Question

A+in=√1+i÷1-i,prove that a^2+b^2=1​

Answer 1

Answer with Step-by-step explanation:

We are given that

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

We have to prove that

$a^2+b^2=1$

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

Using rationalization property

$a+ib=\sqrt{\frac{(1+i)^2}{(1^2-i^2)}}$

Using the property

$(a+b)(a-b)=a^2-b^2$

$a+ib=\sqrt{\frac{(1+i)^2}{(1-(-1))}}$

Using

$i^2=-1$

$a+ib=\frac{1+i}{\sqrt{2}}$

$a+ib=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$

By comparing we get

$a=\frac{1}{\sqrt{2}}, b=\frac{1}{\sqrt{2}}$

$a^2+b^2=(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2$

$a^2+b^2=\frac{1}{2}+\frac{1}{2}$

$a^2+b^2=\frac{1+1}{2}$

$a^2+b^2=\frac{2}{2}$

$a^2+b^2=1$

Hence, proved.