**Example Problems on Powers of Imaginary Unit i :**

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

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 0 1 2 3 |
Value of the given exponent 1 i -1 -i |

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

**Question 1 :**

Simplify the following

i^{1947} + i^{1950}

**Solution :**

i^{1947}

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

So, the value of i^{1947 }is -i

i ^{1950}

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

So, the value of i^{1950 }is -1

i^{1947} + i ^{1950 } = -i - 1

i^{1947} + i ^{1950 }= -i - 1

Hence the answer us -1 - i.

(ii) Simplify the following

i^{1948} - i^{-1869}

**Solution :**

i^{1948} - i^{-1869 } = 1 - (1/i)

= [(i - 1)/i] [i/i]

= (i^{2} - i)/i^{2}

= (-1 - i)/(-1)

= 1 + i

Hence the answer is 1 + i.

(iii)

**Solution :**

If n = 1, then i^{n} = i^{1} = i

If n = 2, then i^{n} = i^{2} = -1

If n = 3, then i^{n} = i^{3} = i

If = 4, then i^{n} = i^{4} = 1

= i + i^{2} + i^{3} + i^{4} + ...................... + i^{12}

= (i - 1 + i + 1) + ...................... + i^{12}

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) i^{59} + (1/i^{59})

**Solution :**

= i^{59} + (1/i^{59})

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

= -i + (1/(-i))

= -i - (1/i)

= -i^{2} - 1

= -(-1) - 1

= 1 - 1

= 0

(v) i i^{2 }i^{3} .............i^{2000}

**Solution :**

i i^{2 }i^{3} .............i^{2000}

= i ^{(1 + 2 + 3 +..............+ 2000)}

= i ^{(1 + 2 + 3 +..............+ 2000)}

n = 2000, a = 1, Sn = (n/2)[a + l]

S_{2000} = (2000/2) [1 + 2000]

= 1000 [2001]

= 2001000

= i ^{2001000}

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

Hence the answer is 1.

(vi)

**Solution :**

If n = 1, then i^{n + 50} = i^{51}

If n = 2 , then i^{n + 50} = i^{52..........}

If n = 10 , then i^{n + 50} = i^{60}

= i^{51 }+ i^{52}+ i^{53 + ............... + }+ i^{60}

= i^{51 }[1 + i + i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10}]

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

= -i^{ }[i - 1]

= -i^{2} + 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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