## Loser66 one year ago What is the real and imaginary parts of $$i^n, 2\leq n\leq 8$$ Please, help

1. 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?

2. 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?

3. anonymous

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

4. anonymous

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

5. Loser66

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

6. anonymous

$i ^{5}=i ^{4+1}= i ^{4}*i ^{1}= 1* i=i$

7. Loser66

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

8. Loser66

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

9. IrishBoy123

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

10. 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?

11. ganeshie8

$i^n\equiv i^{n\pmod{4}}$

12. 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

13. phi

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

14. 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

15. Loser66

16. 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|

17. ganeshie8

all the red lines are of unit length ^

18. Loser66

oh, you convert to cis, right?

19. phi

Irish posted the form you should use

20. Loser66

I got it. Thanks a lot, friends :)

21. 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

22. Loser66

Yes, Sir.

23. 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?

24. ganeshie8

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

25. ganeshie8

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

26. Loser66

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

27. Loser66

I meant the expanding form.

28. 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?

29. ganeshie8

one sec for magnitude, we just use the property $$|z^n| = |z|^n$$

30. 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

31. ganeshie8

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

32. 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?

33. ganeshie8

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

34. Loser66

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

35. Loser66

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

36. ganeshie8

oh ok, are you allowed to use binomial thm ?

37. ganeshie8

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

38. 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.