
Find the multiplicative inverse of \[\left( {1 + i} \right)\].
Answer
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Hint: Multiplicative inverse is nothing but just the reciprocal of the given quantity. But if we want to find the multiplicative inverse of a complex number then we have to do some more steps also so that we can eliminate the imaginary part from the denominator.
Complete step-by-step solution:
In the given question, we have
\[\left( {1 + i} \right)\] is a complex number.
Therefore, the multiplicative inverse of \[\left( {1 + i} \right)\] is $\dfrac{1}{{\left( {1 + i} \right)}}$.
Now,
On simplification,
$ = \dfrac{1}{{\left( {1 + i} \right)}}$
Multiplying the numerator and denominator by \[\left( {1 - i} \right)\].
$ = \dfrac{1}{{\left( {1 + i} \right)}} \times \dfrac{{\left( {1 - i} \right)}}{{\left( {1 - i} \right)}}$
Using formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$ = \dfrac{{\left( {1 - i} \right)}}{{\left( {{1^2} - {i^2}} \right)}}$
We know that $\left( {{i^2} = - 1} \right)$,
$ = \dfrac{{\left( {1 - i} \right)}}{{\left( {1 - \left( { - 1} \right)} \right)}}$
On calculation, we get
$ = \dfrac{{\left( {1 - i} \right)}}{{\left( {1 + 1} \right)}}$
On adding, we get
$ = \left( {\dfrac{{1 - i}}{2}} \right)$
Taking common $\dfrac{1}{2}$.
$ = \dfrac{1}{2}\left( {1 - i} \right)$
Therefore, the multiplicative inverse of \[\left( {1 + i} \right)\] is $\dfrac{1}{2}\left( {1 - i} \right)$.
Note: The multiplicative inverse of any complex number $a\, + \,bi$ is $\dfrac{1}{{a + bi}}$. However, since “i” is a radical and in the denominator of a fraction, many teachers will ask you to rationalize the denominator. To rationalize the denominator just multiply by the complex conjugate of the original complex number (which is now in the denominator). The multiplicative inverse of a number, say, N is represented by $\dfrac{1}{N}\,or\,{N^{ - 1}}$. It is also called reciprocal, derived from a Latin word ‘reciprocus‘. The meaning of inverse is something which is opposite. The reciprocal of a number obtained is such that when it is multiplied with the original number the value equals identity 1. In other words, it is a method of dividing a number by its own to generate identity $1$, such as $\dfrac{N}{N} = 1$.
Complete step-by-step solution:
In the given question, we have
\[\left( {1 + i} \right)\] is a complex number.
Therefore, the multiplicative inverse of \[\left( {1 + i} \right)\] is $\dfrac{1}{{\left( {1 + i} \right)}}$.
Now,
On simplification,
$ = \dfrac{1}{{\left( {1 + i} \right)}}$
Multiplying the numerator and denominator by \[\left( {1 - i} \right)\].
$ = \dfrac{1}{{\left( {1 + i} \right)}} \times \dfrac{{\left( {1 - i} \right)}}{{\left( {1 - i} \right)}}$
Using formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$ = \dfrac{{\left( {1 - i} \right)}}{{\left( {{1^2} - {i^2}} \right)}}$
We know that $\left( {{i^2} = - 1} \right)$,
$ = \dfrac{{\left( {1 - i} \right)}}{{\left( {1 - \left( { - 1} \right)} \right)}}$
On calculation, we get
$ = \dfrac{{\left( {1 - i} \right)}}{{\left( {1 + 1} \right)}}$
On adding, we get
$ = \left( {\dfrac{{1 - i}}{2}} \right)$
Taking common $\dfrac{1}{2}$.
$ = \dfrac{1}{2}\left( {1 - i} \right)$
Therefore, the multiplicative inverse of \[\left( {1 + i} \right)\] is $\dfrac{1}{2}\left( {1 - i} \right)$.
Note: The multiplicative inverse of any complex number $a\, + \,bi$ is $\dfrac{1}{{a + bi}}$. However, since “i” is a radical and in the denominator of a fraction, many teachers will ask you to rationalize the denominator. To rationalize the denominator just multiply by the complex conjugate of the original complex number (which is now in the denominator). The multiplicative inverse of a number, say, N is represented by $\dfrac{1}{N}\,or\,{N^{ - 1}}$. It is also called reciprocal, derived from a Latin word ‘reciprocus‘. The meaning of inverse is something which is opposite. The reciprocal of a number obtained is such that when it is multiplied with the original number the value equals identity 1. In other words, it is a method of dividing a number by its own to generate identity $1$, such as $\dfrac{N}{N} = 1$.
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