Show that one value of #(1+i)^(1/2)-(1-i)^(1/2) = i sqrt{2(sqrt2-1)}#?

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Answer 1 · 2017-12-29T09:45:40

Please refer to the Explanation.

We have, #{(1+i)^(1/2)-(1-i)^(1/2)}^2#,

#={(1+i)^(1/2)}^2-2(1+i)^(1/2)(1-i)^(1/2)+{(1-i)^(1/2)}^2#,

#=(1+i)-2(1-i^2)^(1/2)+(1-i)#,

#=2-2{1-(-1)}^(1/2)#,

#=2-2(2)^(1/2)#,

#=2-2sqrt2#,

#=2(1-sqrt2)#,

#=2(-1)(sqrt2-1)#,

#=i^2{sqrt(2(sqrt2-1))}^2#.

# rArr (1+i)^(1/2)-(1-i)^(1/2)=i{sqrt(2(sqrt2-1))}#.

Answer 2 · 2017-12-29T10:45:53

See the proof below

Apply Demoivre's theorem

#sqrt(1+i)=sqrt2(cos(pi/8)+isin(pi/8))#

#sqrt(1-i)=sqrt2(cos(pi/8)-isin(pi/8))#

Therefore,

#sqrt(1+i)-sqrt(1-i)=sqrt2(cos(pi/8)+isin(pi/8))-sqrt2(cos(pi/8)-isin(pi/8))#

#=2sqrt2isin(pi/8)#

#cos(pi/4)=1-2sin^2(pi/8)#

#sin(pi/8)=sqrt((1-cos(pi/4))/(2))=sqrt((1-1/sqrt2)/(2))#

So,

#sqrt(1+i)-sqrt(1-i)=2sqrt2isqrt((1-1/sqrt2)/(2))#

#=isqrt(2(sqrt2-1))#

#QED#