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Value of i – Power of i – Solved Examples

Value of i

Iota ‘i’ is an imaginary number having a value equal to the square root of (-1). This is used in complex numbers and problems related to them. 

i=\sqrt{-1}

Power of i

When i is multiplied by itself, the product may be an imaginary number or real number based on the number of times it is multiplied by itself.

Let us find the values for different powers of i.

i=\sqrt{-1}

i^{2}=(\sqrt{-1})^{2}=(-1)

 

i^{3}=(\sqrt{-1})^{3}=(-1) \sqrt{-1}=(-i)

 

i^{4}=(\sqrt{-1})^{4}=(-1)(-1)=1

 

i^{5}=(\sqrt{-1})^{5}=\sqrt{-1}

 

i^{6}=(\sqrt{-1})^{6}=(-1)

 

i^{7}=(\sqrt{-1})^{7}=(-1) \sqrt{-1}=(-i)

 

i^{8}=(\sqrt{-1})^{8}=1

After finding out the above values, we can notice that the values for different powers of i follows a series. The series (\sqrt{-1},(-1),(-i), 1) keeps on repeating. Seeing this we can conclude the following points:

i^{4 n}=1  

i^{4 n+1}=\sqrt{(-1)}=i

 

i^{4 n+2}=(-1)

 

i^{4 n+3}=(-i)

Here ‘n’ is any integer.

Solved examples

  • Find the value of i^{20}.

Solution:

20 can be written as 20=4 \times 5

i^{20}=i^{4 \times 5}=i^{4 n}

And we know that i^{4 n}=1.

Hence,i^{20}=1.

  • Find the value of i+4+i^{9}

Solution:

i^{9}=i^{4(2)+1}=i^{4 n+1}=i

i+4+i^{9}=i+4+i=2 i+4=4+2 \sqrt{-1}

Hence, the value of i+4+i^{9} is equal to 4+2 \sqrt{-1}.

 

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Frequently Asked Questions 

    Q1. What is the value of iota (i)?

    Ans:  Iota ‘i’ is an imaginary number having a value equal to the square root of (-1).

    i=\sqrt{-1}

    Q2. What is the value of i^{2}?

    Ans: The value of i^{2} \text { is }(-1).