Question

Prove the multiplicative inverse property of complex numbers.

Answer 1

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 $\dfrac{b}{a}$ .