Question

If \( z = x+iy \) and \(x^2+y^2=1\), then \( \frac{1+x+iy}{1+x-iy} = \)

• A. \( \bar{z} \)
• B. \( z \)
• C. \( z+1 \)
• D. \( z-1 \)

Hint:

Given \(x^2+y^2=1 \implies |z|=1 \implies \bar{z}=1/z\). Expression \(E = \frac{1+(x+iy)}{1+(x-iy)} = \frac{1+z}{1+\bar{z}}\). Substitute \(\bar{z}=1/z\): \(E = \frac{1+z}{1+1/z} = \frac{1+z}{(z+1)/z}\). If \(z \neq -1\), \(E = (1+z) \frac{z}{1+z} = z\).

Answer 1

The Correct Option is B

Given \(z = x+iy\) and \(x^2+y^2=1\).
The condition \(x^2+y^2=1\) implies \(|z|^2=1\), so \(|z|=1\).
Since \(|z|=1\), we know that \(z\bar{z} = |z|^2 = 1\), which gives \( \bar{z} = \frac{1}{z} \) (for \(z \neq 0\)).
The expression is \( E = \frac{1+x+iy}{1+x-iy} \).
Substitute \(x+iy = z\) and \(x-iy = \bar{z}\): \[ E = \frac{1+z}{1+\bar{z}} \] Now substitute \(\bar{z} = \frac{1}{z}\) into the expression for \(E\): \[ E = \frac{1+z}{1+\frac{1}{z}} \] To simplify the denominator: \(1+\frac{1}{z} = \frac{z}{z} + \frac{1}{z} = \frac{z+1}{z}\).
So, \[ E = \frac{1+z}{\frac{z+1}{z}} \] Assuming \(1+z \neq 0\) (i.
e.
, \(z \neq -1\)), we can multiply by the reciprocal of the denominator: \[ E = (1+z) \cdot \frac{z}{z+1} \] \[ E = z \] This identity holds for all \(z\) such that \(|z|=1\) and \(z \neq -1\).
\[ \boxed{z} \]