math qu 's ?... and i will give u a best?

Question

Perform the indicated operations. write the answer in standard form a + Bi.

8i/1+i




find the power of i .

i ^ 29

Answer 1

Let 's do the second one first.
Recall that i^4 = 1.
So i^29 = (i^4)^7 *i = i
First one:
8i/(1+i)
To simplify this multiply numerator and denominator by
1-i, the conjugate of 1+1.
We get 8i(1-i)/ ( (1+i)(1-i) ) = 8i/2* (1-i)
= 4i(1-i) = 4 + 4i.

Answer 2

Assuming that you mean 8i/(1+i), the math is as follows:

8i/(1+i)=8i/1+8i/i=8i+8 I imagine this is what they were looking for.


If they were looking for (8i/1)+i, it would simply be 9i (+0)


In the case of i^29, this can be simplified like so:

i=sqrt(-1) square this to get:

i^2=-1 square this to get:

i^4=1


So, if i^4=1, then you can take i's away in powers of 4, because anything multiplied by 1 is itself.

i^29=i^25=i^21=....=i^5=i

So, i^29 is equal to i.

Answer 3

8i/(i+1):
multiply top/bottom by (1-i) (multiplying by (i-1) will work too)
8i(1-i)/(i+1)(1-i) = (8i+8)/(1+0i+1) = (8+8i)/2 = 4+4i

i^29:
note that i^2 = -1
i^29 = i*i^28 = i*(i^2)^14 = i(-1)^14
(a neg to a pos power is pos)
= i*1 = i

Answer 4

8i/1+i = 8

Answer 5

8i/1+i

multiply top and bottom by conjugate

so, 4 + 4i

i^2 = -1
(i^2)^14*i = -sqr(1)

Answer 6

i dont know math always had F