Question

a
57. A is the set of all non-zero complex numbers of the form a + ib. The binary operation * is defined such
that Z1 * Z2 = Z1 Z2 where Z1, Z2 €A. The inverse of a +ib is

A) a -ib

B) a - i b
______ ______
a^2 + b^2 a^2 + b^2

C) a + i b
______ ______
a^2 + b^2 a^2 + b^2

D) 1 + i0

E) 2 + i0
​

Answer 1

Given:

A is the set of all non-zero complex numbers of the form a + ib. The binary operation * is defined such

that Z1 * Z2 = Z1 Z2 where Z1, Z2 €A

To Find:

The inverse of a +ib

Solution:

let z be a complex number,  $z^{-1}$ be the inverse of z

Let $z^{-1}$= u+ iv be the inverse of z= a+ i b

We have,  $zz^{-1}$ = 1

That is ,

            ( a+ i b)( u +  iv )  = 1

     (au - bv)   + i (av+ub)  = 1+ i0

Equating real and imaginary parts we get

Real part          :   au -bv  =  1

Imaginary part :  av + ub = 0

Solving the above system of simultaneous equations in u and v

we get,

       $u=\frac{a}{a^{2} +b^{2} }$     &    $v=\frac{-b}{a^{2} +b^{2} }$                      (z is non-zero⇒${a^{2} +b^{2} }$> 0  )

If z =a+ ib , then

 $z^{-1}$ =    $\frac{a}{a^{2} +b^{2} }$+i   $\frac{-b}{a^{2} +b^{2} }$                                 (  $z^{-1}$is not defined when z = 0 )

The inverse of a+ib is   $\frac{a}{a^{2} +b^{2} }$+i   $\frac{-b}{a^{2} +b^{2} }$    

Answer 2

Given:

A is the set of all non-zero complex numbers of the form a + ib. The binary operation * is defined such

that Z1 * Z2 = Z1 Z2 where Z1, Z2 €A

To Find:

The inverse of a +ib

Solution:

let z be a complex number, $z^{-1}$  be the inverse of z

Let = u+ iv be the inverse of z= a+ i b

We have,  $zz^{-1}=1$

That is ,

$= > ( a+ i b)( u + iv ) = 1\\ = > (au - bv) + i (av+ub) = 1+ i0$

Equating real and imaginary parts we get

Real part          :   au -bv  =  1

Imaginary part :  av + ub = 0

Solving the above system of simultaneous equations in u and v

we get,

 $u=\frac{a}{a^{2}+b^{2} }$        &      $v=\frac{-b}{a^{2}+b^{2} }$                    (z is non-zero⇒ 0  )

If z =a+ ib , then

$z^{-1}=\frac{a}{a^{2}+b^{2} }+i\frac{-b}{a^{2} +b^{2} }$                                   ( $z^{-1}$ is not defined when z = 0 )

The inverse of a+ib is $z^{-1}=\frac{a}{a^{2}+b^{2} }+i\frac{-b}{a^{2} +b^{2} }$

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