Question

Use the binomial theorem to express (a+ib)^n, given that {a,b} are in the domain of Real Numbers, i^2 = -1, n is in the domain of Natural Numbers, and n odd.

Answer 1

\[(a+ib)^n\]

Answer 2

It says to put it in the form of A+iB where A, B are real numbers expressed as summations. I've been having alot of problems with the idea of this to the nth degree.

Answer 3

\[(\color{blue}{a}+\color{red}{ib})^n = \sum\limits_{k=0}^n \dbinom{n}{k} \color{blue}{a}^{n-k} (\color{red}{ib})^k\]

Did you get this far ?

Answer 4

It also states I can use the following result: \[\sum_{k=0}^{n}a_{k} = \sum_{j=0}^{\frac{ n-1 }{ 2 }}(c_{2j}c_{2j+1})\]

Answer 5

No I really am completely clueless on this topic.

Answer 6

I struggle alot with summations.

Answer 7

How do I get to that point?

Answer 8

What are \(C_{2j}, C_{2j+1}\) ?
If possible could you take a screenshot of the entire question and attach please ?

Answer 9

Its not a problem thats online. IT just says that I am allowed to use that result. The exact thing it says I can use is in that equation there.

Answer 10

I still don't get what are \(C_{2j}, C_{2j+1}\) ?

Answer 11

Im not sure. I find this very difficult to comprehend in class. How did you get that result that you got earlier?

Answer 12

What I have earlier is the binomial theorem
http://latex.artofproblemsolving.com/4/9/4/494769588adb2ac3700e25b086fe2b7d41bba70a.png

Answer 13

And I can just plug in my results into that theorem? Let me do that and i'll post what I get.

Answer 14

Oh I see. So I just replace all instances of a with a and all of b with ib? So ib would just be the imaginary part "b", right?

Answer 15

How would I simplify that big equation to the form of A+iB?

Answer 16

Yeah, I'm trying to make sense of the result you have posted earlier. Do you have any more knowledge on that result ?

Answer 17

I feel like the fact that i^2 = -1 is going to come in handy here, I just dont know how it helps solve this. Really sorry to use up so much of your time, I know you have other people to help too.

Answer 18

I think that the result shown previously is the result of the summation from k=1 to n of x^n

Answer 19

Gotcha! You made a typo in your result

Answer 20

Oh I did?

Answer 21

oh is it the +?

Answer 22

Yes

\[\sum_{k=0}^{n}a_{k} = \sum_{j=0}^{\frac{ n-1 }{ 2 }}(a_{2j} + a_{2j+1})\]

Answer 23

They are simply splitting the sum into two halves

Answer 24

How does that help?

Answer 25

Notice the pattern

i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i

Observe that alternate powers give you real numbers 1, -1

Answer 26

Yeah. So then i^4 would be 1 again.

Answer 27

and i^5 would be i

Answer 28

Do I need to find the remainder of k/4 then? Somehow?

Answer 29

Let's split the sum using your earlier result :

\[\begin{align}(\color{blue}{a}+\color{red}{ib})^n &= \sum\limits_{k=0}^n \dbinom{n}{k} \color{blue}{a}^{n-k} (\color{red}{ib})^k\\~\\
&= \sum\limits_{j=0}^{(n-1)/2} \dbinom{n}{2j} \color{blue}{a}^{n-2j} (\color{red}{ib})^{2j} +\dbinom{n}{2j+1} \color{blue}{a}^{n-(2j+1)} (\color{red}{ib})^{2j+1} \\~\\
\end{align}\]

Answer 30

See if above step makes sense. I've just used your result..

Answer 31

Yes that makes sense, so both parts of that are the two parts shown in the result, right?

Answer 32

Yes, if you get this far, rest should be easy

Answer 33

Like I said, nothing about this topic is easy to me. For some reason I struggle on this particular topic alot.

Answer 34

Let's split the sum using your earlier result :

\[\begin{align}(\color{blue}{a}+\color{red}{ib})^n &= \sum\limits_{k=0}^n \dbinom{n}{k} \color{blue}{a}^{n-k} (\color{red}{ib})^k\\~\\
&= \sum\limits_{j=0}^{(n-1)/2} \dbinom{n}{2j} \color{blue}{a}^{n-2j} (\color{red}{ib})^{2j} + \sum\limits_{j=0}^{(n-1)/2}\dbinom{n}{2j+1} \color{blue}{a}^{n-(2j+1)} (\color{red}{ib})^{2j+1} \\~\\
\end{align}\]

Answer 35

Look at the expressions that have `i`

Answer 36

oh so we can get rid of the i and replace it with -1^j, right?

Answer 37

and the second one with j+1/2

Answer 38

\[\large (ib)^{2j} = i^{2j}b^{2j} =(i^2)^{j}b^{2j} =?\]

Answer 39

(-1)^jb^2j

Answer 40

But now it depends on whether or not J is odd...

Answer 41

That shouldn't be a problem, all we want is to express it in the form A + iB

Answer 42

\[\large (ib)^{2j+1} = i^{2j+1}b^{2j+1} = i*i^{2j}b^{2j+1} =i*(i^2)^{j}b^{2j} =?\]

Answer 43

So the i^2 = -1, right?

Answer 44

Yes, plugging them in we have

\[\begin{align}(\color{blue}{a}+\color{red}{ib})^n &= \sum\limits_{k=0}^n \dbinom{n}{k} \color{blue}{a}^{n-k} (\color{red}{ib})^k\\~\\
&= \sum\limits_{j=0}^{(n-1)/2} \dbinom{n}{2j} \color{blue}{a}^{n-2j} (\color{red}{ib})^{2j} + \sum\limits_{j=0}^{(n-1)/2}\dbinom{n}{2j+1} \color{blue}{a}^{n-(2j+1)} (\color{red}{ib})^{2j+1} \\~\\

&= \sum\limits_{j=0}^{(n-1)/2} \dbinom{n}{2j} \color{blue}{a}^{n-2j} \color{red}{(-1)^jb^{2j}} + \sum\limits_{j=0}^{(n-1)/2}\dbinom{n}{2j+1} \color{blue}{a}^{n-(2j+1)} \color{red}{i(-1)^jb^{2j+1}} \\~\\

&= \sum\limits_{j=0}^{(n-1)/2} \dbinom{n}{2j} \color{blue}{a}^{n-2j} \color{red}{(-1)^jb^{2j}} + \color{red}{i}\sum\limits_{j=0}^{(n-1)/2}\dbinom{n}{2j+1} \color{blue}{a}^{n-(2j+1)} \color{red}{(-1)^jb^{2j+1}} \\~\\

\end{align}\]

Answer 45

Looks we're done ?

Answer 46

Ohhhhh, I see, so you just move the i to the other side of the equation and it becomes A+iB where A and B are summations.... THanks a ton man!

Answer 47

Np :) Do you get why below splitting works ?


\[\sum_{k=0}^{n}a_{k} = \sum_{j=0}^{\frac{ n-1 }{ 2 }}(a_{2j} + a_{2j+1})\]

Answer 48

Yes

Answer 49

If you have time could you explain to me

Answer 50

Thanks so much for your help man

Answer 51

I don't unfortunately, It's getting late and I need to go to bed. I'll tag you tomorrow morning, in the meantime I gotta help this other guy out. Thanks so much for your help.

Answer 52

Np.. Have god sleep :)