solve and put back in a+bi form, i^7^777

Question

This is dealing with complex numbers and conjugates.  I'm not sure where to start or how I would out it back in a+bi form.  The question is about i^7^777

Answer 1

Hi Sarah,

 

The key to this problem is understanding the definition of the number i and what happens when you raise it to any power.

 

By definition,

i² = -1

 

Using the product property of exponents,

i³ = i*i² = -i

i⁴ = i²*i² = (-1)(-1) = 1

 

For any power greater than 4, you can factor out i⁴, and this factor always reduces to 1.

i⁵ = i*i⁴ = i

i⁶ = i²*i⁴ = i² = -1

i⁷ = i³*i⁴ = i³ = -i

i⁸ = i⁴*i⁴ = 1*1 = 1

 

We can simplify this problem by determining what the remainder of the exponent is after factoring out (dividing) all possible 4s from it.

 

(i⁷)⁷⁷⁷

= i⁵⁴³⁹

 

Multiples of 100 are always divisible by 4 (since 100 = 4*25), so we can pull them out

= i⁵⁴⁰⁰*i³⁹

= i³⁹

 

Now you need to find the largest multiple of 4 that is less than your exponent. In this case 4*9=36

= i³⁶*i³

= i³

 

And from above,

= -i