Phase factor in quantum mechanics

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Shreya
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Homework Statement
I watched a lecture in which the professor called the circle expression as a pure phase and took its absolute value to be 1. I don't understand how it's absolute value is 1.
Please refer the image for the circled expression.
Relevant Equations
Wave function at x,t
Please be kind to help.I would be grateful
Handwritten wave packet expansion: psi(x,t) = e^(-iω(k0)t) e^(ik0 dω/dk|k0 t) times the integral of dk phi(k) e^(i(kx - ω t)), with the first exponential factor circled in red.
 
Any complex number of the form ##e^{i\theta}## has absolute value (modulus) ##1##.
 
Shreya said:
Can you please explain why is it so?
The norm of a complex number can be calculated using the number ##z## and ##z^*=\bar{z}## the complex conjugate.
So, given ##z=e^{i\theta}, \quad \theta \in \mathbb{R}##, you should be able to compute ##z^*## and its norm.
You will see that the norm is 1 independent on ##\theta##.
 
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Shreya said:
Can you please explain why is it so?
Any complex number can be written ##re^{i\theta}##, where ##r## is the modulus and ##\theta## is the angle in the complex plane. That's fairly elementary, I'm sorry to say!