Cauchy Riemann complex function real and imaginary parts

[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.

 

[SammyS]

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 $\dfrac{-1}{(1+z)^2}$ ?

 

[Redwaves]

Does if match $\dfrac{-1}{(1+z)^2}$ ?

As far as I can tell, I don't get the same answer.

 

[SammyS]

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 $\dfrac{1+x}{(1+x)^2 +y^2} - \dfrac{iy}{(1+x^2) + y^2}$

 

[Redwaves]

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.

 

[SammyS]

I have the same on paper, typo here, sorry.

Does that solve your problem ?

 

[Redwaves]

Does that solve your problem ?

I still don't get the same answer.

 

[SammyS]

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.

 

[Redwaves]

The thing is I don't get 1 at numerator.
$\frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{-(1+x)^2 + y^2}{((1+x)^2 + y^2)^2} + \frac{i2y(1+x)}{((1+x)^2 + y^2)^2}$

 

[vela]

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.

 

[SammyS]

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.

 

[Redwaves]

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.

 

[vela]

Did you try what @SammyS suggested?

 

[Redwaves]

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?

 

[vela]

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.

 

[PeroK]

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 $x, y$.

 

[SammyS]

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.)

 

[PeroK]

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.

 

[SammyS]

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 $(1+x)^4$ .

 

[PeroK]

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.