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bahrom7893

  • 3 years ago

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

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

  2. bahrom7893
    • 3 years ago
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    @satellite73

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

  4. bahrom7893
    • 3 years ago
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    @phi

  5. swissgirl
    • 3 years ago
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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

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

  7. bahrom7893
    • 3 years ago
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    well thanks everyone

  8. bahrom7893
    • 3 years ago
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    back to studying..

  9. anonymous
    • 3 years ago
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    start with \[e^{-2}e^i\] then oh, you get what you wrote

  10. anonymous
    • 3 years ago
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    since \(e^{ix}=\cos(x)+i\sin(x)\)

  11. kc_kennylau
    • 2 years ago
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    I've recently dealt with this kinda things before... let me think...

  12. kc_kennylau
    • 2 years ago
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    Wait oh satellite has already solved the problem, a=0 and b=\(\large e^{-2}\)

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