Simplify each power of i.

Question

2i(5-1)^2 

 

 

and 

 

 

3(5+2i^3)-i(7i^2+4i) 

 

 

 

im not sure how to simplify these.

Answer 1

As a grad student in math, one of my specialties was complex numbers. 

 

But they are not at all "complex" or difficult.

 

The easiest and best thing you can do for now is to just consider the "letter" i as just that, another letter or variable that is no different than ANY other letter.

 

Then,  REMEMBER that    i=the square root of -1     and all of this thus follows as a result of that DEFINITION:

 

i^2=-1

i^3=-i

i^4=1     simply because i^4 is really [i^2]^2 or (-1)(-1)

 

i^5=i

i^6=-1

i^7=-i

i^8=1

 

i^9=i

i^10=-1

i^11=-i

i^12=1

 

and this pattern continues forever..........

 

This should help you.

Answer 2

In math, i is used to represent the imaginary number √-1.  Then i² = -1.

 

So I think Arthur is probably right.  But maybe the i in your problem is a variable.

 

Then for 2i(5-1)² we have 2i (4)² = 2i (16)  =  32i.

So if i = 10 then 32 i = 32 times 10 = 320.

 

For 3(5 +  2i³) - i(7i²  +  4i):

 

=   3(5 +  2i³) - i(7i²  +  4i)

 

=   15   + 6i³   - 7i³    -  4i²

 

=   1 5   -       i³         -  4i²

 

I think.  But let's continue from here as if i was now given as √-1.   Then....

 

=   15    +      i           +  4 

 

=   19    +      i.   same as Arthur.

 

 

 

 

Answer 3

2i(5-i)^2

2i(25-10i+i²)

2i(25-10i-1)   (i²=-1)

2i(24-10i)

48i-20i²

48i+20

 

3(5+2i^3)-i(7i^2+4i)

3(5-2i)-i(-7+4i) because i^3=i*i^2=i*(-1)=-i

15-6i+7i-4i²

15-6i+7i+4 because i^2=-1

19+i