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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
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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
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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.

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