Absolute value of complex exponent

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Mondo

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I know that the modulus for [imath]|e^{i\theta}| = 1[/imath] that is by definition of complex number in polar form. But I wonder why wolframalpha says the exact result is [imath]|e^{i\theta}| = e^{-Im(\theta)}[/imath] - what is [imath]Im{\theta}[/imath]? [imath]\theta[/imath] is a real number so it will always be 0.
 
Mondo said:
I know that the modulus for [imath]|e^{i\theta}| = 1[/imath] that is by definition of complex number in polar form. But I wonder why wolframalpha says the exact result is [imath]|e^{i\theta}| = e^{-Im(\theta)}[/imath] - what is [imath]Im{\theta}[/imath]? [imath]\theta[/imath] is a real number so it will always be 0.
Click to expand...
Here is what they say:

1689025290092.png​

Notice that last bit: they do know what the answer is if you assume [imath]\theta[/imath] 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 [imath]|e^{iz}|[/imath], where z could be anything.

Yes, we have a tendency to make assumptions based on common usage, that are not inherently necessary!
 
@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 [imath]e^{-Im(\theta)}[/imath]? More so it appears to contradict the true result - let's take any complex number i.e [imath]20i[/imath] then [imath]e^{-Im(20i)} = e^{-20} \ne 1[/imath]
Additionally why there is a minus sign in the expeconent? For real numbers it is not needed as it will always evaluate to [imath]e^{0} = 1[/imath]
 
Mondo said:
@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 [imath]e^{-Im(\theta)}[/imath]? More so it appears to contradict the true result - let's take any complex number i.e [imath]20i[/imath] then [imath]e^{-Im(20i)} = e^{-20} \ne 1[/imath]
Additionally why there is a minus sign in the expeconent? For real numbers it is not needed as it will always evaluate to [imath]e^{0} = 1[/imath]
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If [imath]\theta = 20i[/imath] then [imath]|e^{i\theta}| = \left|e^{i\cdot 20i}\right| = \left|e^{-20}\right| = e^{-20} = e^{-Im (20i)}[/imath]
And no, minus sign is not needed for real numbers, but it is needed in the general, i.e. non-trivial, case.
 
Mondo said:
@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 [imath]e^{-Im(\theta)}[/imath]? More so it appears to contradict the true result - let's take any complex number i.e [imath]20i[/imath] then [imath]e^{-Im(20i)} = e^{-20} \ne 1[/imath]
Additionally why there is a minus sign in the exponent? For real numbers it is not needed as it will always evaluate to [imath]e^{0} = 1[/imath]
Click to expand...
It seems that you missed the point that [imath]e^{iz}=1[/imath] only when z is real. There is no contradiction!

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

Here is the general case: [math]|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)}[/math]
 
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