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June 16th, 2019, 09:42 AM   #1
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A calculation

Hello all,

Calculate $\displaystyle [\cos{(x^2)}+i\sin{(x^2)}]^x$.

All the best,

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June 16th, 2019, 11:14 AM   #2
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$\displaystyle e^{i \theta} = \cos{(\theta)} + i \sin{(\theta)} $
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June 16th, 2019, 01:52 PM   #3
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Quote:
Originally Posted by Integrator View Post
Hello all,

Calculate $\displaystyle [\cos{(x^2)}+i\sin{(x^2)}]^x$.

All the best,

Integrator
We need some restrictions here. I'm presuming that x is a real number so we have to be careful about $\displaystyle cos( x^2 ) \leq 0$ for various values of x.

-Dan
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June 16th, 2019, 02:00 PM   #4
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$\cos(x^2)+i \sin(x^2) = e^{i x^2}$

$\left(e^{i x^2}\right)^x = e^{i x^3}$
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June 16th, 2019, 03:18 PM   #5
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$\left(e^{i x^2}\right)^x = e^{i x^3}$
I'm afraid complex exponentiation doesn't work like that.
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June 16th, 2019, 05:38 PM   #6
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I'm afraid complex exponentiation doesn't work like that.
Okay you got me, too. Why doesn't it?

-Dan
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June 16th, 2019, 07:36 PM   #7
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I'm afraid complex exponentiation doesn't work like that.
and you're just going to leave it at that?

How about being useful and showing us how it works then.

Sheesh.
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June 16th, 2019, 09:42 PM   #8
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Quote:
Originally Posted by romsek View Post
$\cos(x^2)+i \sin(x^2) = e^{i x^2}$

$\left(e^{i x^2}\right)^x = e^{i x^3}$
Hello,

I do not understand!I think that $\displaystyle \left(e^{i x^2}\right)^{x} =e^{i^x\cdot x^{2x}}$ where $\displaystyle x\in \mathbb R , x>0$ is an identity and so $\displaystyle \left(e^{i x^2}\right)^{x} = e^{i x^3}$ is an equation.
-----------------------
How do we calculate $\displaystyle [\cos(x^2)+i \sin(x^2)]^x$?

All the best,

Integrator
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