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

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

Mathematics
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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?
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?
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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if u get this, u can understand the whole concept.........
@lall Thanks for the answer. I got it. All I want is to classify the cases to not list them up.
\[i ^{5}=i ^{4+1}= i ^{4}*i ^{1}= 1* i=i\]
How about this question: Find Real and Imaginary part of \(\left (\dfrac{1+i}{\sqrt2}\right )^n, 2\leq n\leq 8\)
Again, I don't want to list the cases. Please, help me or tell me "No choice for you" :)
tried the exponential form? so \(\large e^{i \ n\frac{\pi}{4}}\) is handy notation
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?
\[i^n\equiv i^{n\pmod{4}}\]
  • 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
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
Show me, please.
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|
all the red lines are of unit length ^
oh, you convert to cis, right?
  • phi
Irish posted the form you should use
I got it. Thanks a lot, friends :)
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
Yes, Sir.
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?
with the number \(1+i\), the length won't stay the same when you rise it to a power
\[z= (1+i)^6\] \(|z| = \sqrt{2} \implies |z^6| = |z|^6 = (\sqrt{2})^6\)
How to find the simpliest form of (1+i)^6 ?
I meant the expanding form.
Like if we have \(z = (1+i)^2 = 2i\) ,then \(|z|= 2\), but how to work if it is z ^6 or z^8?
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
do you want to expand (1+i)^6 just for finding the magnitude or for some other reason ?
@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?
you can prove that little thing as a one line lemma at the start of ur answer its not a huge thingy :)
And in class, he expanded everything in simplest form a+bi to find the length, real, imaginary part, and conjugate.
That is why I don't want to use cis. :)
oh ok, are you allowed to use binomial thm ?
if no geometry / no polar form, then i guess binomial thm is the way to go : \[(1+i)^6 = 1+\binom{6}{1}i + \binom{6}{2}i^2+\binom{6}{3}i^3+\binom{6}{4}i^4+\binom{6}{5}i^5+\binom{6}{6}i^6 \]
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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