Question

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

Answer 1

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

Answer 2

@satellite73

Answer 3

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

Answer 4

@phi

Answer 5

\( e^{i\theta} = Cos\theta + iSin\theta\)
Ya thats what i Have seen around so I am assuming ur answer is correct

Answer 6

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

Answer 7

well thanks everyone

Answer 8

back to studying..

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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