How can evaluate?(1+i)^2008=?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(1+i)^2008.

To simplify, first we will rewrite the exponent.

==> ( 1+ i)^2008 = (1+ i)^(2*1004)

Now we know from exponent properties that x^ab= (x^a)^b

==> (1+i)^(2*1004) =[ (1+i)^2]^1004

Now we will expand the brackets.

==> (1+i)^2]^1004 = (1 + 2i+ i^2) ^1004

But we know that i^1 = -1

==> (1+2i+i^2)^1004 = (2i)^1004

==> (2i)^(2*502) =[ (2i)^2]^502 = (4i^2)^502

==> (4*-1)^502 = = 4^502

==> (1+i)^2008 = 2^(1004)

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To evaluate (1+i)^(2008).

(1+i) = 2^(1/2){1/2^1/2 + (1/2^(1/2) i}

(1+i)^2008 = {2^(1/2) (cos pi/4 +isin pi/4)}^2008.

(1+i)^2008 = 2^(2008/2){cos 2008pi/4+isin 2008/4}, by De Moivre's theorem which states that (cos x+isin x)^n = cosnx+isinnx for all n.

=> (1+i)^2008 = 2^1004{cos 502ppi+isin 502pi}

=> (1+i)^2008 = 2^1004 {1+0), as cos 2npi = 1 and sin 2npi = 0, for any integer n.

Therefore (1+i)^2008 = 2^1004.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll calculate (1+i)^2 = 1 + 2i + i^2

We recall the fact that i^2 = -1

(1+i)^2 = 1 + 2i - 1

We'll eliminate like terms:

(1+i)^2 = 2i

We notice that 2008 = 2*1004

We'll write (1+i)^2008 = [(1+i)^2]^1004

We'll substitute (1+i)^2 = 2i:

[(1+i)^2]^1004 = (2i)^1004 = 2^1004*i^1004

We know that i^4 = 1 and we notice hat i^1004 = (i^4)^251.

i^1004 = 1^251

i^1004 = 1

After evaluation, we'll get: (1+i)^2008 = 2^1004.

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