OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Exercise 9 (E)

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Question 1. Illustrate in the complex plane, the set of points satisfying the following conditions. Explain your answer:
(i) \( |z| \leq 3 \)
(ii) \( \arg (z – 2) = \frac { \pi }{ 3 } \)
(iii) \( | i – 1 – 2z | > 9 \)
Answer:
(i) Let \( z = x + iy \)
So, \( |z| \leq 3 \implies |x + iy| \leq 3 \)
This means \( x^2 + y^2 \leq 9 \).
This equation describes all points inside and on a circle. The center of this circle is at \( (0, 0) \) and its radius is 3 units. So, the condition represents all points within or on the boundary of this circle.

(ii) We are given \( \arg (z – 2) = \frac { \pi }{ 3 } \)
Let \( z = x + iy \).
So, \( \arg (x + iy – 2) = \frac { \pi }{ 3 } \)
This can be written as \( \tan^{-1} \left( \frac{y}{x-2} \right) = \frac{\pi}{3} \)
So, \( \frac{y}{x-2} = \tan \frac{\pi}{3} \implies \frac{y}{x-2} = \sqrt{3} \)
This gives us the equation \( y = \sqrt{3}(x – 2) \).
This equation represents a straight line. This line passes through the point \( (2, 0) \) on the x-axis and makes an angle of \( 60^\circ \) (which is \( \frac{\pi}{3} \) radians) with the positive x-axis. The diagram shows the ray starting from \( (2,0) \) and extending in this direction.

(iii) Given \( | i – 1 – 2z | > 9 \)
Let \( z = x + iy \). Substitute this into the inequality:
\( | i – 1 – 2(x + iy) | > 9 \)
\( | i – 1 – 2x – 2iy | > 9 \)
Group the real and imaginary parts:
\( | (-1 – 2x) + i(1 – 2y) | > 9 \)
Now, find the magnitude:
\( \sqrt{(-1 – 2x)^2 + (1 – 2y)^2} > 9 \)
Square both sides:
\( (-1 – 2x)^2 + (1 – 2y)^2 > 81 \)
\( (1 + 2x)^2 + (1 – 2y)^2 > 81 \)
\( 1 + 4x + 4x^2 + 1 – 4y + 4y^2 > 81 \)
\( 4x^2 + 4y^2 + 4x – 4y + 2 > 81 \)
Divide by 4:
\( x^2 + y^2 + x – y + \frac{1}{2} > \frac{81}{4} \)
Rearrange to complete the square:
\( (x^2 + x) + (y^2 – y) > \frac{81}{4} – \frac{1}{2} \)
\( (x^2 + x + \frac{1}{4}) + (y^2 – y + \frac{1}{4}) > \frac{81}{4} – \frac{2}{4} \)
\( (x + \frac{1}{2})^2 + (y – \frac{1}{2})^2 > \frac{79}{4} \)
This represents the region *exterior* to a circle. The center of this circle is \( \left(-\frac{1}{2}, \frac{1}{2}\right) \) and its radius is \( \sqrt{\frac{79}{4}} = \frac{\sqrt{79}}{2} \). The points satisfying the condition lie outside this circle.
In simple words: For part (i), the points are inside or on a circle centered at (0,0) with a radius of 3. For part (ii), the points form a straight line that starts at (2,0) and goes up at a 60-degree angle. For part (iii), the points are outside a circle centered at (-1/2, 1/2) with a radius of \( \frac{\sqrt{79}}{2} \). These plots help us visualize complex number conditions as shapes on a flat surface.

🎯 Exam Tip: When illustrating regions in the complex plane, always clearly identify the center and radius (or starting point and angle) of the geometric shape, and ensure the shading accurately reflects whether the region is inside/outside, above/below, or on the boundary. Always convert absolute value inequalities to equations of circles and argument conditions to lines or rays.

