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bahrom7893

write e^(-2+i) in a+bi form

  • one year ago
  • one year ago

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  1. bahrom7893
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    Is it: (1/e^2)*(Cos1 + iSin1) ?

    • one year ago
  2. bahrom7893
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    @satellite73

    • one year ago
  3. bahrom7893
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    a complex number in form: z = r*e^(i*theta) = r(Cos(theta)+iSin(theta))

    • one year ago
  4. bahrom7893
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    @phi

    • one year ago
  5. swissgirl
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    \( e^{i\theta} = Cos\theta + iSin\theta\) Ya thats what i Have seen around so I am assuming ur answer is correct

    • one year ago
  6. bahrom7893
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    I guess.. it's just that I thought it would look somewhat neater

    • one year ago
  7. bahrom7893
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    well thanks everyone

    • one year ago
  8. bahrom7893
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    back to studying..

    • one year ago
  9. satellite73
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    start with \[e^{-2}e^i\] then oh, you get what you wrote

    • one year ago
  10. satellite73
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    since \(e^{ix}=\cos(x)+i\sin(x)\)

    • one year ago
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