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#