Answer
398.7k+ views
Hint:We know that$i = \sqrt { - 1} $. So by using this basic equation and by finding the values of other powers of $i$we can solve this question. Also we can use ${x^{ - m}} = \dfrac{1}{{{x^m}}}$. Here we have to convert our question in such a way that it can be expressed using various powers of $i$ whose values are known to us.
Complete step by step solution:
Given
${i^{ - 6}}.........................................\left( 1 \right)$
We know that
$\;i = \sqrt { - 1} \;\;\; \Rightarrow {i^2} = - 1.........................\left( 2 \right)$
Also since here negative power is used, we know that ${x^{ - m}} = \dfrac{1}{{{x^m}}}$.
Using the above equation to find the value of ${i^{ - 2}}$, such that:
${i^{ - 2}} = \dfrac{1}{{{i^2}}} = \dfrac{1}{{\left( { - 1} \right)}} = - 1..............................\left( 3 \right)$
So from (3) we can find the value of ${i^{ - 4}}$, which is
${i^{ - 4}} = \left( {{i^{ - 2}}} \right) \times \left( {{i^{ - 2}}} \right) = \left( { - 1} \right) \times \left( { - 1}
\right) = 1......................\left( 4 \right)$
Now our aim is to find ${i^{ - 6}}$, which can be expressed in terms of ${i^{ - 4}}\;and\;{i^{ - 6}}$.
i.e. ${i^{ - 6}} = \left( {{i^{ - 4}}} \right) \times \left( {{i^{ - 2}}} \right)$
So from (3) and (4) we can write the values of${i^{ - 4}}\;and\;{i^{ - 6}}$.
\[ \Rightarrow {i^{ - 6}} = \left( 1 \right) \times \left( { - 1} \right) = - 1\]
Therefore \[{i^{ - 6}} = - 1\].
Note:
Formulas useful for solving these types of questions:
$
i = \sqrt { - 1} \\
{i^2} = - 1\; \\
{i^3} = - i \\
{i^4} = 1 \\
{i^{4n\;}} = 1 \\
{i^{4n - 1}} = - i \\
$
Also when negative powers come in a question we can use ${x^{ - m}} = \dfrac{1}{{{x^m}}}$.
So for finding higher or lower powers of $i$ we have to express its power in terms of the above powers and directly substitute the values associated with it since any power can be expressed in terms of the above equations. Another property widely popular and used for solving imaginary powers are the exponential properties, which form the basis for solving many problems with imaginary numbers.
Complex numbers which are numbers expressed in the form of $a + bi$ where ‘a’ is the real part and ‘b’ is the imaginary part is the field where imaginary numbers are important and are of great use.
Complete step by step solution:
Given
${i^{ - 6}}.........................................\left( 1 \right)$
We know that
$\;i = \sqrt { - 1} \;\;\; \Rightarrow {i^2} = - 1.........................\left( 2 \right)$
Also since here negative power is used, we know that ${x^{ - m}} = \dfrac{1}{{{x^m}}}$.
Using the above equation to find the value of ${i^{ - 2}}$, such that:
${i^{ - 2}} = \dfrac{1}{{{i^2}}} = \dfrac{1}{{\left( { - 1} \right)}} = - 1..............................\left( 3 \right)$
So from (3) we can find the value of ${i^{ - 4}}$, which is
${i^{ - 4}} = \left( {{i^{ - 2}}} \right) \times \left( {{i^{ - 2}}} \right) = \left( { - 1} \right) \times \left( { - 1}
\right) = 1......................\left( 4 \right)$
Now our aim is to find ${i^{ - 6}}$, which can be expressed in terms of ${i^{ - 4}}\;and\;{i^{ - 6}}$.
i.e. ${i^{ - 6}} = \left( {{i^{ - 4}}} \right) \times \left( {{i^{ - 2}}} \right)$
So from (3) and (4) we can write the values of${i^{ - 4}}\;and\;{i^{ - 6}}$.
\[ \Rightarrow {i^{ - 6}} = \left( 1 \right) \times \left( { - 1} \right) = - 1\]
Therefore \[{i^{ - 6}} = - 1\].
Note:
Formulas useful for solving these types of questions:
$
i = \sqrt { - 1} \\
{i^2} = - 1\; \\
{i^3} = - i \\
{i^4} = 1 \\
{i^{4n\;}} = 1 \\
{i^{4n - 1}} = - i \\
$
Also when negative powers come in a question we can use ${x^{ - m}} = \dfrac{1}{{{x^m}}}$.
So for finding higher or lower powers of $i$ we have to express its power in terms of the above powers and directly substitute the values associated with it since any power can be expressed in terms of the above equations. Another property widely popular and used for solving imaginary powers are the exponential properties, which form the basis for solving many problems with imaginary numbers.
Complex numbers which are numbers expressed in the form of $a + bi$ where ‘a’ is the real part and ‘b’ is the imaginary part is the field where imaginary numbers are important and are of great use.
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