
Find the value of $\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}}$
Answer
567.3k+ views
Hint: We have given an expression in ‘$i$ ’ where ‘$i$’ is called IOTO. ‘$i$’ is used to express the complex numbers. We know that value of $i$ = $\sqrt { - 1} $. So to solve this expression we put the value of ‘$i$’. Before this, we factorize the expression and cancel the terms common in numerator and denominator. Then we put the value of $i,{i^2},{i^3},{i^4}$ and solve it.
Complete step-by-step answer:
We have given
$\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}}$
The denominator is $i$ \[{i^2} + {i^2}\]it can be written as\[{i^2}(i + 1)\].
The numerator is \[{i^6} + {i^7} + {i^8} + {i^9}\]. It can be written as \[{i^2}({i^4} + {i^5} + {i^6} + {i^7})\].
So \[\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} = \dfrac{{{i^2}({i^4} + {i^5} + {i^6} + {i^7})}}{{{i^2}(1 + i)}}\]
=\[\dfrac{{({i^4} + {i^5} + {i^6} + {i^7})}}{{1 + i}}\]
Again numerator is \[{i^4} + {i^5} + {i^6} + {i^7}\]
It can be written as \[{i^4}(1 + i) + {i^6}(1 + i)\]taking \[(1 + i)\]common, we get\[({i^4} + {i^6})(1 + i)\]
So the expression can be written as
\[\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} = \dfrac{{({i^4} + {i^6})(1 + i)}}{{(1 + i)}} = {i^4} + {i^6}\]
As we know that \[i = \sqrt { - 1} \]
So \[{i^2} = i \times i = \sqrt { - 1} \times \sqrt { - 1} = {\left( {\sqrt { - 1} } \right)^2} = - 1\]
So \[{i^4} = {i^2} \times {i^2} = ( - 1) \times ( - 1) = 1\]
\[{i^6} = {i^4} \times {i^2} = 1 \times ( - 1) = - 1\]
Therefore \[{i^4} + {i^6} = 1 + ( - 1) = 1 - 1 = 0\]
Therefore $\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} = 0$.
Note: The complex number in mathematics are those numbers which can be written in $a + ib$ the form where ‘$i$’ is the imaginary number called iota and has the value $\sqrt { - 1} $ e.g. $2 + 3i$ is a complex number where $2$ is its real part and $3i$ is its imaginary part. The combination of both real and imaginary parts is called a complex number. The main application of these numbers is that they represent periodic motion such as water waves alternating current light waves etc.
There are four types of algebraic expressions that we can apply to a complex number. These operations are addition, subtraction, multiplication, and division.
Complete step-by-step answer:
We have given
$\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}}$
The denominator is $i$ \[{i^2} + {i^2}\]it can be written as\[{i^2}(i + 1)\].
The numerator is \[{i^6} + {i^7} + {i^8} + {i^9}\]. It can be written as \[{i^2}({i^4} + {i^5} + {i^6} + {i^7})\].
So \[\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} = \dfrac{{{i^2}({i^4} + {i^5} + {i^6} + {i^7})}}{{{i^2}(1 + i)}}\]
=\[\dfrac{{({i^4} + {i^5} + {i^6} + {i^7})}}{{1 + i}}\]
Again numerator is \[{i^4} + {i^5} + {i^6} + {i^7}\]
It can be written as \[{i^4}(1 + i) + {i^6}(1 + i)\]taking \[(1 + i)\]common, we get\[({i^4} + {i^6})(1 + i)\]
So the expression can be written as
\[\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} = \dfrac{{({i^4} + {i^6})(1 + i)}}{{(1 + i)}} = {i^4} + {i^6}\]
As we know that \[i = \sqrt { - 1} \]
So \[{i^2} = i \times i = \sqrt { - 1} \times \sqrt { - 1} = {\left( {\sqrt { - 1} } \right)^2} = - 1\]
So \[{i^4} = {i^2} \times {i^2} = ( - 1) \times ( - 1) = 1\]
\[{i^6} = {i^4} \times {i^2} = 1 \times ( - 1) = - 1\]
Therefore \[{i^4} + {i^6} = 1 + ( - 1) = 1 - 1 = 0\]
Therefore $\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} = 0$.
Note: The complex number in mathematics are those numbers which can be written in $a + ib$ the form where ‘$i$’ is the imaginary number called iota and has the value $\sqrt { - 1} $ e.g. $2 + 3i$ is a complex number where $2$ is its real part and $3i$ is its imaginary part. The combination of both real and imaginary parts is called a complex number. The main application of these numbers is that they represent periodic motion such as water waves alternating current light waves etc.
There are four types of algebraic expressions that we can apply to a complex number. These operations are addition, subtraction, multiplication, and division.
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