Why is it like this?

[slim ridge]

If s is a complex number and sigma is the real part of s, then why is it like this?

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[slim ridge]

Shouldn't it be like so:

|e^-t t^s-1| = | e^-t t^(sigma + itau)-1 | = |e^-t (t^sigma-1) * (t^itau)| = (e^-t t^sigma-1) * | t^itau |

Where does the factor of |t^itau| go?

[astral vigil]

Is t a real number?

[astral vigil]

It does look like it is the case

[spice pewter]

$|e^{-t}t^{s-1}|$

$|e^{-t}e^{\ln(t)(\sigma + iIm(s) - 1)}|$

$e^{-t}|e^{\ln(t)(\sigma-1)}e^{\ln(t)iIm(s)}|$

$e^{-t}t^{\sigma-1}|e^{\ln(t)iIm(s)}|$

$e^{-t}t^{\sigma-1}$

[slim ridge]

s is complex

[astral vigil]

Yes but what about t

I assume that t is real and nonnegative

[spice pewter]

modulus of e^ik = 1

[slim ridge]

ah yes

t is real

[spice pewter]

yeah I'm assuming so too

[astral vigil]

I am not entirely convinced about the ln part

[slim ridge]

where is e^ik in the expression?

Ravi

You forgot the i

[spice pewter]

$e^{\ln(t)Im(s)}$

shit i forgot the i

[slim ridge]

when you write s as sigma + Im(s)

[plush onyx]

Ravi #GTFOanemia#NoLifer

[slim ridge]

bruh xd

[spice pewter]

yeah-

[brisk bison]

to the title "why is it like this?" I'd like to answer : because of the axioms of the complex field

[spice pewter]

you get it right?

[slim ridge]

no

[spice pewter]

bruhhhh

[slim ridge]

Im s can be a real number

[astral vigil]

Because the exponential function is extended to complex numbers, so I am not sure if it still holds that t^s = exp(s ln(t))

[plush onyx]

Ravi #GTFOanemia#NoLifer

[slim ridge]

t^s = exp(ln(t)) is the definition

for s complex

Ahhhhhhh

[astral vigil]

The s disappeared

[slim ridge]

NOW I SEE

[astral vigil]

But otherwise then yes

[slim ridge]

exp(ln(s)*t)

[spice pewter]

yep

[astral vigil]

The other way around

exp(s ln(t))

[slim ridge]

WAIT

No

I still dont see??

e^(ln(t)iIm(s)) can be 1 only if ln(t)*Im(s)) is a multiple of 2pi

but this can be anything?

[astral vigil]

The modulus

Is 1

[slim ridge]

nvm you're right..

+close