I agree that the algebra isn't too difficult. Just trying to avoid having OP multiply out the ##(1+x)^2## worse yet, multiplying out ##(1+x)^4## .
You don't have to do any of that.
 
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