Question

I have i^(-n) and my final answer should be cos((pi n)/2)- i sin((pi n)/2). How do I do the work to get my answer. Thanks

Answer 1

you i^(-n) - very well. you have to get something as a final answer - okay. Where did the main question vanish though?

Answer 2

It says to compute i^(-n) and then i have to put it in the form a+ib

Answer 3

you have to devide n into 4 partition3 that are n = 4x,4x +1,4x+2,4x+

Answer 4

and you will have 4 answers

Answer 5

a = 0, b = 1 --> r = 1
tan (theta) = 1/0 => theta = pi/2
--> i^ (-n) = r cis n (theta) = cos ( -n pi/2) + sin ( - n pi/2)
=> i^ (-n) = cos((pi n)/2)- i sin((pi n)/2)

Answer 6

I am still a little confused. Here is a problem we did in class and the work. Can you help explain this to me in relation to the problem I am qeustioning.
i(i^2)i^3......i^n
=i^((n(n+1))/2)
=e^((i pi)/2)((n(n+1))
=e
=(cos((n(n+1)pi)/4)+i(sin((n(n+1)pi)/4)
=cos(pi/2)+i sin(pi/2)
=e^(i(pi/2))

Answer 7

Also, why does theta =pi/2

Answer 8

tan (theta) = 1/0 = infinitive
theta = pi/2 ( use unit circle)

Answer 9

I guess what I am asking is how does tan(theta)=1/0 and then =pi/2

Answer 10

a = 0, b = 1
tan (theta) = b/a = 1/0 ( formula)

Answer 11

Okay, I see how yo get tan(theta) =b/a=1/0. What do you mean by formula and how do I plot that on a unit circle. I always thought that if there was a 0 in the denominator it is indeterminate.