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idku

  • one year ago

Maybe it is too early for me to ask this question, but why .... ?

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  1. idku
    • one year ago
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    \(\large\color{black}{ \displaystyle \left|1+i\right|=\sqrt{2} }\)

  2. idku
    • one year ago
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    (and so \(\large\color{black}{ \displaystyle \left|1-i\right|=\sqrt{2} }\) )

  3. idku
    • one year ago
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    can someone explain this?

  4. anonymous
    • one year ago
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    both \(1+i\) and\(1-i\) are \(\sqrt2\) units away from \(0\)

  5. idku
    • one year ago
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    Is that because i is a rotation?

  6. idku
    • one year ago
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    A 45 degree rotation?

  7. anonymous
    • one year ago
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    |dw:1442713983303:dw|

  8. anonymous
    • one year ago
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    just like both \((1,1)\) and \((1,-1)\) have distance \(\sqrt2\) from \((0,0)\)

  9. anonymous
    • one year ago
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    or if you prefer, just like the diagonal of any square with side 1 has length \(\sqrt2\)

  10. idku
    • one year ago
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    Wait, so i is a diagonal?

  11. anonymous
    • one year ago
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    |dw:1442714099587:dw|

  12. idku
    • one year ago
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    |dw:1442714123626:dw|

  13. anonymous
    • one year ago
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    ok i didn't mean to confuse you \(i\) is not the diagonal the absolute value of a complex number is its distance from 0

  14. idku
    • one year ago
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    and that would mean that \(\large\color{black}{ \displaystyle \left|a+i\right|=a\sqrt{2} }\) \(\forall a\in\mathbb{R}\)

  15. idku
    • one year ago
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    wait, sorry for a>0 too

  16. anonymous
    • one year ago
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    oh no

  17. idku
    • one year ago
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    oh, that is not right I guess. (I just heard someone said that i is sort of a rotation)

  18. anonymous
    • one year ago
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    \[|a+i|=\sqrt{a^2+1}\]

  19. anonymous
    • one year ago
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    what is true is that \[|a+bi|=|a-bi|=\sqrt{a^2+b^2}\]

  20. anonymous
    • one year ago
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    it has no bearing on the sign of \(a\) or \(b\) since you are going to square them in any case

  21. idku
    • one year ago
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    oh, ok.... I guess I will explore that a little more. For now I will just have the rules at least;)

  22. idku
    • one year ago
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    yes as long as a, b are Real...

  23. anonymous
    • one year ago
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    in \(a+bi\) \(a\) and \(b\) are always real

  24. idku
    • one year ago
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    ok...

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is replying to Can someone tell me what button the professor is hitting...

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