square root of i +square root of -i = sq. root 2.explain ?

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Given problem is

`sqrt(i)+sqrt(-i)=sqrt(2)`

LHS=`((sqrt(i)+sqrt(-i))^2)^(1/2)`

because

`x=(x^m)^(1/m) ` for all `m !=0`

Thus

LHS=`(i-i+2sqrt(-i^2))^(1/2)` ,because `(a+b)^2=a^2+b^2+2ab`

=`(2sqrt(-i^2))^(1/2)` since `-i^2=1`

therefore

`LHS=(2sqrt(1))^(1/2)`

`=(2xx1)^(1/2)`

`=(2)^(1/2)`

`=sqrt(2)`

`=RHS`

Hence proved.

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