. . Powers of imaginary unit i We note that, for any integer n , i n has only four possible values: they correspond to values of n when divided by leave the remainders , , , and .That is when the integer n or using division algorithm, n can be written as n q q are integers and we write q k q q q ( ) ( ) Example . Simplify the following (i) i (ii) i (iii) i − + i (iv) i n (v) i i (i) i ; (ii) i (iii) i ( ) ( ) x' y' - - - O ( ) f x x' y' - - - O ( ) f = , i = i i i i i Fig.
. Equation has two real solutions, x and x = . We know that solving an equation in x is equivalent to finding the x -intercepts of a graph of f x crosses the x -axis at ( , ) − and ( , ) . Fig.
. By the same logic, equation has no real solutions since the graph of f x does not cross the x -axis; we can see this by looking at the graph of f x . - - (iv) = i = i … = i (What is this number?) (v) i i Result: Sum of four consecutive powers of i is zero. That is i n + i n+ + i n+ + i n+ = ∀ n ∈ Note (i) ab a b valid only if at least one of a b is non-negative.
For example, )( ) ( ) = ( )( ) i , a contradiction. Since we have taken )( ) ( , we arrived at a contradiction. Therefore ab a b valid only if at least one of a b is non-negative. (ii) For y ∈ , y ≥ Therefore, )( )( ) ( ) ( iy yi