Question

If (a+ib)=(1+i)/(1-i) then prove that (a^(2)+b^(2))=1

Answer 1

αɳรωεɾ

Given →

$a + ib = \frac{1 + i}{1 - i}$

To prove →

a²+b² = 1

Proof→

Taking LHS of given equation , that is :-

$\frac{1 + i}{1 - i}$

Multiplying and dividing by conjugate of ( 1-i ) that is (1+i)

$\frac{1 + i}{1 - i} \times \frac{1 + i}{1 - i}$

$\frac{( {1 + i)}^{2} }{( {1)}^{2} - i ^{2} }$

As we know that value of i² = -1

$\frac{1 - 1 + 2i}{1 + 1}$

$\frac{2i}{2} = 0 + 1i$

Now comparing RHS with LHS

a+ib = 0 +1i

So , from here we got :-

a = 0 and b = 1

Now the given equation whose value we have to find out is :-

→ a² + b²

→ 0² +1²

→ 1

Hence proved