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February 27th, 2019, 12:54 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87  Equation with complex variable
$\displaystyle z=\ln x +i \pi x =0$ . x=? , $\displaystyle x\in \mathbb{R}$ . Last edited by idontknow; February 27th, 2019 at 01:00 PM. 
February 27th, 2019, 01:33 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,256 Thanks: 926 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle ln(x) = 0$ $\displaystyle \pi x = 0$ Thus there is no solution for x. So x has to be complex. Let $\displaystyle x = r e^{i \theta + 2i k \pi}$ where r, $\displaystyle \theta$ are real and k is any integer. Then $\displaystyle ln(x) = ln \left ( r e^{i \theta + 2i \pi k} \right ) = ln(r) + i \theta + 2i \pi k$. Now your equation becomes $\displaystyle z = ln \left ( r e^{ i \theta + 2i k \pi} \right ) + i \pi r e^{i \theta + 2i k \pi} = 0$ Now equate the real and imaginary parts. Dan  
February 27th, 2019, 01:38 PM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87 
That is the reason I posted it . For x=1 it holds true but cannot find the method . 
February 27th, 2019, 03:14 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,256 Thanks: 926 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
All right. So with r > 0: $\displaystyle ln(r) = ln(r) + i \pi$ So the original equation is $\displaystyle ln(r) + i \pi (r) = ln(r) + i \pi  i \pi r = 0$ Thus equating real and complex parts: $\displaystyle ln(r) = 0$ $\displaystyle \pi  \pi r = 0$ So the only solution is r = 1, thus x = 1. (You can do this my way and check to see if there is more than one solution, but I doubt there will be so I'm not going to bother.) Dan  
February 27th, 2019, 03:31 PM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87 
This may also work . $\displaystyle x( \frac{\ln x }{x} +i\pi )=0 \; $ and x cannot be 0 . $\displaystyle e^{\frac{\ln x }{x} }=e^{i\pi }=1$ or $\displaystyle x^{\frac{1}{x}}=1$ . Last edited by idontknow; February 27th, 2019 at 03:37 PM. 
February 27th, 2019, 04:12 PM  #6 
Senior Member Joined: Aug 2012 Posts: 2,355 Thanks: 737 
The quick way is to just remember the special case of Euler's formula, $e^{i \pi} = 1$. It follows by definition that $log 1 = i \pi$ plus periodicity. [$\log$ is natural log, that's actually the convention past calculus class]. But I think you have to say something about choosing a branch. I'm not actually sure what the rules are for writing equations when one of the terms actually has multiple values as in this case. I may have missed that day in complex variables. I don't think they ever mentioned it. Last edited by Maschke; February 27th, 2019 at 04:17 PM. 

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