First slide

Algebra of complex numbers

Question

If the complex number $\frac{z - 1}{z + 1}$ is purely imaginary, then

Moderate

A

$| z | = 1$
B

$| z | < 1$
C

$| z | > 1$
D

$| z | \geq 2$

Solution

$\frac{z - 1}{z + 1} =$ = it where $t \in R . \Rightarrow \frac{\bar{z} - 1}{\bar{z} + 1} = - i t$
$\begin{array}{l} \Rightarrow \frac{z - 1}{z + 1} + \frac{\bar{z} - 1}{\bar{z} + 1} = 0 \\ \Rightarrow (z - 1) (\bar{z} + 1) + (\bar{z} - 1) (z + 1) = 0 \\ \Rightarrow 2 (z \bar{z} - 1) = 0 \Rightarrow | z | = 1 . \end{array}$

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