DMT (de moive's theorem) q. (1 Viewer)

[astj]

I've converted 1+i and 1-i into mod-arg form and subbed the n in but I don't know how to get the RHS as well as part ii. Thanks in advance

 

[cossine]

View attachment 42144
I've converted 1+i and 1-i into mod-arg form and subbed the n in but I don't know how to get the RHS as well as part ii. Thanks in advance

I think what is going on they are applying binomial theorem.

Basically you should try expand (1+i)^n and (1-i)^n.

I will need to double check to make sure.

 

[WeiWeiMan]

View attachment 42144
I've converted 1+i and 1-i into mod-arg form and subbed the n in but I don't know how to get the RHS as well as part ii. Thanks in advance

1+i=√2cisπ/4
1-i = √2cis(-π/4)

(√2cisπ/4)^n+(√2cis(-π/4))^n

= √2^n (cis(nπ/4)+cis(-nπ/4))

=2(√2)^n cos(nπ/4) QED as req blah blah blah

for part ii do as cossine said (use binomial expansion) and see what happens

 

[astj]

1+i=√2cisπ/4
1-i = √2cis(-π/4)

(√2cisπ/4)^n+(√2cis(-π/4))^n

= √2^n (cis(nπ/4)+cis(-nπ/4))

=2(√2)^n cos(nπ/4) QED as req blah blah blah

for part ii do as cossine said (use binomial expansion) and see what happens

okk cool thanks, how would you do part ii then?

 

[Lith_30]

part ii)
The question states that we should use the result from part i. Lets use the statement that n is divisible by 4, that is we suppose that .
Then we put that into the result from part i, convert it back into terms of n, .

Now the LHS of that question contains a bunch of n choose expressions, which suggests binomial expansion, so we will expand
which simplifies to

now equating these two new expressions gives us the answer.