simplifying help please??

Question

1. i^5

2. sqrt - 32 (remember to deal with the negative under the radical first)

3. sqrt - 25 - sqrt - 49

4. (-3i)(2i)

5. (7 sqrt -3)(2 sqrt -27)

6. sqrt -24 / sqrt - 2

thank you all so much. if anyone solves all of them i will most definetly award a best answer.

Answer 1

1. since we know that i^2 = -1 we can simplify...

(-1)(-1)(i)= i

2. 4i sqrt 2

3. 5i-7i = -2i

4. multiply to get -6i^2 then since i^2= -1 substitute -6(-1)=6

5. First simplify on both parenthesis before taking the product.

(7i sqrt 3) (6i sqrt 3)

multiply to get
42 i^2 *3 again simplify using -1 =i^2

-126

6. first put the fraction all under the radical. sqrt( -24/-2)
in this case we don't need to worry about the -'s because they will just cancel.

so you have sqrt 12 which simplifies into 2sqrt 3


And as someone said above you really need to go back through the book or ask for help. Otherwise you will be lost on a test. Believe me I know I'm a teacher.

Answer 2

1. i^2 = -1 therefore i^4 = 1, therefore i^5 = i^4 x i = i

2. sqrt (-32) = sqrt (-1) x sqrt (16) x sqrt (2) = 4 i sqrt (2)

3. sqrt (-25) - sqrt (-49) = 5i-7i = -2i

4. -6i^2 = -6 x -1 = 6

5. 7 x sqrt (-1) x sqrt (3) x 2 x sqrt (-1) x sqrt (9) x sqrt (3)
= 14 x i x 3 x 3 = 126i

6 sqrt (4) x sqrt (-1) x sqrt (6) / sqrt (-1) x sqrt (2)
= 2 x i x sqrt (6) / i x sqrt (2)
=2 sqrt (6/2) = 2 sqrt (3)

Answer 3

You need to master the material and do your own homework

1. i^5 = i(-1)(-1) = i

2 sqrt(-32) = i squrt (32) = 4i sqrt2

3. 5i - 7i = -2i

4. 6

5. (7i sqrt 3)(6i sqrt3) and you can take it from there

by now you should be able to do # 6 if you UNDERSTAND how the above were done; I suggest you go back and restudy complex variables as you clearly do not have even very fundamental understanding.