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

Hint: 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.