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AmTran_Bus

  • one year ago

Find the real and imaginary parts of this

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  1. AmTran_Bus
    • one year ago
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    |dw:1440206784952:dw|

  2. AmTran_Bus
    • one year ago
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    \[e ^{-2 +i \pi/2}\]

  3. AmTran_Bus
    • one year ago
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    can I use Euler's formula?

  4. Zarkon
    • one year ago
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    yes

  5. AmTran_Bus
    • one year ago
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    Hum. I just don't see it though.

  6. Zarkon
    • one year ago
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    \[\Large e ^{-2 +i \pi/2}=e ^{-2} e^{i \pi/2}\]

  7. AmTran_Bus
    • one year ago
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    I actually have that. But can you walk me through the steps please? Like where did the sin and cos from the formula go?

  8. Zarkon
    • one year ago
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    \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\] \[e^{i\pi/2}=\cos(\pi/2)+i\sin(\pi/2)\]

  9. AmTran_Bus
    • one year ago
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    Good. There was a problem just like that I literally just worked and got that answer. But how does that carry on to this problem? I'm so sorry, I just can't see it (this is math reveiew in a p chem book).

  10. Zarkon
    • one year ago
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    \[\Large e ^{-2 +i \pi/2}=e ^{-2} e^{i \pi/2}=e^{-2}[\cos(\pi/2)+i\sin(\pi/2)]\] \[\Large=e^{-2}[0+i\times1]=e^{-2}i\]

  11. AmTran_Bus
    • one year ago
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    Ahhhh! Duh!!!!!! Basic unit circle!

  12. Zarkon
    • one year ago
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    yes

  13. AmTran_Bus
    • one year ago
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    I owe you a million dollars. Thanks.

  14. Zarkon
    • one year ago
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    np

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