Question

Sa se calculeze :a)(2+2i)^16
b)(1/2-i$\sqrt{3}$/2)^9

Answer 1

2 + 2i = 2 ( 1 +i) 
( 1 + i)² = 1 + 2i + i² = 1 + 2i + ( -1) = 2i 
( 2 + 2i) ¹⁶  = [2(1 +i) ]¹⁶ =2¹⁶ ( 1+ i )¹⁶ =2¹⁶[( 1 +i )² ]⁸ = 2¹⁶ (2i)⁸ = 2¹⁶2⁸ i⁸  = 2¹⁶⁺⁸ = 
i⁸ = ( i⁴ )² = 1² = 1 
 = 2²⁴
b.  ( 1 / 2 - i√3 / 2 )⁹ = [ cos5π/3 + i sin5π/3 ] ⁹  = [  cos 5 ·9π /3 + isin5·9π/3 ]=
1 /2 = cos5π /3 
- √3 /2  =  sin5π /3 
5π /3 = 300° 
| z | = ( 1 /2)² + (√3 /2)² = 1 /4 + 3 /4 =4 /4 = 1
          = cos15π + isin15π = cos[ 7 ·2π +  π ] + isin [ 7 ·2π + π]= 
           = cosπ + isinπ =  - 1