How do you simplify $ {i^{41}} $

Hint: Here we will split the power given on the imaginary on the function with respect to the initial four multiples of it and then will find the value for the required expression by using values of i raise to 4n will be equal to 1.

Complete step-by-step answer:
To solve for the given expression, first let us calculate for the low of powers of “I”
We know that, $ {i^2} = - 1 $
And $ {i^3} = {i^2}.i = - 1.i = - i $
Similarly, $ {i^4} = {i^3}.i = - i.i = - ( - 1) = 1 $ …. (A)
Now, the given power using the law of power and exponent can be re-written as -
  $ {i^{41}} = {i^{40}}.i $ (using the concept that powers are added when bases are same and in multiplicative)
 Similarly, Again using the multiplicative law for the powers and exponent -
  $ \Rightarrow {i^{41}} = {({i^4})^{10}}.i $
 Place the value from equation (A) in the above expression –

 $ \Rightarrow {i^{41}} = {(1)^{10}}.i $
Simplify the above equation –
 $
   \Rightarrow {i^{41}} = (1).i \\
   \Rightarrow {i^{41}} = i \;
  $
This is the required solution.
So, the correct answer is “ $ {i^{41}} $ = i ”.

Note: The complex number consists of the real part and an imaginary part and is denoted by “Z”. It can be expressed as $ z = a + ib $ where “a” is the real part and “b” is the imaginary part. The modulus of the complex number is the length of the vector and can be expressed as $ r = \sqrt {{a^2} + {b^2}} $ .
The above example can be solved using another method which follows the pattern such as $ i,( - 1),( - i),1 $ and the same pattern is repeated from the fifth term and so on, we can find the required value finding the multiple of four and then consecutive terms after it.