Question 2. Illustrate and explain the region of the Argand's plane represented by the inequality \( |z + i| \geq |z + 2| \).
Answer:
Let \( z = x + iy \).
The given inequality is \( |z + i| \geq |z + 2| \).
Substitute \( z = x + iy \):
\( |x + iy + i| \geq |x + iy + 2| \)
Group real and imaginary parts:
\( |x + i(y + 1)| \geq |(x + 2) + iy| \)
Now, calculate the magnitudes:
\( \sqrt{x^2 + (y + 1)^2} \geq \sqrt{(x + 2)^2 + y^2} \)
Square both sides to remove the square roots:
\( x^2 + (y + 1)^2 \geq (x + 2)^2 + y^2 \)
Expand the squared terms:
\( x^2 + y^2 + 2y + 1 \geq x^2 + 4x + 4 + y^2 \)
Subtract \( x^2 \) and \( y^2 \) from both sides:
\( 2y + 1 \geq 4x + 4 \)
Rearrange the terms to get the equation of a line:
\( 2y \geq 4x + 3 \)
\( y \geq 2x + \frac{3}{2} \)
This inequality represents all points that lie on or above the line \( y = 2x + \frac{3}{2} \). This line intersects the y-axis at \( \left(0, \frac{3}{2}\right) \) and the x-axis at \( \left(-\frac{3}{4}, 0\right) \). So the region is the half-plane above this line, including the line itself.

In simple words: This means we are looking for all points \( z \) in the Argand plane that are further away from the point \( -i \) than from the point \( -2 \), or are equally far from both. When we do the math, it turns out to be all the points that are on or above the straight line \( y = 2x + \frac{3}{2} \).

🎯 Exam Tip: When dealing with \( |z - z_1| \geq |z - z_2| \), remember that it represents a half-plane separated by the perpendicular bisector of the line segment connecting \( z_1 \) and \( z_2 \). The inequality sign determines which side of the line is included.

Question 3. Illustrate and explain the set of points z in the Argand diagram, which represents \( |z – z_1| \leq 3 \) where \( z_1 = 3 – 2i \).
Answer:
We are given the condition \( |z – z_1| \leq 3 \), where \( z_1 = 3 – 2i \).
Let \( z = x + iy \).
Substitute \( z \) and \( z_1 \) into the inequality:
\( | (x + iy) – (3 – 2i) | \leq 3 \)
Group the real and imaginary parts:
\( | (x – 3) + i(y + 2) | \leq 3 \)
Now, calculate the magnitude of the complex number:
\( \sqrt{(x – 3)^2 + (y + 2)^2} \leq 3 \)
Square both sides:
\( (x – 3)^2 + (y + 2)^2 \leq 3^2 \)
\( (x – 3)^2 + (y + 2)^2 \leq 9 \)
This equation describes all points that are inside or on the boundary of a circle. The center of this circle is \( (3, -2) \) and its radius is 3 units. So, the condition represents all points within or on the boundary of this circle.

In simple words: The problem asks us to show all complex numbers \( z \) that are a distance of 3 units or less from the complex number \( z_1 = 3 - 2i \). This creates a filled circle on the Argand plane. The center of this circle is at the point \( (3, -2) \), and its edge is 3 units away from the center in any direction.

🎯 Exam Tip: Remember that \( |z - z_1| \leq r \) always represents a closed disk (circle and its interior) with center \( z_1 \) and radius \( r \). For full marks, draw the diagram clearly, label the center, and indicate the radius.

Question 4. If \( \omega = \frac{1-zi}{z-i} \), then \( |\omega| = 1 \) implies that in the complex plane
(a) z lies on the imaginary axis
(b) z lies on the real axis
(c) z lies on the unit circle
(d) None of the options
Answer: (b) z lies on the real axis
Let \( z = x + iy \).
We are given \( \omega = \frac{1-zi}{z-i} \) and \( |\omega| = 1 \).
So, \( \left| \frac{1-zi}{z-i} \right| = 1 \)
This means \( |1-zi| = |z-i| \).
Substitute \( z = x + iy \):
\( |1-(x+iy)i| = |(x+iy)-i| \)
\( |1-xi-i^2y| = |x+i(y-1)| \)
Since \( i^2 = -1 \):
\( |1-xi+y| = |x+i(y-1)| \)
Group real and imaginary parts on the left side:
\( |(1+y) - xi| = |x + i(y-1)| \)
Now, find the magnitude of both sides:
\( \sqrt{(1+y)^2 + (-x)^2} = \sqrt{x^2 + (y-1)^2} \)
Square both sides:
\( (1+y)^2 + x^2 = x^2 + (y-1)^2 \)
Subtract \( x^2 \) from both sides:
\( (1+y)^2 = (y-1)^2 \)
Expand both sides:
\( 1 + 2y + y^2 = y^2 - 2y + 1 \)
Subtract \( y^2 \) and 1 from both sides:
\( 2y = -2y \)
\( 4y = 0 \)
This means \( y = 0 \).
The condition \( y = 0 \) implies that \( z \) lies on the real axis (the x-axis) in the complex plane. If the imaginary part is zero, the complex number is purely real.
In simple words: When we are told that the absolute value of omega is 1, and we replace z with \( x + iy \), the whole equation simplifies. After solving it, we find that the imaginary part, \( y \), must be zero. This means the complex number \( z \) can only be found on the real axis in the complex plane.

