Absolute value of complex exponent

[Mondo]

I know that the modulus for $|e^{i\theta}| = 1$ that is by definition of complex number in polar form. But I wonder why wolframalpha says the exact result is $|e^{i\theta}| = e^{-Im(\theta)}$ - what is $Im{\theta}$? $\theta$ is a real number so it will always be 0.

 

[Dr.Peterson]

I know that the modulus for $|e^{i\theta}| = 1$ that is by definition of complex number in polar form. But I wonder why wolframalpha says the exact result is $|e^{i\theta}| = e^{-Im(\theta)}$ - what is $Im{\theta}$? $\theta$ is a real number so it will always be 0.

Here is what they say:

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Notice that last bit: they do know what the answer is if you assume $\theta$ is real! But you didn't tell them it is, so they can't assume that!

In your world (writing a complex number in polar form), theta is always an angle, and therefore real. But it is just a variable, and could be anything. What they've told you is one way to simplify $|e^{iz}|$, where z could be anything.

Yes, we have a tendency to make assumptions based on common usage, that are not inherently necessary!

 

[Mondo]

@Dr.Peterson, I agree with you that I didn't specify if my argument is real or imaginary or complex. But the big question is why they say it is $e^{-Im(\theta)}$? More so it appears to contradict the true result - let's take any complex number i.e $20i$ then $e^{-Im(20i)} = e^{-20} \ne 1$
Additionally why there is a minus sign in the expeconent? For real numbers it is not needed as it will always evaluate to $e^{0} = 1$

 

[blamocur]

@Dr.Peterson, I agree with you that I didn't specify if my argument is real or imaginary or complex. But the big question is why they say it is $e^{-Im(\theta)}$? More so it appears to contradict the true result - let's take any complex number i.e $20i$ then $e^{-Im(20i)} = e^{-20} \ne 1$
Additionally why there is a minus sign in the expeconent? For real numbers it is not needed as it will always evaluate to $e^{0} = 1$

If $\theta = 20i$ then $|e^{i\theta}| = \left|e^{i\cdot 20i}\right| = \left|e^{-20}\right| = e^{-20} = e^{-Im (20i)}$
And no, minus sign is not needed for real numbers, but it is needed in the general, i.e. non-trivial, case.

 

[Dr.Peterson]

@Dr.Peterson, I agree with you that I didn't specify if my argument is real or imaginary or complex. But the big question is why they say it is $e^{-Im(\theta)}$? More so it appears to contradict the true result - let's take any complex number i.e $20i$ then $e^{-Im(20i)} = e^{-20} \ne 1$
Additionally why there is a minus sign in the exponent? For real numbers it is not needed as it will always evaluate to $e^{0} = 1$

It seems that you missed the point that $e^{iz}=1$ only when z is real. There is no contradiction!

You just made a false assumption based on the fact that $e^{i\theta}$ is a complex number with modulus 1 when $\theta$ is real.

Here is the general case: $$|e^{iz}|=|e^{i(x+iy)}|=|e^{-y+ix}|=|e^{-y}e^{ix}|=|e^{-y}||\cos(x)+i\sin(x)|=e^{-y}=e^{-Im(z)}$$

 

[Mondo]

@Dr.Peterson, @blamocur , thank you. All is clear now

 

[Dr.Peterson]

It seems that you missed the point that $e^{iz}=1$ only when z is real. There is no contradiction!

For the record, I meant to say

It seems that you missed the point that $|e^{iz}|=1$ only when z is real. There is no contradiction!​