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

Mathematics
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Is it: (1/e^2)*(Cos1 + iSin1) ?
a complex number in form: z = r*e^(i*theta) = r(Cos(theta)+iSin(theta))

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Other answers:

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

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