🎯 Exam Tip: The condition \( |z_1| = |z_2| \) for complex numbers \( z_1 \) and \( z_2 \) means that \( z \) lies on the perpendicular bisector of the line segment joining the points representing \( z_1 \) and \( z_2 \). For MCQs, simplifying the expression quickly and identifying the geometric representation is key.

Question 5. Find the locus of a complex number z such that \( \arg \left(\frac{z-2}{z+2}\right) = \frac { \pi }{ 3 } \).
Answer:
We are given \( \arg \left(\frac{z-2}{z+2}\right) = \frac { \pi }{ 3 } \).
Using the property \( \arg \left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) \):
\( \arg(z-2) - \arg(z+2) = \frac { \pi }{ 3 } \)
Let \( z = x + iy \).
\( \arg(x+iy-2) - \arg(x+iy+2) = \frac { \pi }{ 3 } \)
\( \arg((x-2)+iy) - \arg((x+2)+iy) = \frac { \pi }{ 3 } \)
We know that \( \arg(a+ib) = \tan^{-1}\left(\frac{b}{a}\right) \). So:
\( \tan^{-1}\left(\frac{y}{x-2}\right) - \tan^{-1}\left(\frac{y}{x+2}\right) = \frac { \pi }{ 3 } \)
Use the formula for \( \tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \):
\( \tan^{-1}\left( \frac{\frac{y}{x-2} - \frac{y}{x+2}}{1 + \left(\frac{y}{x-2}\right)\left(\frac{y}{x+2}\right)} \right) = \frac { \pi }{ 3 } \)
Take \( \tan \) of both sides:
\( \frac{\frac{y(x+2) - y(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2) + y^2}{(x-2)(x+2)}} = \tan \frac { \pi }{ 3 } \)
\( \frac{yx+2y - yx+2y}{x^2-4 + y^2} = \sqrt{3} \)
\( \frac{4y}{x^2+y^2-4} = \sqrt{3} \)
Rearrange the terms:
\( 4y = \sqrt{3}(x^2+y^2-4) \)
Divide by \( \sqrt{3} \) and move terms to one side:
\( x^2+y^2 - \frac{4}{\sqrt{3}}y - 4 = 0 \)
This is the equation of a circle. To find the center and radius, complete the square for the y-terms:
\( x^2 + \left(y^2 - \frac{4}{\sqrt{3}}y + \left(\frac{2}{\sqrt{3}}\right)^2\right) - \left(\frac{2}{\sqrt{3}}\right)^2 - 4 = 0 \)
\( x^2 + \left(y - \frac{2}{\sqrt{3}}\right)^2 - \frac{4}{3} - 4 = 0 \)
\( x^2 + \left(y - \frac{2}{\sqrt{3}}\right)^2 = 4 + \frac{4}{3} \)
\( x^2 + \left(y - \frac{2}{\sqrt{3}}\right)^2 = \frac{12+4}{3} \)
\( x^2 + \left(y - \frac{2}{\sqrt{3}}\right)^2 = \frac{16}{3} \)
This represents a circle with center \( \left(0, \frac{2}{\sqrt{3}}\right) \) and radius \( \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}} \). The locus is an arc of this circle, as the argument condition restricts the angle.

