First slide

Algebra of complex numbers

Question

If $|z_{1}| = |z_{2}| = 1, z_{1} z_{2} \neq - 1$ and $z = \frac{z_{1} + z_{2}}{1 + z_{1} z_{2}}$

Moderate

A

z is a purely real number
B

z is a purely imaginary number
C

|z| = 1
D

none of these

Solution

$\bar{z} = \frac{{\bar{z}}_{1} + {\bar{z}}_{2}}{1 + {\bar{z}}_{1} {\bar{z}}_{2}} = \frac{1 / z_{1} + 1 / z_{2}}{1 + 1 / z_{1} z_{2}} = \frac{z_{1} + z_{2}}{1 + z_{1} z_{2}} = z$
Thus, z is purely real.

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