Question

The region of Argand plane defined by \[\left| {z - 1} \right| + \left| {z + 1} \right| \le 4\;\] is
A) Interior of an ellipse
B) Exterior of an ellipse
C) Interior and boundary of an ellipse
D) None of these

Answer 1

Hint: in this question we have to find the region of Argand plane lie on which part of ellipse. First write the given complex number as a combination of real and imaginary number. . Put z in form of real and imaginary number into the equation.



Formula Used:Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one



Complete step by step solution:Given: A complex equation
Now we have complex equation equal to\[\left| {z - 1} \right| + \left| {z + 1} \right| \le 4\;\]
\[|z - 1{|^2} + |z + 1{|^2} + 2|z - 1||z + 1| \le 16\]
\[(z - 1)(\overline z - 1) + (z + 1)(\overline z + 1) + 2|(z - 1)(z + 1)| \le 16\]
\[2|z{|^2} + 2 + 2|{z^2} - 1| \le 16\]
\[|z{|^2} + |{z^2} - 1| \le 7\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number

Put this value in complex equation\[|z{|^2} + |{z^2} - 1| \le 7\]
\[|x + iy{|^2} + |{(x + iy)^2} - 1| \le 7\]
On simplification we get
\[\dfrac{{{x^2}}}{4} + \dfrac{{{y^2}}}{3} \le 1\]
This equation represents the equation of ellipse.
The region of Argand plane defined by \[\left| {z - 1} \right| + \left| {z + 1} \right| \le 4\;\] is represented interior and boundary of an ellipse.



Option ‘C’ is correct


Note: Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.