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 Complex Analysis Complex Analysis Math Forum

 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
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Quote:
 Originally Posted by idontknow $\displaystyle z=\ln x +i \pi x =0$ . x=? , $\displaystyle x\in \mathbb{R}$ .
If we assume x is real there is a problem. We equate real and imaginary parts on both sides of the equation, so
$\displaystyle ln(x) = 0$

$\displaystyle \pi x = 0$

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 . Thanks from topsquark February 27th, 2019, 03:14 PM   #4
Math Team

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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by idontknow That is the reason I posted it . For x=-1 it holds true but cannot find the method .
You are right, I didn't consider x < 0.

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. Thanks from topsquark Last edited by Maschke; February 27th, 2019 at 04:17 PM. Tags complex, equation, variable Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Robart Complex Analysis 1 February 3rd, 2017 11:59 AM mhhojati Linear Algebra 3 January 20th, 2017 04:41 AM Adam Ledger Number Theory 7 May 2nd, 2016 02:42 AM dpwheelwright Algebra 5 August 19th, 2012 12:48 AM Kosta Algebra 2 December 23rd, 2008 03:04 PM

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