
How do you simplify ${i^3}$?
Answer
540.3k+ views
Hint: Here, we are required to simplify ${i^3}$ when we know that iota $i$ is an imaginary number. Thus, using exponential identity and writing the cube of iota as a product of iota and iota square and substituting the values, we will be able to simplify the given imaginary number.
Formula used:
${a^m} \times {a^n} = {a^{m + n}}$
Complete step by step solution:
As we know,
Complex numbers are those numbers that can be written in the form of $a + bi$, where $a$ and $b$ are real numbers including 0 and $i$ is an imaginary number.
Here, an imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit called iota or $i$.
Here, $i$ is used to solve the questions when we have square roots with negative values inside them.
Here, $i = \sqrt { - 1} $
Thus, when we are given that the square of any number is negative, then, we use this imaginary number iota to solve it further.
According to the question, in order to simplify ${i^3}$, we can write this as:
${i^3} = i \cdot {i^2}$
This is because of the identity ${a^m} \times {a^n} = {a^{m + n}}$
Here, substituting $i = \sqrt { - 1} $, we get,
${i^3} = \sqrt { - 1} \cdot {\left( {\sqrt { - 1} } \right)^2}$
Cancelling put the square with the square root and hence, we get,
${i^3} = \sqrt { - 1} \cdot \left( { - 1} \right)$
But, rewriting $\sqrt { - 1} = i$, we get,
${i^3} = - i$
Therefore, we can simplify ${i^3}$ as $ - i$
This is the required answer.
Note:
Since complex numbers include both real and imaginary numbers, we should know which numbers are real numbers and which are imaginary. Any number which is present in a number system no matter whether it is positive, negative, zero, integer, rational, irrational, etc. are real numbers. Whereas, imaginary numbers are those numbers that give us a negative result when they are squared. Imaginary numbers are denoted by Iota or $i$. Also, we have $i = \sqrt { - 1} $. Squaring this on both sides, we get, ${i^2} = - 1$. But as we know, squares of any number can never be negative. Hence, these are called ‘imaginary numbers’.
Formula used:
${a^m} \times {a^n} = {a^{m + n}}$
Complete step by step solution:
As we know,
Complex numbers are those numbers that can be written in the form of $a + bi$, where $a$ and $b$ are real numbers including 0 and $i$ is an imaginary number.
Here, an imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit called iota or $i$.
Here, $i$ is used to solve the questions when we have square roots with negative values inside them.
Here, $i = \sqrt { - 1} $
Thus, when we are given that the square of any number is negative, then, we use this imaginary number iota to solve it further.
According to the question, in order to simplify ${i^3}$, we can write this as:
${i^3} = i \cdot {i^2}$
This is because of the identity ${a^m} \times {a^n} = {a^{m + n}}$
Here, substituting $i = \sqrt { - 1} $, we get,
${i^3} = \sqrt { - 1} \cdot {\left( {\sqrt { - 1} } \right)^2}$
Cancelling put the square with the square root and hence, we get,
${i^3} = \sqrt { - 1} \cdot \left( { - 1} \right)$
But, rewriting $\sqrt { - 1} = i$, we get,
${i^3} = - i$
Therefore, we can simplify ${i^3}$ as $ - i$
This is the required answer.
Note:
Since complex numbers include both real and imaginary numbers, we should know which numbers are real numbers and which are imaginary. Any number which is present in a number system no matter whether it is positive, negative, zero, integer, rational, irrational, etc. are real numbers. Whereas, imaginary numbers are those numbers that give us a negative result when they are squared. Imaginary numbers are denoted by Iota or $i$. Also, we have $i = \sqrt { - 1} $. Squaring this on both sides, we get, ${i^2} = - 1$. But as we know, squares of any number can never be negative. Hence, these are called ‘imaginary numbers’.
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