Question

What is the real and imaginary parts of \(i^n, 2\leq n\leq 8\)
Please, help

Answer 1

What I don't get is when n is odd, like 3, 7, then \(i^3 = -i\) and \(i^7 = -i\) also.
But for n =5, we have \(i^5 = i\) , hence, we can classify n is odd as a common case to determine Real and Imaginary, right?

Answer 2

Similarly to n is even, if n =2, then \(i^2 =-1\), but if n =4, 8 \(i^4 =1 ~~and~~ i^8 =1 \)
We have to break it into many cases and I don't want it. Is there any way to combine them to general cases? Because if the limits are \(2\leq n\leq 100\), how can I list all of the cases out?

Answer 3

what's i?\[i=\sqrt{-1}\] which is totally imaginary....
\[i ^{2}=(\sqrt{-1})^{2} =-1\] but \[i ^{4} = (\sqrt{-1})^{4}= i^{2}*i ^{2}= (-1)*(-1)= 1\]

Answer 4

if u get this, u can understand the whole concept.........

Answer 5

@lall Thanks for the answer. I got it. All I want is to classify the cases to not list them up.

Answer 6

\[i ^{5}=i ^{4+1}= i ^{4}*i ^{1}= 1* i=i\]

Answer 7

How about this question:
Find Real and Imaginary part of \(\left (\dfrac{1+i}{\sqrt2}\right )^n, 2\leq n\leq 8\)

Answer 8

Again, I don't want to list the cases. Please, help me or tell me "No choice for you" :)

Answer 9

tried the exponential form? so \(\large e^{i \ n\frac{\pi}{4}}\) is handy notation

Answer 10

Is it not that it is NOT a cis problem? If we convert it to cis, then we have to convert back to complex to find R and Im, right?

Answer 11

\[i^n\equiv i^{n\pmod{4}}\]

Answer 12

I assume you noticed for the first question
\[ i^n= i^{n\ mod\ 4} \]
for the second, it is easier to convert to polar (or cis ) form, do the calculation, then convert back to rectangular

Answer 13

also, notice the answers will be equally spaced on the unit circle, starting at i

Answer 14

i feel the second problem can be done easily by looking at geometry :

notice that the given complex number lies exactly halfway between real and imaginary axes and recall the fact that angles add up when you multiply two complex numbers

Answer 15

Show me, please.

Answer 16

let \(z=\dfrac{1+i}{\sqrt2}\)
it is of unit length, so each power increments the angle by 45 degrees but the length stays same :

Answer 17

all the red lines are of unit length ^

Answer 18

oh, you convert to cis, right?

Answer 19

Irish posted the form you should use

Answer 20

I got it. Thanks a lot, friends :)

Answer 21

np, you must have noticed \(\left (\dfrac{1+i}{\sqrt2}\right )^n, 1\leq n\leq 8\) give you the eight roots of the equation \(z^8-1=0\) and they also form a group under regular multiplication of complex numbers

Answer 22

Yes, Sir.

Answer 23

Question @ganeshie8 I have to find |z|, I confused.
\(z= (1+i)^6\)
My logic: \(|1+i|= \sqrt 2\rightarrow |1+i|^6 =8\)
If we convert to cis, like what you did, then
What is wrong?