bahrom7893
write e^(2+i) in a+bi form



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

bahrom7893
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@satellite73

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

bahrom7893
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@phi

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

bahrom7893
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I guess.. it's just that I thought it would look somewhat neater

bahrom7893
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well thanks everyone

bahrom7893
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back to studying..

anonymous
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start with
\[e^{2}e^i\] then oh, you get what you wrote

anonymous
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since \(e^{ix}=\cos(x)+i\sin(x)\)

kc_kennylau
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I've recently dealt with this kinda things before... let me think...

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