Some imaginary numbers problems?

Question

Just a couple.

1. (-5-i)+(6+2i)

2. (2-5i)-(-3+2i)

3. (1-3i)*(3+i)

4. (5-2i)^2

5. i/(3i+4)

6. Last one-
2*sqroot(5x+1)-8=0

Thses are all I missed. I need to see where I made my error.

Answer 1

1) (-5-i)+(6+2i)
=(2i-i)+(-5+6)
=i+1

2) (2-5i)-(-3+2i)
=(2-5i)+(3-2i)
=(-5i-2i)+(2+3)
=-7i+5

3) (1-3i)(3+i) use FOIL
=(3*1)+(1*i)+(-3i*3)+(-3i*i) note: i*i = -1
=3+i-9i+(-3*-1)
=3-8i+3
=-8i+6

Answer 2

1. (- 5 - i) + (6 + 2i) =
6 - 5 + 2i - i =
1 + i

2. (2 - 5i) - (- 3 + 2i) =
2 - 5i + 3 - 2i =
5 - 7i

3. (1 - 3i)*(3 + i) =
3 + i - 3i(3 + i) =
3 + i - 9i - 3i^2 =
- 8i + 6

4. (5 - 2i)^2 =
25 - 20i + 4i^2 =
21 - 20i

5. i / (3i + 4) =
i(3i - 4) / (3i + 4)(3i - 4) =
(- 3i - 4) / (- 9 - 16) =
(4 + 3i) / 25

6. Last one-
2*sqroot(5x + 1) - 8 = 0
2*sqroot(5x + 1) = 8
sqroot(5x + 1) = 4
5x + 1 = 16
5x = 15
x = 3

Answer 3

Keep in mind that "i" = sqrt(-1) and i^2 = -1
I will show you the first and 3rd, you can do the rest...

1. -5-i+6+2i = i + 1

3. (1-3i)(3+i) = 3 + i - 9i -3i^2 = 3 + i - 9i + 3 = -8i+6

Answer 4

i see that others have done these out for you, but I wanted to make some comments that might help with future ones:

1. when dealing with complex numbers (numbers with a real and an imaginary component) you can only add the reals to the reals and the imaginaries to the imaginaries
2. if you are trying to remove a complex number (or imaginary) from a denominator, then multiply both the numerator and the denominator by the complex conjugate

i.e. the complex conjugate of 6+4i is 6-4i

this simplifies since, like you more than likely already know i=sqrt(-1) so i^2= -1, etc. this simplification can also be noted the other problems.