Question

Sa se calculeze
a) i^1000
b) i^1004
c) i^4005
d) i^5423
e) (-i)^334
f) i+i^2+i^3+...+i^1000
g) (1+i)^8
h) (1+i)^36

Answer 1

Explicație pas cu pas:

Stim ca i²=-1.

a) i^1000=(i^2)^500=(-1)^500=1^500=1

b) i^1004=(i^2)^502=(-1)^502=1^502=1

c) i^4005=i*i^4004=i*i(i^2)^2002=i*(-1)^2002=i*1^2002=i*1=i

d) i^5423=i*i^5422=i*i(i^2)^2711=i*(-1)^2711=i*(-1)=-i

e) (-i)^334=i^334=(i^2)^167=(-1)^167=-1

f) S=i+i^2+i^3+....+i^1000

Observam ca termenii sumei sunt in progresie geometrica cu primul termen b1=i si de ratie q=i. Suma are 1000 de termeni

S=b1*(q^n-1)/(q-1)=i*(i^1000-1)/(i-1)=i*(1-1)/(i-1)=i*0=0

g) (1+i)^8=[(1+i)^2]^4=(1+2i+i^2)^4=(1+2i-1)^4=(2i)^4=2^4*i^4=16*(i^2)^2=16*(-1)^2=16*1=16

h) (1+i)^36=[(1+i)^2]^18=(1+2i+i^2)^18=(1+2i-1)^18=(2i)^18=2^18*i^18=2^18*(i^2)^9=2^18*(-1)^9=2^18*(-1)=-2^18