Question

"Find (1 + i)^2 and conclude that w^2 = i. Why does this show that w^8 = 1?"

Can someone help me with the first part? I know that (1 + i)^1 = 2i, but I can't figure out how to "conclude" that w^2= i.

I know that if w^2 does equal i, then w^2 x w^2 x w^2 x w^2 = 1 because i x i = -1, so i^2 x i^2 = 1.

Answer 1

*(1 + i )^2, my bad!

Answer 2

what is \(\omega\)?

Answer 3

I don't know. This is where I'm getting stuck.

Answer 4

well I am not quite sure how you can say anything about w unless you know what it is....

Answer 5

I've tried experimenting with 2i to see if it shows anything, but I'm truly stuck. I'm sure it must be something with (1 + i)^2, but I just can't see it.

Answer 6

Is \[w=\frac{1}{\sqrt2}+\frac{i}{\sqrt2}=\frac{1}{\sqrt2}\left(1+i\right)?\]That's really the only thing that makes sense to me.

Answer 7

I guess that must be it. I have no answer key, so I don't have the answer with me to check. The thing that confuses me is where does the 2i come in then?

Answer 8

Well, the \(2i\) comes from the fact that \((1+i)^2=2i\). To demonstrate this, just start multiplying\[(1+i)(1+i)=1+i+i+i^2=1+2i-1=2i.\]

Answer 9

Mm, yes. I see that.

Answer 10

I guess this is about as far as I can get with this problem. Thank you so much for your help.

Answer 11

I guess I should at least ask if you see why it follows that \[\left(\frac{1}{\sqrt2}\left(1+i\right)\right)^2=i?\]

Answer 12

Yes, I do. The 1/√2 becomes 1/2, and (1 + i) becomes 2i. When multiplied together, the 2 cancels out and leaves me with i. Is that correct?

Answer 13

Looks right to me.

Answer 14

Ah, that's good! Thank you again so much.

Answer 15

You're welcome.