My Math Forum i don't know how to google is -3=2^x a possible complex no.?

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 May 28th, 2012, 05:42 AM #1 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 i don't know how to google is -3=2^x a possible complex no.? i know complex number got square root -3 such that [ i ] x [ square root +3 ] but is $-3=2^x$ got imagery figures as solutions?
 May 28th, 2012, 06:14 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: i don't know how to google is -3=2^x a possible complex $2^x=-3=3i^2\Rightarrow \log_{2}2^x=\log_{2}\left(3i^2\right)\Rightarrow x=\frac{\ln\left(3i^2\right)}{\ln(2)}=\frac{\ln(3) +\ln(i^2)}{\ln(2)}=\frac{\ln(3)+2\ln(i)}{\ln(2)}=\ frac{\ln(3)+\pi i (2n+1)}{\ln(2)}\;\;for\;\;n\in\mathbb{Z}$
 July 5th, 2012, 10:56 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: i don't know how to google is -3=2^x a possible complex To extend what Zardoz said, we can represent the number x+ iy as the point (x, y) in the complex plane and then change to polar coordinates r, $\theta$ with $x= r cos(\theta)$ and $y= r sin(\theta)$. Further, since we can show that $e^{i\theta}= cos(\theta)+ i sin(\theta)$, $x+ iy= re^{i\theta}$. Taking the logarithm of that, $ln(x+ iy)= ln(r)+ ln(e^i\theta)= ln(r)+ i \theta$. But sine and cosine are periodic with period $2\pi$ so that $x+ iy= r e^{i\theta+ 2n\pi i}$ for any integer n. And then $ln(x+ iy)= ln(r)+ i\theta+ 2n\pi$. That's the reason for the "$\i i(2n+1)$".

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