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 4

Value of the given exponent

0

1

2

3

1

i

-1

-i

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

Question 2 :

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

Question 3 :

Simplify the following :

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.

Question 4 :

Simplify the following :

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

Question 5 :

Simplify the following :

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 [2001]

  =  2001000

  =  2001000

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

Hence the answer is 1.

Question 6 :

Simplify the following :

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

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