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Can somebody help me With imaginary numbers?

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4 Answers

(3-i)+(22+4i) combine like terms
 
(3+22)+(4i-i) = 25 + 3i
 
 
last term 
 
3 + 4i / 16 -2i this needs to be in the form x + iy therefore the i needs to be taken out of the denominator
to do this requires that 16 - 2i be multiplyed by its complex congregate which is 16 + 2i but that also must be multiplied in the numerator
 
in the denominator 16 -2i x 16 + 2i = 16x16=256 -2i x 2i = -4 x -1 =4  for the mildde term it is -32i +32i = 0
so the denominator becomes 260
 
since the numerator must also be multiplied by 16 + 2i  now the numerator is 3+4i x 16 + 2i
3x16=48  4i x 2i = 8 x -1  = -8 the middle term is 64i + 6i = 70i
 
in the numerator collecting terms is 40 + 70i and the denominator is 260
 
40 + 70i / 260 = 40/260 + 70i/260
1.
( 3 - i ) + ( 22 + 4i ) = (3 + 22) + (-1 + 4) i = 25 + 3i

2.
i^31 = ?
 
i := √(-1) where := means "is defined as"
 
i^0 =1 because i^0 = i^(1-1) = i/i = 1
i^1 = i
i^2 = -1 because (√(a))^2 = a
i^3 = -i
i^4 = 1 because -i(i) = -i^2 = -(-1) = 1
and if you keep going you will see that the values of powers of i cycle through the four numbers 1, i, -1, -i.
 
So i^x = i^(remainder(x/4)), and
 
i^31 = i^3 = -i

3.
(3+4i) / (16-2i) =
(3+4i) / (2(8-i)) =
((3+4i) / (2(8-i)))*((8+i)/(8+i)) =
(3+4i)(8+i) / (2(8-i)(8+i)) =
(24 + 3i + 32i +4i^2) / (2(64 + 8i - 8i - i^2)) =
(24 + 35i - 4) / (2(64 + 1)) =
(20 + 35i) / (2(65)) =
(20 + 35i) / 130 =
20/130 + (35/130) i =
 
2/13 + (7/26) i
 
i ^31 = (i ^4) ^7 . i^3 = -i
 
( 3 - i ) + ( 22 +4i) = 25+ 3i
 
 
( 3 +4i)  *   (16 + 2i)  =
16 - 2i        16 + 2i
 
 
48 +64i +6i -8  =   40  + 70 i
  256 +4              260    260 
                       =  2    i 
                            13    26
 
   Exponents of i
 
    i^2 = -1
   i ^3 = i^2 . i = -i
    i^4 = (i^2) ^2 = ( -1) ^2 = 1
 
     For numbers n > 4
 
        i^n = i ^ (4n + 1 ) = i ^ 4n . i = i
 
                 i ^ ( 4n + 2 )  = i ^4n . i^2 = -1
 
                 i ^ ( 4n + 3 )= i ^4n . i ^3 = - i  
 
(3-i)+(22+4i)= 3-i+22+4i
 
In this case, treat i like a variable and just combine like terms= 25+3i
 
 
 
i^31
 
i^2 = -1, so i^31= i^30 x i^1
 
i^30= -1x15= -15
 
i^30 x i^1= -15i
 
 
3+4i/16-2i = (3+4i)(16+2i)/(16-2i)(16+2i)
 
(3+4i)(16+2i) = 48+6i+24i+8(i)^2
(16-2i)(16+2i) = 256-4(i)^2
 
48+30i+8(-1) / 256-4(-1)
 
48+30i-8/256+4
 
48+30i-8/260
 
56+30i/260
 
divide the whole equation by 2 = 28+15i/130
 
 
 

Comments

Crystal, two of your answers are wrong and your conceptual understanding of powers of i is wrong.

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