If #y=sqrt(x^2+6x+8)#,show that one value of #sqrt(1+iy)+sqrt(1-iy)=sqrt(2x+8)#?

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Answer 1 · 2017-12-30T09:17:08

Kindly find a Proof in the Explanation.

#y=sqrt(x^2+6x+8)#.

#:. y^2=x^2+6x+8#.

Adding #1# to both sides, # y^2+1=x^2+6x+9=(x+3)^2#.

Taking square root, #sqrt(y^2+1)=x+3, or, x=sqrt(y^2+1)-3#

#"Therefore, "2x+8,#

#=2{sqrt(y^2+1)-3}+8#,

#=2sqrt(y^2+1)+2#,

#=2+2sqrt(1-(-y^2))#,

#=2+2sqrt(1-(iy)^2)#,

#=2+2sqrt{(1+iy)(1-iy)#,

#=(1+iy)+(1-iy)+2sqrt{(1+iy)(1-iy)#,

#=(sqrt(1+iy))^2+(sqrt(1-iy))^2+2sqrt{(1+iy)(1-iy)}, i.e., #,

#2x+8={sqrt(1+iy)+sqrt(1-iy)}^2#.

Taking square root, we get,

#sqrt(2x+8)=sqrt(1+iy)+sqrt(1-iy),# as desired!

Q.E.D.

Enjoy Maths.!

Answer 2 · 2017-12-30T09:52:11

see below

#sqrt(1+iy)+sqrt(1−iy)=sqrt(2x+8)#

#1+iy+2sqrt(1+iy)sqrt(1−iy)+1-iy=2x+8#

#2+2sqrt(1+iy)sqrt(1−iy)=2x+8#

#sqrt((1+iy)*(1−iy))=(2x+6)/2#

#(sqrt(1−(iy)^2))^2=(x+3)^2#

#1+y^2=x^2+6x+9#

#y^2=x^2+6x+8#

#y=sqrt(x^2+6x+8)#

Now we know it is true for any number from interval: #x in (-oo,-4]uuu[-2,oo)#