Real parameter given complex variable modulus

[jubbles]

Homework Statement

Suppose z is complex number with |z| = 1 (also assume that z is not 1 + 0*i)
Let z = a + b*i where a and b are real numbers.

Find a real parameter, t, such that

z = (t-i)/(t+i), where i = sqrt(-1)

Homework Equations

|z| = sqrt(a^2+b^2)

The Attempt at a Solution

Since we are given |z| = 1, this implies that a^2 + b^2 = (a+b*i)*(a-b*i) = 1.
Also, (t-i)/(t+i) can be expressed as [1/(t^2+1)]*[(t^2-1)+(-2*t)*i].

Despite my efforts, I end up at two deadends: I reduce an equation to a tautology like 0 = 0 or I reduce to an expression for t that involves i (thereby making t imaginary instead of real).

Any help is greatly appreciated.

 

[hunt_mat]

First off find c(t),d(t) such that:
<br /> c(t)+d(t)i=\frac{t-i}{t+i} <br />
Then use the fact that c(t)^{2}+d(t)^{2}=1

 

[SammyS]

The Attempt at a Solution

Since we are given |z| = 1, this implies that a^2 + b^2 = (a+b*i)*(a-b*i) = 1.
Also, (t-i)/(t+i) can be expressed as [1/(t^2+1)]*[(t^2-1)+(-2*t)*i].

Despite my efforts, I end up at two deadends: I reduce an equation to a tautology like 0 = 0 or I reduce to an expression for t that involves i (thereby making t imaginary instead of real).

Any help is greatly appreciated.

Write

\frac{1}{t^2+1}\,\left[(t^2-1)+(-2t)i\right]

as

\frac{(t^2-1)}{t^2+1}-\frac{2t}{t^2+1}\,i\right]