1. ## modulus

i was wondering how do you remove the modulus of a function e.g. if you had modulus of y how do you find y.
i thought you would just do the inverse of (A^2+B^2)^0.5 but i was told this was wrong?? any help would be appreciated

Hello hmmmm
Originally Posted by hmmmm
i was wondering how do you remove the modulus of a function e.g. if you had modulus of y how do you find y.
i thought you would just do the inverse of (A^2+B^2)^0.5 but i was told this was wrong?? any help would be appreciated
Not quite sure what you mean here. Can you give an example?

3. aorry it wasnt that great an explanation if you integrated 1/z dz=2x+4you would end up with ln|z|=x^2+4x+c then getting rid of the ln you would have
|z|=e^(x^2+4x+c) i was wondering how you would write this for z=??

4. Hello,
Originally Posted by hmmmm
aorry it wasnt that great an explanation if you integrated 1/z dz=2x+4you would end up with ln|z|=x^2+4x then getting rid of the ln you would have
|z|=e^(x^2+4x) i was wondering how you would write this for z=??
Since you talk about modulus, I assume that z is complex, then you can write $\ln(z)$, and this defined a value that exists.

Note that if z and z' are complex numbers, we have :
$\ln(z)=\ln(z') \Leftrightarrow \text{Arg}(z)=\text{Arg}(z')+2k\pi$ and $|z|=|z'|$

If you're talking about absolute values, and z is a real number, then you indeed have $\ln|z|$
So $|z|=e^{x^2+4x}$
Hence z can be equal to $+e^{x^2+4x}$ or $- e^{x^2+4x}$

Also, don't forget the integration constant.

5. Originally Posted by hmmmm
aorry it wasnt that great an explanation if you integrated 1/z dz=2x+4you would end up with ln|z|=x^2+4x then getting rid of the ln you would have
|z|=e^(x^2+4x) i was wondering how you would write this for z=??
Once you exponentiate you can get rid of the modulus since $e^{x^2+4x}$ is always positive.

By the way, you forgot the '+ C' in your solution .... The final solution is $z = e^{x^2+4x + C} = A e^{x^2+4x}$.

6. Originally Posted by mr fantastic
Once you exponentiate you can get rid of the modulus since $e^{x^2+4x}$ is always positive.

7. ## thanks

thanks, if it is an absolute value is this always how you would remove the modulus?

8. Originally Posted by hmmmm
thanks, if it is an absolute value is this always how you would remove the modulus?
Yes.

If a > 0, then if $|x|=a$, $x=\pm a$

9. Originally Posted by Moo
I think it's reasonable to assume that x is real. In which case the exponential is always going to be greater than zero.

The real problem is that the OP has not specified what z and x represent nor what the broader context of the question is .....

10. Originally Posted by mr fantastic
Once you exponentiate you can get rid of the modulus since $e^{x^2+4x}$ is always positive.

By the way, you forgot the '+ C' in your solution .... The final solution is $z = e^{x^2+4x + C} = A e^{x^2+4x}$.
You missed a step, just for the benefit of the OP...

Let's just remember that if

$|x| = a$ then $x = a$ or $x = -a$.

This makes sense because the modulus is really the SIZE or LENGTH or MAGNITUDE of what is inside it.

So if we had, as above

$|z| = e^{x^2 + 4x + C}$

we use the index law $a^{m + n} = a^m\,a^n$ to get

$|z| = e^C\,e^{x^2 + 4x}$

So $z = \pm e^C\,e^{x^2 + 4x}$

But since $e^C$ is an arbitrary constant, so is $-e^C$. Therefore we could write is as another letter, say A.

So $z = Ae^{x^2 + 4x}$.

Make sense?

11. thinik so, are you saying that as |z| is the magnitude we must do the inverse of what we did to get the magnitude?? this is what i origonally thought however i was told this was wrong by my teacher??

12. Originally Posted by hmmmm
thinik so, are you saying that as |z| is the magnitude we must do the inverse of what we did to get the magnitude?? this is what i origonally thought however i was told this was wrong by my teacher??
Yes that's correct in essentials.

I think your teacher is just getting a bit pedantic - technically speaking if you deal with mods you have to deal with conditions, but if dealing with integration you're assumed to already know this and so can do the "inverse modding" as you say.

13. thanks alot