help plz, 2 "complex number" questions?

Question

1.Find the real numbers x and y such that

(x+yi) (3-5i)= 34

The answer is supposed to be x=3, y=5
I didn’t get it.



2. Express in the form a+bi, the following complex numbers

[1+root of(3i)] / 2 + 2 / [1+root of(3i)]

/ means divide

The answer is supposed to be 1+0i

I didn’t get it.

Thanks a lot in Advance

Answer 1

Question 1
3x - 5i x + 3i y + 5y = 34
(3x + 5y) + (3y - 5x) i = 34

3x + 5y = 34
- 5x + 3y = 0

15x + 25y - 170
-15x + 9y = 0
34y = 170
y = 5

3x + 25 = 34
3x = 9
x = 3

x = 3 , y = 5

Question 2
(1 + √3 i) / 2 + 2 / (1 + √3 i)
[(1 + √3 i)(1 + √3 i) + 4 ] / 2(1 + √3 i)
[1 + 2√3 i - 3 + 4 ] / 2(1 + √3 i)
[ 2 + 2√3 i] / 2(1 + √3 i)
(1 + √3 i) / (1 + √3 i) = 1 = 1 + 0 i

Answer 2

1) (x+yi)(3-5i) =
3x + 3yi - 5xi - 5y *i^2
= 3x +3yi - 5xi +5y (since i^2=-1)
= (3x+5y) - (5x-3y)i = 34
So,
3x + 5y = 34,
5x - 3y = 0
5x = 3y
x = .6y
Subsitute x into very first equation,
1.8 y + 5y = 34
6.8y = 34
y = 5
Subsitute y to 5x = 3y,
5x = 3*5
x = 3
So, x=3, y=5

2) I'll make 1+root of (3i) as p.
p/2 + 2/p =
[p^2 + 4]/2p
p^2 = [1+root of 3i]^2 = 1+3i+2*root of 3i
[5 + 3i + 2*root of 3i] / [1+root of 3i]
I can just go here...

Answer 3

1. 3x + 3yi - 5xi -5yi^2 = 34, so x(3 - 5i) + y(3i +5) = 34. the i terms need to be eliminated, so -5xi + 3yi = 0 and 3x + 5y = 34. y = (5/3)x, and 3x + (25/3)x = 34, or 34/3x = 34, and x = 3.
y = 5 follows from 3x+5y =34

Answer 4

1 is above

2. multiply top and bottom of 1st term by 1 + root(3)i to get

(1 + root(3)i)(1+root(3)i)) / 2(1+root(3)i)

Multiply top and bottom of second term by 2 to get

4 / 2 (1 + root(3))

Use FOIL on the top part of the 1st term to get:

(1 + 2root(3)i + 3(-1) + 4)/2(1 + root(3)i) + 4 / 2(1 + root(3)i)

Simplify the top of the first term

2 + 2root(3)i / 2(1 + root(3)i)

Which simplifies to 1

Answer 5

root(3i) = 3i^1/2 = (3 which are different
Taking one This is sqrt6/2(1+i)
(1+sqrt3i)^2= 1+3i+2sqrt(3i)
[1+i(3+2sqrt(3i)+4]/2(1+sqrt6+i*sqrt6/2)Follow from here

Answer 6

expand the (x+iy)(3-5i) out to get:
3x-5ix+3iy-5yi^2
=3x-5ix+3iy+5y (i^2=-1)
equate the real and imaginary parts:
real:3x+5y=34
imaginary:-5x+3y=0

solve the equations simultaneously by multiplying the real equation by 5 and the imaginary equation by 3 and adding them together.
therefore 34y=170 so y=5
sub y=5 into one of the equations and u get x=3.

Answer 7

in my view I easily have in no way achieved that. yet I easily have replied the question with a ineffective answer. it must be becasue the guy would not understand the respond and that they desire the two factors for dropping their time analyzing the question. Or it must be that they desire factors so as that they circulate to each question they see and supply tht as an asnwer. the two way it works. besides the reality that its no longer very efficient thats why i haven't achieved that. I in no way knew factors mattered, do they?

Answer 8

3x-5xi+3yi+5y=34;
3x+5y+(-5x+3y)i=34+0i;
3y-5x=0;->multiply by 5
5y+3x=34;->multiply by 3
15y-25x=0
15y+9x=102
subtracting
-34x=-102;
x=3;
putting this value in eq. 3y-5x=0
3y-15=0
y=5;