Can somebody help me With imaginary numbers?

Question

(3-i)+(22+4i)

 

I^31=?

 

3+4i / 16-2i

 

please HELP!

 

Answer 1

(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

Answer 2

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

 

Answer 3

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   +  7   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  

 

Answer 4

(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