Question

value of √i +√-i (..i=√-1) solve

Answer 1

√i + √-i = P (let )

we know,
i^2 = -1
-i ^2 = 1
-i = 1/i

so, √i + √-i = √i + √1/i

take square both sides

i + 1/i +2√i ×1/i = P^2

(i + 1/i ) +2 = P^2

(i^2+1)/i + 2 = P^2

(-1+1)/i +2 = P^2

2 = P^2

P = +_√2

Answer 2

√i + √-i  = x 
x² = i - i + 2 √(-i²) = 2
x = √2    taking positive value

i = exp(i π/2)     and     -i = exp(- i π/2)    or    exp(i 3π/2)
√i = exp(i π/4)    and   √-i = exp(- i π/4)   or   exp(i 3π/4)
√i + √-i = exp(i π/4) + exp(-i π/4)   OR  exp(iπ/4) + exp(i 3π/4)
            = 2 cos π/4            OR             i /√2 + i/√2 
            = √2         OR   √2 i

There seem to be two possible answers..purely real and purely imaginary.