In simple words: We are looking for all complex numbers \( z \) where the angle formed by drawing lines from \( -2 \) to \( z \) and from \( 2 \) to \( z \) is exactly \( \frac{\pi}{3} \) (which is 60 degrees). When we work through the algebra, this condition defines a part of a circle. The center of this circle is at \( \left(0, \frac{2}{\sqrt{3}}\right) \) on the y-axis, and its radius is \( \frac{4}{\sqrt{3}} \).

🎯 Exam Tip: The locus for \( \arg \left(\frac{z-z_1}{z-z_2}\right) = \alpha \) is typically an arc of a circle passing through \( z_1 \) and \( z_2 \). Remember to simplify the expression using trigonometric identities for inverse tangents and then complete the square to find the circle's equation, center, and radius.

Question 6. If the amplitude of \( z – 2 – 3i \) is \( \frac { \pi }{ 4 } \), then find the locus of \( z = x +yi \).
Answer:
We are given that the amplitude (or argument) of \( z – 2 – 3i \) is \( \frac { \pi }{ 4 } \).
Let \( z = x + iy \).
Substitute \( z \) into the expression:
\( z – 2 – 3i = (x + iy) – 2 – 3i \)
Group the real and imaginary parts:
\( z – 2 – 3i = (x – 2) + i(y – 3) \)
Now, we are given that \( \arg((x – 2) + i(y – 3)) = \frac { \pi }{ 4 } \).
The argument of a complex number \( a+bi \) is \( \tan^{-1}\left(\frac{b}{a}\right) \). So:
\( \tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac { \pi }{ 4 } \)
Take \( \tan \) of both sides:
\( \frac{y-3}{x-2} = \tan \frac { \pi }{ 4 } \)
Since \( \tan \frac { \pi }{ 4 } = 1 \):
\( \frac{y-3}{x-2} = 1 \)
Multiply both sides by \( (x-2) \):
\( y – 3 = x – 2 \)
Rearrange the terms to get the equation of a line:
\( x – y + 1 = 0 \)
This equation represents a straight line. This line intersects the coordinate axes at \( (-1, 0) \) (when \( y=0 \)) and \( (0, 1) \) (when \( x=0 \)). This is a single straight line, not just a ray, as the question asks for the locus which implies all points satisfying the condition.
In simple words: The problem asks us to find the path that complex number \( z \) follows if its angle (amplitude) when shifted by \( -2-3i \) is always \( 45^\circ \). By replacing \( z \) with \( x + iy \) and solving the equation, we find that \( z \) must lie on a straight line described by the equation \( x - y + 1 = 0 \).

🎯 Exam Tip: The condition \( \arg(z - z_0) = \alpha \) represents a ray starting from \( z_0 \) and making an angle \( \alpha \) with the positive real axis. However, if the question asks for the "locus" and allows the parameter to vary, the general form of the derived equation (a line) is often expected. Double-check if the context specifies a ray or a full line.

Question 7. Find the locus of z if \( \omega = \frac{z}{z-\frac{1}{3} i} = 1 \).
Answer:
We are given \( \omega = \frac{z}{z-\frac{1}{3} i} = 1 \).
This means \( \frac{z}{z-\frac{1}{3} i} = 1 \).
Let \( z = x + iy \).
Substitute \( z \) into the equation:
\( \frac{x+iy}{ (x+iy) - \frac{1}{3} i } = 1 \)
\( \frac{x+iy}{ x + i\left(y - \frac{1}{3}\right) } = 1 \)
This means the numerator and denominator must be equal (and non-zero):
\( x+iy = x + i\left(y - \frac{1}{3}\right) \)
Equating the real parts, \( x = x \), which is always true.
Equating the imaginary parts:
\( y = y - \frac{1}{3} \)
\( 0 = -\frac{1}{3} \)
This result \( 0 = -\frac{1}{3} \) is a contradiction, implying there are no values of \( z \) for which \( \omega = 1 \).
Let's re-examine the condition \( \omega = 1 \). If \( \omega \) is a complex number, \( \omega = 1 \) means its real part is 1 and its imaginary part is 0. However, the question likely implies \( |\omega| = 1 \), which is a common context for locus problems involving \( \frac{z-z_1}{z-z_2} \). If \( \omega = 1 \), it means \( z = z - \frac{1}{3}i \), which simplifies to \( 0 = -\frac{1}{3}i \), an impossibility. The provided solution also makes this assumption of \( |\omega| = 1 \). Let's proceed with \( |\omega| = 1 \).

