Question

The complex number $z = x + iy$ for which ${\log _{\dfrac{1}{2}}}\left| {z - 2} \right| > {\log _{\dfrac{1}{2}}}\left| z \right|$ is given by
A.$\operatorname{Re} \left( z \right) \leqslant 1$
B.$\operatorname{Im} \left( z \right) \leqslant 1$
C.$\operatorname{Re} \left( z \right) > 1$
D.$\operatorname{Im} \left( z \right) > 1$

Answer 1

Hint: In this question, we will form equations by using the properties of log and modulus of complex numbers. Then, we will solve the equations to find the required condition.

Complete step-by-step answer:
We are given that $z = x + iy$ is a complex number. We can observe that $x$ is the real part and $y$ is an imaginary part of the given complex number.
Also, we are given a condition, ${\log _{\dfrac{1}{2}}}\left| {z - 2} \right| > {\log _{\dfrac{1}{2}}}\left| z \right|$
If ${\log _m}\left| a \right| > {\log _m}\left| b \right|$ and $0 < m < 1$, then $\left| a \right| < \left| b \right|$
On solving the given condition, we get,
$\left| {z - 2} \right| < \left| z \right|$
On substituting the value of $z = x + iy$ in the above equation, we will have,
$\left| {x + iy - 2} \right| < \left| {x + iy} \right|$
Combining the real parts together, we get,
$\left| {\left( {x - 2} \right) + iy} \right| < \left| {x + iy} \right|$
Simplify the expression by calculating the modulus of complex number:
If $a + ib$ is any complex number, then the modulus is given by $\sqrt {{a^2} + {b^2}}$
Therefore, we have,
$\sqrt {{{\left( {x - 2} \right)}^2} + {y^2}} < \sqrt {{x^2} + {y^2}}$
On squaring both sides, we get,
${\left( {x - 2} \right)^2} + {y^2} < {x^2} + {y^2}$
On simplifying the above expression and solving it further, we have,
${\left( {x - 2} \right)^2} < {x^2} \\\ {x^2} - 4x + 4 < {x^2} \\\ \- 4x + 4 < 0 \\\ 4x > 4 \\\ x > 1 \\\$
The $x$is the real part of the complex number $z$. Thus $x > 1$is equivalent to $\operatorname{Re} (z) > 1$.
Thus option C is the correct answer.

Note: It is known for the inequality ${\log _m}\left| a \right| > {\log _m}\left| b \right|$, it can be reduced to $\left| a \right| < \left| b \right|$ if and only if $0 < m < 1$,otherwise the inequality is reduced to $\left| a \right| > \left| b \right|$.