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# Can somebody help me With imaginary numbers?

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

I^31=?

3+4i / 16-2i

### 4 Answers by Expert Tutors

Robert S. | Robert Math PhysicsRobert Math Physics
0
(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
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
0
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

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
0
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

Crystal H. | Chemistry and English tutorChemistry and English tutor
4.7 4.7 (93 lesson ratings) (93)
0
(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