If a,b are real and a^2+b^2=1 then show that the equation {sqrt(1+x)-isqrt(1-x)}/{sqrt(1+x)+isqrt(1-x)}=a-ib is satisfy by a real value of x?

{sqrt(1+x)-isqrt(1-x)}/{sqrt(1+x)+isqrt(1-x)}=a-ib√1+x−i√1−x√1+x+i√1−x=a−ib

1 Answer
Cesareo R.
Jan 1, 2018

See below.

Explanation:

Using the facts

u + i v = sqrt(u^2+v^2) e^(i phi)u+iv=√u2+v2eiϕ with phi = arctan(v/u)ϕ=arctan(vu)
e^(i phi) = cos phi + i sin phieiϕ=cosϕ+isinϕ

we have

(u-iv)/(u+iv) = e^(-2i phi) = cos 2phi - i sin 2 phiu−ivu+iv=e−2iϕ=cos2ϕ−isin2ϕ

now calling

a = cos 2 phia=cos2ϕ
b = sin 2 phib=sin2ϕ
u = sqrt(1+x)u=√1+x
v = sqrt(1-x)v=√1−x

all the conditions are satisfied.

Another approach.

Considering

u = sqrt(1+x)u=√1+x
v = sqrt(1-x)v=√1−x

(u-iv)/(u+iv) = (u-iv)^2/((u+iv)(u-iv)) = (u^2-v^2-2i u v)/(u^2+v^2)u−ivu+iv=(u−iv)2(u+iv)(u−iv)=u2−v2−2iuvu2+v2 or

(1+x-(1-x)-2i sqrt(1-x^2))/(1+x+1-x) = (2x-2isqrt(1-x^2))/2 = x-i sqrt(1-x^2)1+x−(1−x)−2i√1−x21+x+1−x=2x−2i√1−x22=x−i√1−x2

so finally

a = xa=x and b = sqrt(1-x^2)b=√1−x2

is satisfied for all abs x le 1|x|≤1

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