Recent content by Brian_D
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Finding the nth roots of a complex number
Thank you, annuhere3. Very clear and comprehensive.- Brian_D
- Post #23
- Forum: Calculus and Beyond Homework Help
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Compton shift problem
Got it, thank you PeroK.- Brian_D
- Post #10
- Forum: Introductory Physics Homework Help
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Finding the nth roots of a complex number
Duly noted, thank you.- Brian_D
- Post #22
- Forum: Calculus and Beyond Homework Help
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Finding the nth roots of a complex number
OK, but the precision of the approximation is irrelevant to what I was talking about.- Brian_D
- Post #20
- Forum: Calculus and Beyond Homework Help
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Compton shift problem
Important reminder, thank you PeroK.- Brian_D
- Post #8
- Forum: Introductory Physics Homework Help
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Compton shift problem
Yes, I see mistake mistake now, and with this correction I also get 18%. Thank you, renormalize.- Brian_D
- Post #7
- Forum: Introductory Physics Homework Help
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Compton shift problem
Yes, thank you Steve4Physics, my mistake was using the wrong value for cos 110 degrees.- Brian_D
- Post #6
- Forum: Introductory Physics Homework Help
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Compton shift problem
(1-cos 110 degrees) is .342, so the Compton shift is .00243 * 10^-9 * .342 = 8.31 * 10^-13 m. The wavelength of the scattered photon is the Compton shift plus the initial wavelength, that is, (8.31 * 10^-13) + (15 * 10^-12), which equals 1.58 * 10^-11 m. So the energy of the scattered photon...- Brian_D
- Thread
- Replies: 9
- Forum: Introductory Physics Homework Help
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Finding the nth roots of a complex number
Yes. I said that the cube root of 2*sqrt(2) is 1.41, which is also correct.- Brian_D
- Post #18
- Forum: Calculus and Beyond Homework Help
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Finding the nth roots of a complex number
The signs are wrong because -2 +2i and 2 -2i have the same absolute value. Everything is crystal clear to me now. Thank you, FactChecker!- Brian_D
- Post #14
- Forum: Calculus and Beyond Homework Help
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Finding the nth roots of a complex number
I see now that there was a mistake in my previous post. If the modulus is supposed to be the absolute value of -2 + 2i, then that should be 2*sqrt(2), and the cube root of that is 1.41, not 1.26. With this correction, the formula gives 1.36-.365i and the cube of that is 2-2i. For some reason...- Brian_D
- Post #13
- Forum: Calculus and Beyond Homework Help
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Finding the nth roots of a complex number
I posted a reply, but somehow it got lost so I'll try again. Thank you, FactChecker, got it. So let's see how that works. For k=0, the formula now gives us 1.26*(.966-.259i)=(1.22-.326i). But when we cube this result, we get 1.43-1.42i, not -2+2i. So I'm still not getting a correct solution...- Brian_D
- Post #12
- Forum: Calculus and Beyond Homework Help
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Finding the nth roots of a complex number
Thank you, Renormalize. However, if you check Mathematica's (rectangular) solutions by cubing each one, only the first solution produces the original expression -2+2i. I tried the same problem with Maple, and, like Mathematica, two of Maple's solutions were in an oddly complicated form, but...- Brian_D
- Post #10
- Forum: Calculus and Beyond Homework Help