Absolute value of exponetial?

1. Mar 16, 2013

arierreF

I want to calculate $|e^{a^{2} + \frac{it}{m\hbar}}|^{2}$

i is imaginary unit.

my trie:

$a^{2} + \frac{it}{2m\hbar}$ is a complex number so its module is:

$\sqrt{a^{4} + \frac{t^{2}}{m^{2}\hbar^{2}}}$

= $\sqrt{a^{4}(1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}})}$

$a^2\sqrt{1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}}}$

So the solution is:

$(e^{a^2\sqrt{1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}}}})^{2}$

=
$e^{2a^2\sqrt{1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}}}}$

My friend said that is is definitely wrong. And i actually think that it is wrong.

can somebody tell me where?

2. Mar 16, 2013

Fredrik

Staff Emeritus
Is there a theorem that says that $|e^z|=e^{|z|}$ for all complex numbers z? Since I'm asking, you can guess that the answer is no. I suggest that you verify this by finding a counterexample.

Do you know anything about the exponential function that would allow you to rewrite $e^{a^2+\frac{it}{m\hbar}}$ in a different way?

3. Mar 16, 2013

arierreF

yup

$e^{(x+b)} =e^{x}.e^{b}$

so i am going to have a e^(real part)e^(imaginary part)

absolute value of e^(imaginary) = 1 right (euler's formula)

so the solution is $|e^{(real part)}|^2 = e^{(2(a^{2}))}$

Last edited: Mar 16, 2013
4. Mar 16, 2013

Dick

Yes, correct.

5. Mar 16, 2013

arierreF

ok thanks for help