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?

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${sqrt(1+x)-isqrt(1-x)}/{sqrt(1+x)+isqrt(1-x)}=a-ib$

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Answer 1 · 2018-01-01T15:13:37

See below.

Using the facts

$u + i v = sqrt(u^2+v^2) e^(i phi)$ with $phi = arctan(v/u)$
$e^(i phi) = cos phi + i sin phi$

we have

$(u-iv)/(u+iv) = e^(-2i phi) = cos 2phi - i sin 2 phi$

now calling

$a = cos 2 phi$
$b = sin 2 phi$
$u = sqrt(1+x)$
$v = sqrt(1-x)$

all the conditions are satisfied.

Another approach.

Considering

$u = sqrt(1+x)$
$v = sqrt(1-x)$

$(u-iv)/(u+iv) = (u-iv)^2/((u+iv)(u-iv)) = (u^2-v^2-2i u v)/(u^2+v^2)$ or

$(1+x-(1-x)-2i sqrt(1-x^2))/(1+x+1-x) = (2x-2isqrt(1-x^2))/2 = x-i sqrt(1-x^2)$

so finally

$a = x$ and $b = sqrt(1-x^2)$

is satisfied for all $abs x le 1$