Answer with Step-by-step explanation:
We are given that
[tex]a+ib=\sqrt{\frac{1+i}{1-i}}[/tex]
We have to prove that
[tex]a^2+b^2=1[/tex]
[tex]a+ib=\sqrt{\frac{(1+i)(1+i)}{(1-i)(1+i)}}[/tex]
Using rationalization property
[tex]a+ib=\sqrt{\frac{(1+i)^2}{(1^2-i^2)}}[/tex]
Using the property
[tex](a+b)(a-b)=a^2-b^2[/tex]
[tex]a+ib=\sqrt{\frac{(1+i)^2}{(1-(-1))}}[/tex]
Using
[tex]i^2=-1[/tex]
[tex]a+ib=\frac{1+i}{\sqrt{2}}[/tex]
[tex]a+ib=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}[/tex]
By comparing we get
[tex]a=\frac{1}{\sqrt{2}}, b=\frac{1}{\sqrt{2}}[/tex]
[tex]a^2+b^2=(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2[/tex]
[tex]a^2+b^2=\frac{1}{2}+\frac{1}{2}[/tex]
[tex]a^2+b^2=\frac{1+1}{2}[/tex]
[tex]a^2+b^2=\frac{2}{2}[/tex]
[tex]a^2+b^2=1[/tex]
Hence, proved.
