Question

A+in=√1+i÷1-i,prove that a^2+b^2=1
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Answer 1

Answer with Step-by-step explanation:

We are given that

`a+ib=\sqrt{(1+i)/(1-i)}`

We have to prove that

`a^2+b^2=1`

`a+ib=\sqrt{((1+i)(1+i))/((1-i)(1+i))}`

Using rationalization property

`a+ib=\sqrt{((1+i)^2)/((1^2-i^2))}`

Using the property

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

`a+ib=\sqrt{((1+i)^2)/((1-(-1)))}`

Using

`i^2=-1`

`a+ib=(1+i)/(√(2))`

`a+ib=(1)/(√(2))+i(1)/(√(2))`

By comparing we get

`a=(1)/(√(2)), b=(1)/(√(2))`

`a^2+b^2=((1)/(√(2)))^2+((1)/(√(2)))^2`

`a^2+b^2=(1)/(2)+(1)/(2)`

`a^2+b^2=(1+1)/(2)`

`a^2+b^2=(2)/(2)`

`a^2+b^2=1`

Hence, proved.