Assuming \( |\omega| = 1 \):
\( \left| \frac{z}{z-\frac{1}{3} i} \right| = 1 \)
So, \( |z| = \left|z - \frac{1}{3} i\right| \)
Let \( z = x + iy \).
\( |x+iy| = \left|x+iy - \frac{1}{3} i\right| \)
\( |x+iy| = \left|x + i\left(y - \frac{1}{3}\right)\right| \)
Calculate the magnitudes:
\( \sqrt{x^2+y^2} = \sqrt{x^2 + \left(y - \frac{1}{3}\right)^2} \)
Square both sides:
\( x^2+y^2 = x^2 + \left(y - \frac{1}{3}\right)^2 \)
Subtract \( x^2 \) from both sides:
\( y^2 = \left(y - \frac{1}{3}\right)^2 \)
Expand the right side:
\( y^2 = y^2 - \frac{2}{3}y + \frac{1}{9} \)
Subtract \( y^2 \) from both sides:
\( 0 = - \frac{2}{3}y + \frac{1}{9} \)
Now, solve for \( y \):
\( \frac{2}{3}y = \frac{1}{9} \)
\( y = \frac{1}{9} \times \frac{3}{2} \)
\( y = \frac{1}{6} \)
This equation \( y = \frac{1}{6} \) represents a horizontal straight line parallel to the x-axis, at a distance of \( \frac{1}{6} \) units from it. This is a common method for finding the locus where a ratio of distances from two fixed points is constant.

In simple words: The problem asks for the path of complex number \( z \) if the absolute value of the given fraction is 1. If we assume the condition is about the magnitude (absolute value), this means that \( z \) is equally far from the origin \( (0,0) \) and the point \( (0, \frac{1}{3}) \) on the imaginary axis. This leads to the equation \( y = \frac{1}{6} \), which is a horizontal line.

🎯 Exam Tip: Always be clear whether the condition is \( \omega = 1 \) or \( |\omega| = 1 \). If \( |\omega| = 1 \) for \( \omega = \frac{z-z_1}{z-z_2} \), it means \( |z-z_1| = |z-z_2| \), which always gives the perpendicular bisector of the line segment joining \( z_1 \) and \( z_2 \). For a purely "equals 1" condition, check for contradictions first.

Question 8. A variable complex number z is such that the amplitude of \( \frac{z-1}{z+1} \) is always equal to \( \frac { \pi }{ 4 } \). Illustrate the locus of z in the Argand plane.
Answer:
We are given \( \arg\left(\frac{z-1}{z+1}\right) = \frac { \pi }{ 4 } \).
Using the property \( \arg \left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) \):
\( \arg(z-1) - \arg(z+1) = \frac { \pi }{ 4 } \)
Let \( z = x + iy \).
\( \arg((x-1)+iy) - \arg((x+1)+iy) = \frac { \pi }{ 4 } \)
Using the formula \( \arg(a+ib) = \tan^{-1}\left(\frac{b}{a}\right) \):
\( \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right) = \frac { \pi }{ 4 } \)
Apply the \( \tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \) formula. This formula is valid if \( AB > -1 \).
\( \tan^{-1}\left( \frac{\frac{y}{x-1} - \frac{y}{x+1}}{1 + \left(\frac{y}{x-1}\right)\left(\frac{y}{x+1}\right)} \right) = \frac { \pi }{ 4 } \)
Take \( \tan \) of both sides:
\( \frac{\frac{y(x+1) - y(x-1)}{(x-1)(x+1)}}{\frac{(x-1)(x+1) + y^2}{(x-1)(x+1)}} = \tan \frac { \pi }{ 4 } \)
\( \frac{yx+y - yx+y}{x^2-1 + y^2} = 1 \)
\( \frac{2y}{x^2+y^2-1} = 1 \)
Rearrange the terms:
\( 2y = x^2+y^2-1 \)
Move all terms to one side:
\( x^2+y^2-2y-1 = 0 \)
To find the center and radius of this circle, complete the square for the y-terms:
\( x^2 + (y^2-2y+1) - 1 - 1 = 0 \)
\( x^2 + (y-1)^2 = 2 \)
This represents a circle with center \( (0, 1) \) and radius \( \sqrt{2} \). The locus is an arc of this circle since the argument condition limits the region.

