# EXAMPLE PROBLEMS ON POWERS OF IMAGINARY UNIT I

Example Problems on Powers of Imaginary Unit i :

Here we are going to see some example problems on powers of imaginary unit i.

## Value of i to the Power n

For any integer n , in has only four possible values: they correspond to values of n when divided by 4 leave the remainders 0, 1, 2, and 3.

That is when the integer n  −4 or n ≥0 4 , using division algorithm, n can be written as n = 4q + k, 0 k < 4, k and q are integers and we write

 Remainder by dividing the exponent by 40123 Value of the given exponent1i-1-i

## Example Problems on Powers of Imaginary Unit i - Questions

Let us look into some practice problems based on the above solution.

Question 1 :

Simplify the following

i1947 + i1950

Solution : i1947

By dividing 1947 by 4, we get the remainder 3.

So, the value of i1947 is -i

1950

By dividing 1950 by 4, we get the remainder 2.

So, the value of i1950 is -1

i1947 + i 1950  =  -i - 1

i1947 + i 1950  =  -i - 1

Hence the answer us -1 - i.

(ii)  Simplify the following

i1948 - i-1869

Solution : i1948 - i-1869  =  1 - (1/i)

=  [(i - 1)/i] [i/i]

=  (i2 - i)/i2

=  (-1 - i)/(-1)

=  1 + i

Hence the answer is 1 + i.

(iii) Solution :

If n = 1, then in  =  i1  =  i

If n = 2, then in  =  i2  =  -1

If n = 3, then in  =  i3  =  i

If  = 4, then in  =  i4  =  1

=  i + i2 + i3 + i4 + ...................... + i12

=  (i - 1 + i + 1) + ...................... + i12

The above series will have 12 elements, it can be divided into three equal groups.

Each group will contain the above four elements. Hence the answer is 0.

(iv)  i59 + (1/i59)

Solution :

=  i59 + (1/i59

By dividing 59 by 4, we get 3 as remainder.

=  -i + (1/(-i))

=  -i - (1/i)

=  -i2 - 1

=  -(-1) - 1

=  1 - 1

=  0

(v)  i ii3 .............i2000

Solution :

i ii3 .............i2000

=  i (1 + 2 + 3 +..............+ 2000)

=  (1 + 2 + 3 +..............+ 2000)

n = 2000, a = 1, Sn  =  (n/2)[a + l]

S2000  =  (2000/2) [1 + 2000]

=  1000 

=  2001000

=  2001000

The power is 2001000 is exactly divisible by 4, we get 0 as remainder.

(vi) Solution :

If n = 1, then in + 50  =  i51

If n = 2 , then in + 50  =  i52..........

If n = 10 , then in + 50  =  i60

=  i51 + i52+ i53 + ............... + + i60

=  i51 [1 + i + i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10]

By grouping the first and the next four terms, we get 0.

=  -i [i - 1]

=  -i2 + i

=  1 + i After having gone through the stuff given above, we hope that the students would have understood, "Example Problems on Powers of Imaginary Unit i".

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