Find the value of {sqrt(i) + sqrt(-i)} ?

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Answer 1 · 2018-08-01T10:00:33

$sqrt(i) + sqrt(-i)$ is $sqrt2i$ or $-sqrt2i$

Let $sqrt(i)=x$ , then $sqrt(-i)=sqrt(i^3)=isqrt(i)=ix$

then $sqrt(i) + sqrt(-i)=x(1+i)$

Let $sqrti=a+bi$ , then squaring we get

$i=(a^2-b^2)+2abi$

i.e. $a^2-b^2=0$ and $2ab=1$

meaning $a^2+b^2=sqrt((a^2-b^2)^2+4a^2b^2)=sqrt1=1$

Hence $a^2=1/2$ and $b^2=1/2$ i.e. $a=1/sqrt2$ and $b=1/sqrt2$ or $a=-1/sqrt2$ and $b=-1/sqrt2$

and $sqrti=1/sqrt2+1/sqrt2i$ or $-1/sqrt2-1/sqrt2i$

and $sqrti+sqrt(-i)=(1/sqrt2+1/sqrt2i)(1+i)$

= $sqrt2i$

or $sqrti+sqrt(-i)=(-1/sqrt2-1/sqrt2i)(1+i)$

= $-sqrt2i$