Simplify the following...?

1) -1

2) -1

3) 113

Hope i helped! ;P

P.S. the guy below me is wrong when saying i to an even number is -1

the rule is:

i to an even number divisible by 2 but not by 4 is -1

i to an even number not divisible by 2 but divisible by 4 is -1

This is because i^2 = -1

and

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

Hope this helped more! ;P

We know that i=sqrt(-1)

so i^2= (sqrt(-1))^2

= -1

The next part is where we have to come up with a pattern. Rather than multiplying 78 times, we can conclude the following.

We know

i = i

i^2= -1

i^3=-i (i^2 times i)

i^4=1 (i^2 times i^2, -1x-1=1)

i^5=i (i^4 times i)

If we kept going we can definitely see a repeating cycle, every 4 powers it starts over. So now lets take the power of I, and divide by 4 (with remainder). 78/4= 19 R 2.

The only thing we really care about is the remainder. Basically what we just did was find out how many 1's we have (the whole number, so 1^19=1). This is shown exponentially

(i^4)^19 = 1^19 = 1

Now that we have gotten rid of the i^4 we are left with i^2. i^2=-1.

1x -1= -1

i^78=-1

Third.

Treat this like a quadratic and use the FOIL method to simplify.

49-56i +56i - 64i^2

49 - (-64)

49+64=113

Another thing to do here is notice that these are conjugates (a-bi)(a+bi) so you will just get a^2 +b^2.

1) i^2 = -1 (by definition)

2) i^4 = 1 so write 78 = 19*4 + 2, then i^78 = i^(19*4 + 2) = i^(19*4)*i^2 = (i^4)^19 * i^2 = 1^19 * (-1) = -1

3) (7+8i)(7-8i) = 7^2 - (8i)^2 = 49 - 64*i^2 = 49+64 = 113

i is the square root of -1, so i^2 = -1.

i to an even number is -1. so i^78 = -1.

(7+8i)(7-8i) = 7^2 - (8i)^2

= 49 - 64(-1)

= 49 + 64

= 113

i^2=-1

i^4= (i^2)(i^2)=1

i^6=i^4(i^2)=-1

i^8=i^6 i^2=1

i^78 = 78 is a multiple of 3, so it is -1