bahrom7893
  • bahrom7893
write e^(-2+i) in a+bi form
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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bahrom7893
  • bahrom7893
Is it: (1/e^2)*(Cos1 + iSin1) ?
bahrom7893
  • bahrom7893
@satellite73
bahrom7893
  • bahrom7893
a complex number in form: z = r*e^(i*theta) = r(Cos(theta)+iSin(theta))

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More answers

bahrom7893
  • bahrom7893
@phi
swissgirl
  • swissgirl
\( e^{i\theta} = Cos\theta + iSin\theta\) Ya thats what i Have seen around so I am assuming ur answer is correct
bahrom7893
  • bahrom7893
I guess.. it's just that I thought it would look somewhat neater
bahrom7893
  • bahrom7893
well thanks everyone
bahrom7893
  • bahrom7893
back to studying..
anonymous
  • anonymous
start with \[e^{-2}e^i\] then oh, you get what you wrote
anonymous
  • anonymous
since \(e^{ix}=\cos(x)+i\sin(x)\)
kc_kennylau
  • kc_kennylau
I've recently dealt with this kinda things before... let me think...
kc_kennylau
  • kc_kennylau
Wait oh satellite has already solved the problem, a=0 and b=\(\large e^{-2}\)

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