# Cauchy Riemann complex function real and imaginary parts

• Redwaves
In summary: 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##...Yes, that's a good point. I was thinking more about the case where you have to multiply out just the numerator, in which case the substitution is unnecessary. But if you also have to multiply out the denominator, then substituting might be a good idea.
Redwaves
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.

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.

Last edited:
FactChecker
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}##

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.

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.

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.

Did you try what @SammyS suggested?

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?

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.

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##.

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.)

Last edited:
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.

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## .

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.

## 1. What is the Cauchy-Riemann complex function?

The Cauchy-Riemann complex function is a mathematical concept that describes the behavior of a complex-valued function in terms of its real and imaginary parts. It is based on the Cauchy-Riemann equations, which state that a function is complex differentiable if and only if its real and imaginary parts satisfy certain conditions.

## 2. What are the Cauchy-Riemann equations?

The Cauchy-Riemann equations are a set of two partial differential equations that relate the real and imaginary parts of a complex function. They state that if a complex function is differentiable, then its real and imaginary parts must satisfy these equations at every point in its domain.

## 3. What is the significance of the Cauchy-Riemann complex function?

The Cauchy-Riemann complex function is significant because it helps us understand the behavior of complex functions and their derivatives. It also plays a crucial role in the study of complex analysis, which has many practical applications in fields such as physics, engineering, and economics.

## 4. How can the Cauchy-Riemann complex function be used to solve problems?

The Cauchy-Riemann complex function can be used to solve problems involving complex functions, such as finding the derivative of a complex function or determining if a function is analytic. It can also be used to solve problems in areas such as fluid dynamics, electromagnetism, and signal processing.

## 5. Are there any limitations to the Cauchy-Riemann complex function?

While the Cauchy-Riemann complex function is a powerful tool in complex analysis, it does have some limitations. It can only be applied to functions that are differentiable, and it may not always provide a unique solution to a problem. Additionally, it is not applicable to real-valued functions.

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