The imaginary number i is defined such as i^2=-1. What does i^+ i^2+i^3+...i^23 equal?

Question

I keep getting i^23=i^20+3=i(4x5)+3=-i

 

It should up -1 but I do not know how

Answer 1

It's a geometric series with the ratio i.  The sum is (i-i²⁴)/(1-i)=(i-1)/(1-i)=-1

Answer 2

All odd powers of i are -i and I alternatingly.

For example-   i = i

                    i ^3=-i

                    i^5 = i

 

So, i+ i^3 +i^5+i^7+i^9+i^11      ----------- +I^23 =0

Since there are all 12 values, all gets cancelled out.

 

All even powers of i are = 1 and -1 that is alternating in sign.

i^2=-1

i^4=1

i^6=-1

 

So,

i ^2+i ^4+i^6+------------i ^22

=-1

Since there are 11 values.

The first ten will cancel out.

I ^22=-1

 

So, the answer is -1

 

Answer 3

The powers of i repeat themselves every forth time. Thus i=i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, i^6 = -1, etc.

Since 23/4 = 5r3, i + i^2 + i^3 + i^4 + i^5 + ... + i ^23 = 5(i-1-i+1) + i -1 - i = 5(0) - 1 = 0-1 = -1