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

Note: For a complex number $$z=a+ib$$ , its complex conjugate is denoted by $$\overline{z}=a-ib$$ 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 $${\displaystyle \frac{\overline{z}}{{\left|z\right|}^2}}$$ .
Whereas multiplicative inverse for a real number is defined as a reciprocal for a number x, denoted by $${\displaystyle \frac{1}{x}}$$ or $$x^{-1}$$ , is a number which when multiplied by x yields the multiplicative identity, 1. Similarly, the multiplicative inverse of a fraction $${\displaystyle \frac{a}{b}}$$ is $${\displaystyle \frac{b}{a}}$$ .