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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. Goosline Alfredo Pereira (Goos)‏ @Goosline 20 Mar 2019
      • Report Tweet
      • Report NetzDG Violation

      Goosline Alfredo Pereira (Goos) Retweeted 〈 Berger | Dillon 〉

      No hay una prueba en donde podamos demostrar que i^i es realmente independientemente de la rama?https://twitter.com/InertialObservr/status/1108578964891525120 …

      Goosline Alfredo Pereira (Goos) added,

      〈 Berger | Dillon 〉 @InertialObservr
      Replying to @Goosline
      In a general branch however, 1 = e^{i*2πn} --> 1^i = 1, e^{-2π}, e^{-4π}...
      1 reply 0 retweets 0 likes
    2. 〈 Berger | Dillon 〉‏ @InertialObservr 20 Mar 2019
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      Replying to @Goosline

      No, in general z(θ)= e^{iθ} = e^{iθ + i2πn} implica que z(θ) no tiene un valor único

      2 replies 0 retweets 0 likes
    3. Goosline Alfredo Pereira (Goos)‏ @Goosline 20 Mar 2019
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      Replying to @InertialObservr

      Ya va, lo dije mal, "donde i^i sea real independientemente de la rama" o sea, porque no puede ser que para la rama principal sea real y para otra no :/ ¿o me equivoco?

      1 reply 0 retweets 1 like
    4. 〈 Berger | Dillon 〉‏ @InertialObservr 20 Mar 2019
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      Replying to @Goosline

      así es!

      2 replies 0 retweets 0 likes
      〈 Berger | Dillon 〉‏ @InertialObservr 20 Mar 2019
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      Replying to @InertialObservr @Goosline

      Lo siento.. yo pensé que habías dicho i^i = 1, no i^i ε R

      9:11 PM - 20 Mar 2019
      • 1 Like
      • Goosline Alfredo Pereira (Goos)
      1 reply 0 retweets 1 like
        1. Goosline Alfredo Pereira (Goos)‏ @Goosline 20 Mar 2019
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          Replying to @InertialObservr

          No problem.

          0 replies 0 retweets 1 like
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