Question

The inequality $\left| {z - 4} \right| < \left| {z - 2} \right|$ represent the region given by,
A. $\operatorname{Re} \left( z \right) > 1$
B. $\operatorname{Re} \left( z \right) < 2$
C. $\operatorname{Re} \left( z \right) > 0$
D. None of these

Answer 1

As the inequality is given in the terms of complex number. The general notation of complex number is given as $z = x + iy$. And the modulus of complex number can be given as $\left| z \right| = \sqrt {{x^2} + {y^2}}$. Thus, in above inequality we use the the above mentioned concept. Also remember that x is real part of complex number while y is imaginary part of complex number. And hence choose the correct option.

Complete step by step answer:

As the given inequality is $\left| {z - 4} \right| < \left| {z - 2} \right|$
Substitute the general equation of complex number as, $z = x + iy$.
\Rightarrow $$$$\left| {x + iy - 4} \right| < \left| {x + iy - 2} \right|
Now, taking the real part and imaginary part one side.
\Rightarrow $$$$\left| {x - 4 + iy} \right| < \left| {x - 2 + iy} \right|
Now, take the modulus both side and expand the inequality as,
\Rightarrow $$$${\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}
Now, expanding the above terms as,
\Rightarrow $$$${\left( x \right)^2} - 8x + 16 + {y^2} < {\left( x \right)^2} - 4x + 4 + {y^2}
Hence, on calculating both side
\Rightarrow $$$$ - 8x + 16 < \- 4x + 4
Hence, on further simplifying, we get,
\Rightarrow $$$$16 - 4 < 8x - 4x
On calculating,
\Rightarrow $$$$12 < 4x,
and so $3 < x$
Here, as x is a real part
Hence, $\operatorname{Re} \left( z \right) < 3$
Hence, option (D) is our correct answer.

Note: A complex number is a number that can be expressed in the form $a{\text{ }} + {\text{ }}bi$ , where a and b are real numbers, and i represents the imaginary unit, satisfying the equation ${i^2} =-1$. Because no real number satisfies this equation, i is called an imaginary number.