# Which is the value for (1+i)^20?

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### 3 Answers

(1+i)^20

= (1+i)^2*10

= [(1+i)^2 ]^10

= (2i)10

= (2^10)(i)^10

= (2^10)*(-1)

= -(2^10)= -1024

To find (1+i)^20 =

Solution:

1+i = sqrt2( 1/sqrt2+i/sqrt2) = 2^(1/2)( cospi/4 +isinpi/4)

(1+i )^20= {(2)^1/2 (cospi/4+isinpi/4}^20

=2^10 { cos20*pi/4 +i sin 20*pi/4) , by De Moivre's theorem which states (cosx+isinx)^n = cosnx+isinnx

= 2^10(cos5pi+isin5pi)

=2^10(-1+i*0)

= -2^10.

We'll start from the fact that (1+i)^2 = 2i

We could write (1+i)^20 = (1+i)^2*10 = [(1+i)^2]^10

[(1+i)^2]^10 = (2i)^10 = 2^10 * i^10

We know that:

i^2 =-1

i^10 = i^2*5 = (i^2)^5 = (-1)^5 = -1

**(1+i)^20 = 2^10 * i^10 = -2^10**