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

- Loser66

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

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- Loser66

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?

- Loser66

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?

- anonymous

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

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## More answers

- anonymous

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

- Loser66

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

- anonymous

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

- Loser66

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

- Loser66

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

- IrishBoy123

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

- Loser66

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?

- ganeshie8

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

- phi

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

- phi

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

- ganeshie8

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

- Loser66

Show me, please.

- ganeshie8

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 :
|dw:1440593497736:dw|

- ganeshie8

all the red lines are of unit length ^

- Loser66

oh, you convert to cis, right?

- phi

Irish posted the form you should use

- Loser66

I got it. Thanks a lot, friends :)

- ganeshie8

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

- Loser66

Yes, Sir.

- Loser66

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 |dw:1440595215625:dw|
What is wrong?

- ganeshie8

with the number \(1+i\), the length won't stay the same when you rise it to a power

- ganeshie8

\[z= (1+i)^6\]
\(|z| = \sqrt{2} \implies |z^6| = |z|^6 = (\sqrt{2})^6\)

- Loser66

How to find the simpliest form of (1+i)^6 ?

- Loser66

I meant the expanding form.

- Loser66

Like if we have \(z = (1+i)^2 = 2i\) ,then \(|z|= 2\), but how to work if it is z ^6 or z^8?

- ganeshie8

one sec
for magnitude, we just use the property \(|z^n| = |z|^n\)

- phi

How to find the simpliest form of (1+i)^6 ?
change to polar. find mag^6, and find the new angle = old angle * 6
then change back to rectangular

- ganeshie8

do you want to expand (1+i)^6 just for finding the magnitude
or for some other reason ?

- Loser66

@ganeshie8 The particular problem is \((1+i)^6\). However, I don't want to stop there; I want to generalize the method so that I can solve any problem later.
One more thing: We, the students, use the formulas we proved in class. We are not allowed to use the things we didn't prove yet. Like : \(|\dfrac{z}{w}|=\dfrac{|z|}{|w|}\) is on the book, but if the prof didn't indicate to it, we are not allow to use. :(
@phi No choice for me, right? I have to do all the steps, no shortcut, right?

- ganeshie8

you can prove that little thing as a one line lemma at the start of ur answer
its not a huge thingy :)

- Loser66

And in class, he expanded everything in simplest form a+bi to find the length, real, imaginary part, and conjugate.

- Loser66

That is why I don't want to use cis. :)

- ganeshie8

oh ok, are you allowed to use binomial thm ?

- ganeshie8

if no geometry / no polar form, then i guess binomial thm is the way to go :
\[(1+i)^6 = 1+\binom{6}{1}i + \binom{6}{2}i^2+\binom{6}{3}i^3+\binom{6}{4}i^4+\binom{6}{5}i^5+\binom{6}{6}i^6 \]

- Loser66

We have this property: |v*w|= |v|*|w|. If we consider (1+i) is v, we have \((1+i)^6= v*v*v*v*v*v\), hence
\(|(1+i)^6|=|1+i|*|1+i|*|1+i|*|1+i|*|1+i|*|1+i|=|1+i|^6=\sqrt2^6=8\)
It works, right? hehehe... I feel like I am a middle school student who tries to study multiplication.

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