How do you simplify \[{i^{100}}\] ?
Answer
623.1k+ views
Hint: Given simple can be solved if we know the laws of indices and value of \[i\] if it has an even power as power. 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 \[\sqrt { - 1} = i\] .
We know the law of brackets, that is \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\] . Using this we can solve this.
Complete step-by-step answer:
Given \[{i^{100}}\].
We can write 100 as multiplication of 50 and 2. That is \[100 = 2 \times 50\] . Above becomes
\[{i^{100}} = {i^{2 \times 50}}\]
Using the law of brackets, that is \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\] . We can write it as,
\[{i^{100}} = {\left( {{i^2}} \right)^{50}}\]
We know that the value of \[{i^2} = - 1\]
\[{i^{100}} = {\left( { - 1} \right)^{50}}\]
Since 50 is an even number. We know that even power of any negative number is positive number, hence we have:
\[ \Rightarrow {i^{100}} = 1\] is the required answer.
Therefore, the correct answer is ${i^{100}} = 1$.
Note: We also know that odd power of any negative number is always negative and that even power of any negative number is positive number. In mathematics we should know the laws of indices.
Law of multiplication: If the two terms have the same base and are to be multiplied together their indices are added. That is \[{x^m} \times {x^n} = {x^{m + n}}\] .
Law of division: If the two terms have the same base and are to be divided their indices are subtracted. That is \[\dfrac{{{x^m}}}{{{x^n}}} = {x^{m - n}}\] .
Law of brackets: If a term with a power is itself raised to a power then the powers are multiplied together. That is \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\] . We use this in most of the mathematical problems, remember them.
We know the law of brackets, that is \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\] . Using this we can solve this.
Complete step-by-step answer:
Given \[{i^{100}}\].
We can write 100 as multiplication of 50 and 2. That is \[100 = 2 \times 50\] . Above becomes
\[{i^{100}} = {i^{2 \times 50}}\]
Using the law of brackets, that is \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\] . We can write it as,
\[{i^{100}} = {\left( {{i^2}} \right)^{50}}\]
We know that the value of \[{i^2} = - 1\]
\[{i^{100}} = {\left( { - 1} \right)^{50}}\]
Since 50 is an even number. We know that even power of any negative number is positive number, hence we have:
\[ \Rightarrow {i^{100}} = 1\] is the required answer.
Therefore, the correct answer is ${i^{100}} = 1$.
Note: We also know that odd power of any negative number is always negative and that even power of any negative number is positive number. In mathematics we should know the laws of indices.
Law of multiplication: If the two terms have the same base and are to be multiplied together their indices are added. That is \[{x^m} \times {x^n} = {x^{m + n}}\] .
Law of division: If the two terms have the same base and are to be divided their indices are subtracted. That is \[\dfrac{{{x^m}}}{{{x^n}}} = {x^{m - n}}\] .
Law of brackets: If a term with a power is itself raised to a power then the powers are multiplied together. That is \[{\left( {{x^m}} \right)^n} = {x^{m \times n}}\] . We use this in most of the mathematical problems, remember them.
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