Question

How do you simplify $ {i^{101}} $

Answer 1

Hint: Numbers are of two types – real numbers and imaginary numbers, the numbers that can be plotted on a number line are called real numbers while the imaginary numbers, as the name suggests, cannot be represented on the number line. As we know that the square root of a negative number doesn’t exist so we had to think of a way to represent them that’s why we take $ \sqrt { - 1} = i $ where $ i $ is called iota, so any other negative square root can be expressed in terms of iota, for example $ \sqrt { - 5} = \sqrt 5 i $
And the expressions like $ 2 + \sqrt { - 7} = 2 + \sqrt 7 i $ are called complex numbers as they involve an imaginary number. With the help of the values of $ i,{i^2}and\,{i^3} $ , we can solve the given problem.

Complete step-by-step answer:
Given,
 $ {i^{101}} $
As we know that –
 $
  i = \sqrt { - 1} \\
   \Rightarrow {i^2} = - 1 \\
  {i^3} = {i^2}.i \\
   \Rightarrow {i^3} = - i \;
  $
We simplify $ {i^{101}} $ as follows –
 $
  {i^{101}} = {i^{100 + 1}} \\
  {i^{101}} = {i^{100}} \times i \\
  {i^{101}} = {i^{2 \times 50}} \times i \\
   \Rightarrow {i^{101}} = {({i^2})^{50}} \times i \\
   \Rightarrow {i^{101}} = {( - 1)^{50}} \times i \\
   \Rightarrow {i^{101}} = i \;
  $
Hence the simplified value of $ {i^{101}} $ is $ i $ .
So, the correct answer is “ i”.

Note: When we are multiplying two or more numbers with exponents, we follow certain rules. According to the law of exponents, when we have to multiply two numbers whose base is the same and the exponents are different then keeping the base same, we add up the exponents that is $ {a^x} \times {a^y} = {a^{x + y}} $ . In the solution above we have used the inverse of this law that is we split 101 into 100+1 and then wrote $ {i^{100 + 1}} = {i^{100}} \times {i^1} $ . Also when a base with power has another power then keeping the base same we multiply the powers with each other that is $ {({a^x})^y} = {a^{x \times y}} $ , we have used the inverse of this law in the above solution that is we rewrote 100 as the product of 2 and 50 and then applied the inverse as $ {i^{100}} = {i^{2 \times 50}} = {({i^2})^{50}} $ .