First slide

Algebra of complex numbers

Question

If $z \in C - {0, - 2}$ is such that
$\log_{(1 / 7)} | z - 2 | > \log_{(1 / 7)} | z |$
then

Moderate

A

$Re (z) > 1$
B

$Re (z) < 1$
C

$Im (z) > 1$
D

$Im (z) < 1$

Solution

$\begin{array}{l} \log_{(1 / 7)} | z - 2 | > \log_{(1 / 7)} | z | \\ \Rightarrow | z - 2 | < | z | \end{array}$
But $| z - 2 | = | z |$ represents perpendicular bisector
of the segment joining 0 and 2, that is, represents $| z - 2 | = | z |$ the line. As 0 doe $R e (z) = 1$ is not satisfy we $(1),$ get rep $(1)$ resents $R e (z) > 1$

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