Question

Prove the multiplicative inverse property of complex numbers.

Answer 1

Hint: Multiplicative inverse of a number system is also called reciprocal. Multiplicative inverse in a number system is a number, which when multiplied by that number, gives the identity element. In complex numbers, the identity element is \[1 + {\text{i0}}\] .
To find the multiplicative inverse of a complex number \[a + ib\] , we have to find another complex number \[c + id\] such that \[\left( {a + ib} \right)\left( {c + id} \right) = 1 + i0\]

Complete step-by-step answer:
The complex number is \[a + ib\] .
We have to find the multiplicative inverse of \[a + ib\] .
Let \[c + id\] be the multiplicative inverse of \[a + ib\] .
Then by the definition of multiplicative inverse, we have
\[ \Rightarrow \left( {a + ib} \right)\left( {c + id} \right) = 1 + i0\]
By multiplying the LHS, we get
\[ \Rightarrow ac + iad + \iota bc + {i^2}bd = 1 + i0\]
Since,\[{i^2} = - 1\] ,
Therefore,
\[ \Rightarrow ac + iad + \iota bc + \left( { - 1} \right)bd = 1 + i0\]
Or
\[ \Rightarrow ac + i\left( {ad + bc} \right) - bd = 1 + i0\]
Therefore, we have
\[ \Rightarrow \left( {ac - bd} \right) + i\left( {ad + bc} \right) = 1 + i0\]
Now, comparing LHS and RHS, we can write
\[ \Rightarrow ac - bd = 1\]
And
\[ \Rightarrow ad + bc = 0\]

Now we can write
\[ \Rightarrow ad + bc = 0\]
As
\[ \Rightarrow ad = - bc\]
Therefore
\[ \Rightarrow c = \dfrac{{ - ad}}{b}\]
Now putting \[c = \dfrac{{ - ad}}{b}\] in \[ac - bd = 1\] , we can write
\[ \Rightarrow a\left( {\dfrac{{ - ad}}{b}} \right) - bd = 1\]
So we can write,
\[ \Rightarrow \left( {\dfrac{{ - {a^2}d}}{b}} \right) - \dfrac{{{b^2}d}}{b} = 1\]
That gives us,
\[ \Rightarrow \dfrac{{ - d\left( {{a^2} + {b^2}} \right)}}{b} = 1\]
After simplifying, we get
\[ \Rightarrow d = - \dfrac{b}{{\left( {{a^2} + {b^2}} \right)}}\]
Similarly, since
\[ \Rightarrow ad + bc = 0\]
Then
\[ \Rightarrow ad = - bc\]
Therefore
\[ \Rightarrow d = \dfrac{{ - bc}}{a}\]
Now putting \[d = \dfrac{{ - bc}}{a}\] in \[ac - bd = 1\] , we can write
\[ \Rightarrow ac - b\left( {\dfrac{{ - bc}}{a}} \right) = 1\]
Or
\[ \Rightarrow \dfrac{{{a^2}c}}{a} + \dfrac{{{b^2}c}}{a} = 1\]
Adding, we get
\[ \Rightarrow \dfrac{{\left( {{a^2} + {b^2}} \right)c}}{a} = 1\]
Therefore
\[ \Rightarrow c = \dfrac{a}{{\left( {{a^2} + {b^2}} \right)}}\]
Now putting the values of c and d in the multiplicative inverse \[c + id\] , we have
\[ \Rightarrow c + id = \dfrac{a}{{\left( {{a^2} + {b^2}} \right)}} - \iota \dfrac{b}{{\left( {{a^2} + {b^2}} \right)}}\]
Therefore, the multiplicative inverse of the complex number \[\left( {a + ib} \right)\] is given by \[\dfrac{a}{{\left( {{a^2} + {b^2}} \right)}} - \iota \dfrac{b}{{\left( {{a^2} + {b^2}} \right)}}\] .

Note: For a complex number \[z = a + ib\] , its complex conjugate is denoted by \[\bar z = a - ib\] and the modulus of z is denoted by \[{\left| z \right|^2} = \left( {{a^2} + {b^2}} \right)\] . Hence, we can also write the multiplicative inverse of \[z = a + ib\] as \[\dfrac{{\bar z}}{{{{\left| z \right|}^2}}}\] .
Whereas multiplicative inverse for a real number is defined as a reciprocal for a number x, denoted by \[\dfrac{1}{x}\] or \[{x^{ - 1}}\] , is a number which when multiplied by x yields the multiplicative identity, 1. Similarly, the multiplicative inverse of a fraction \[\dfrac{a}{b}\] is \[\dfrac{b}{a}\] .