Question

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`

Answer 1

See below.

Explanation:

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`