Show that one value of `(sqrti+sqrt-i)` is `sqrt2`

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The value of sqrti is

If y=sqrt(x^(2)+6x+8), show that one value of sqrt(1+iy)+sqrt(1-iy)[i=sqrt(-1)] is sqrt(2x+8)

If y=sqrt(x^(2)+6x+8), show that one value of sqrt(1+iy)+sqrt(1-iy)[i=sqrt(-1)]" "is" sqrt(2x+8).

Find the value of sqrti+sqrt-i .

If y=sqrt(x^2+6x+8) then show that one of the value of sqrt(1+iy)+sqrt(1-iy) is sqrt(2x+8) (i=sqrt(-1))

sqrt(i)-sqrt(-i)=sqrt(2)

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

Show that sqrti+sqrt(-i)=sqrt2

sqrt(-i) = (1-i)/sqrt2