June 16th, 2019, 08:42 AM  #1 
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  A calculation
Hello all, Calculate $\displaystyle [\cos{(x^2)}+i\sin{(x^2)}]^x$. All the best, Integrator 
June 16th, 2019, 10:14 AM  #2 
Member Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry 
$\displaystyle e^{i \theta} = \cos{(\theta)} + i \sin{(\theta)} $

June 16th, 2019, 12:52 PM  #3 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  
June 16th, 2019, 01:00 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 
$\cos(x^2)+i \sin(x^2) = e^{i x^2}$ $\left(e^{i x^2}\right)^x = e^{i x^3}$ 
June 16th, 2019, 02:18 PM  #5 
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 326  
June 16th, 2019, 04:38 PM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  
June 16th, 2019, 06:36 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390  
June 16th, 2019, 08:42 PM  #8  
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  Quote:
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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