My Math Forum  

Go Back   My Math Forum > College Math Forum > Complex Analysis

Complex Analysis Complex Analysis Math Forum


Thanks Tree5Thanks
  • 1 Post By Greens
  • 1 Post By Micrm@ss
  • 1 Post By topsquark
  • 2 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
June 16th, 2019, 08:42 AM   #1
Member
 
Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

A calculation

Hello all,

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

All the best,

Integrator
Integrator is offline  
 
June 16th, 2019, 10:14 AM   #2
Member
 
Greens's Avatar
 
Joined: Oct 2018
From: USA

Posts: 87
Thanks: 59

Math Focus: Algebraic Geometry
$\displaystyle e^{i \theta} = \cos{(\theta)} + i \sin{(\theta)} $
Thanks from topsquark
Greens is offline  
June 16th, 2019, 12:52 PM   #3
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,226
Thanks: 908

Math Focus: Wibbly wobbly timey-wimey stuff.
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
topsquark is offline  
June 16th, 2019, 01:00 PM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,500
Thanks: 1372

$\cos(x^2)+i \sin(x^2) = e^{i x^2}$

$\left(e^{i x^2}\right)^x = e^{i x^3}$
romsek is offline  
June 16th, 2019, 02:18 PM   #5
Senior Member
 
Joined: Oct 2009

Posts: 841
Thanks: 323

Quote:
Originally Posted by romsek View Post
$\left(e^{i x^2}\right)^x = e^{i x^3}$
I'm afraid complex exponentiation doesn't work like that.
Thanks from topsquark
Micrm@ss is offline  
June 16th, 2019, 04:38 PM   #6
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,226
Thanks: 908

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Micrm@ss View Post
I'm afraid complex exponentiation doesn't work like that.
Okay you got me, too. Why doesn't it?

-Dan
Thanks from Greens
topsquark is offline  
June 16th, 2019, 06:36 PM   #7
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,500
Thanks: 1372

Quote:
Originally Posted by Micrm@ss View Post
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.
Thanks from topsquark and Greens
romsek is offline  
June 16th, 2019, 08:42 PM   #8
Member
 
Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

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
Integrator is offline  
Reply

  My Math Forum > College Math Forum > Complex Analysis

Tags
calculation



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
What is this calculation used for? MMath Elementary Math 3 July 7th, 2016 09:13 PM
Integral calculation based on coordinates / Area under Curve calculation joskevermeulen Calculus 1 December 29th, 2015 05:02 AM
A calculation Dacu Elementary Math 9 November 21st, 2014 04:34 PM
I need help with this calculation diegosened Algebra 2 April 7th, 2010 04:53 AM
How can I do the calculation here r-soy Calculus 1 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.