A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
What is the real and imaginary parts of \(i^n, 2\leq n\leq 8\)
Please, help
Loser66
 one year ago
What is the real and imaginary parts of \(i^n, 2\leq n\leq 8\) Please, help

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0What 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
 one year ago
Best ResponseYou've already chosen the best response.0Similarly 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
 one year ago
Best ResponseYou've already chosen the best response.0what's i?\[i=\sqrt{1}\] which is totally imaginary.... \[i ^{2}=(\sqrt{1})^{2} =1\] but \[i ^{4} = (\sqrt{1})^{4}= i^{2}*i ^{2}= (1)*(1)= 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if u get this, u can understand the whole concept.........

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@lall Thanks for the answer. I got it. All I want is to classify the cases to not list them up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[i ^{5}=i ^{4+1}= i ^{4}*i ^{1}= 1* i=i\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0How about this question: Find Real and Imaginary part of \(\left (\dfrac{1+i}{\sqrt2}\right )^n, 2\leq n\leq 8\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Again, I don't want to list the cases. Please, help me or tell me "No choice for you" :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1tried the exponential form? so \(\large e^{i \ n\frac{\pi}{4}}\) is handy notation

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Is 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
 one year ago
Best ResponseYou've already chosen the best response.1\[i^n\equiv i^{n\pmod{4}}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0also, notice the answers will be equally spaced on the unit circle, starting at i

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let \(z=\dfrac{1+i}{\sqrt2}\) it is of unit length, so each power increments the angle by 45 degrees but the length stays same : dw:1440593497736:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1all the red lines are of unit length ^

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, you convert to cis, right?

phi
 one year ago
Best ResponseYou've already chosen the best response.0Irish posted the form you should use

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I got it. Thanks a lot, friends :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1np, you must have noticed \(\left (\dfrac{1+i}{\sqrt2}\right )^n, 1\leq n\leq 8\) give you the eight roots of the equation \(z^81=0\) and they also form a group under regular multiplication of complex numbers

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question @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
 one year ago
Best ResponseYou've already chosen the best response.1with the number \(1+i\), the length won't stay the same when you rise it to a power

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[z= (1+i)^6\] \(z = \sqrt{2} \implies z^6 = z^6 = (\sqrt{2})^6\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0How to find the simpliest form of (1+i)^6 ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I meant the expanding form.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Like 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 year ago
Best ResponseYou've already chosen the best response.1one sec for magnitude, we just use the property \(z^n = z^n\)

phi
 one year ago
Best ResponseYou've already chosen the best response.0How 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
 one year ago
Best ResponseYou've already chosen the best response.1do you want to expand (1+i)^6 just for finding the magnitude or for some other reason ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1you can prove that little thing as a one line lemma at the start of ur answer its not a huge thingy :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0And in class, he expanded everything in simplest form a+bi to find the length, real, imaginary part, and conjugate.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0That is why I don't want to use cis. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1oh ok, are you allowed to use binomial thm ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.0We 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.