# Calculate (1+i)^200=?

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(1+i)^200 =

We know that: x^ab =( x^a)^b

Let us rewrite:

(1+i)^200 = (1+i)^2*100

= [(1+i)^2]^100

= ( 1 + 2i -1)^100

= (2i)^100

= [(2i)^2]^50

= (-4)^50

= 4^50

**==> (1+i)^200 = 4^50**

Because the exponent 200 could be written as 2*100, we'll calculate first (1+i)^2.

We'll expand the binomial using the formula:

(a+b)^2 = a^2 + 2ab + b^2

a = 1 and b = i

(1+i)^2 = 1^2 + 2*1*i + i^2, where i^2 = -1

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

We'll eliminate like terms:

(1+i)^2 = 2i

Now, we'll write (1+i)^200 = [(1+i)^2]^100

[(1+i)^2]^100 = (2i)^100

(2i)^100 = 2^100*i^100

We'll recall the power of i:

i^2 = -1

i^3 = i^2*i = -1*i = -i

i^4 = i^2*i^2 = (-1)*(-1) = 1

We'll write 100 as a multiple of 4:

100 = 4*25

i^100 = (i^4)^25

i^4 = 1

i^100 =1^25 = 1

**(1+i)^200 = 2^100 = (2^4)^25 = 16^25**

(1+i)^200.

We know that 1+i = sqrt2 (1/sqrt2 +i/sqrt2}

Therefore (1+i)^200 = {sqrt2(cos x+isinx) ^200 , wher x = pi/4.

(1+i)^200 = (sqrt)^100 (cosx+isinx}^200

(1+i)^200 = (2^100)(cos200x+isin200x) by De Moivres theorem

(1+i)^200 = (2^100 ){ cos 200*pi/4 +isin 200pi/4)

(1+i)^200 = 2^100 {cos50pi +isin50pi}

(1+i)^200 = 2^100 { cos2pi + sin2pi) as the 50pi is an itegral multiple of 2pi.

**(1+i)^200 = 2^100,** as cos2pi = 1 and sin2pi = 0