In simple words: This question asks for the path of a complex number \( z \) where the angle formed by drawing lines from \( 1 \) to \( z \) and from \( -1 \) to \( z \) is always \( 45^\circ \). The calculations show that these points form a part of a circle. The center of this circle is at \( (0, 1) \) on the y-axis, and its radius is \( \sqrt{2} \).

🎯 Exam Tip: The locus of \( z \) such that \( \arg \left(\frac{z-z_1}{z-z_2}\right) = \alpha \) is an arc of a circle passing through \( z_1 \) and \( z_2 \). The value of \( \alpha \) determines which arc. If \( \alpha > 0 \), the arc will be on one side, and if \( \alpha < 0 \), it will be on the other. Always check the condition \( AB > -1 \) for the \( \tan^{-1} \) formula.

Question 9. Find the radius and centre of the circle \( z\bar { z }+ (1 – i) z + (1 + i) \bar { z } – 7 =0 \).
Answer:
The given equation of the circle is \( z\bar { z }+ (1 – i) z + (1 + i) \bar { z } – 7 =0 \).
Let \( z = x + iy \), then \( \bar{z} = x - iy \).
Substitute these into the equation:
\( (x+iy)(x-iy) + (1-i)(x+iy) + (1+i)(x-iy) – 7 = 0 \)
Expand the terms:
\( (x^2 - (iy)^2) + (x + iy - ix - i^2y) + (x - iy + ix - i^2y) – 7 = 0 \)
Since \( i^2 = -1 \):
\( x^2+y^2 + (x + iy - ix + y) + (x - iy + ix + y) – 7 = 0 \)
Combine real and imaginary parts:
\( x^2+y^2 + x + y + iy - ix + x + y - iy + ix – 7 = 0 \)
The imaginary terms \( (+iy - iy) \) and \( (-ix + ix) \) cancel out.
\( x^2+y^2 + 2x + 2y – 7 = 0 \)
This is the general equation of a circle: \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
Comparing, we have \( 2g = 2 \implies g = 1 \) and \( 2f = 2 \implies f = 1 \) and \( c = -7 \).
The center of the circle is \( (-g, -f) \), which is \( (-1, -1) \).
The radius of the circle is \( \sqrt{g^2+f^2-c} \):
Radius \( = \sqrt{1^2+1^2-(-7)} \)
Radius \( = \sqrt{1+1+7} \)
Radius \( = \sqrt{9} \)
Radius \( = 3 \)
So, the circle has its center at \( (-1, -1) \) and its radius is 3 units. It is useful to recognize the general form of a circle and how complex numbers can represent it.
In simple words: To find the center and radius, we replace the complex number \( z \) with \( x + iy \) and its conjugate \( \bar{z} \) with \( x - iy \) in the given equation. After simplifying, we get a standard equation for a circle. From this, we can easily find that the center of the circle is at \( (-1, -1) \) and its radius is 3.

🎯 Exam Tip: The general equation of a circle in complex form is \( z\bar{z} + \bar{\alpha}z + \alpha\bar{z} + k = 0 \), where \( -\alpha \) is the center and \( \sqrt{|\alpha|^2 - k} \) is the radius. Recognize this form to quickly extract the center and radius without full expansion, but expanding serves as a good check.

Question 10. What is the region represented by the inequality \( 3 < |z - 2 - 3i| < 4 \) in the Argand plane.
Answer:
The given inequality is \( 3 < |z - 2 - 3i| < 4 \).
Let \( z = x + iy \).
Substitute \( z \) into the expression:
\( z - 2 - 3i = (x + iy) - 2 - 3i \)
Group the real and imaginary parts:
\( z - 2 - 3i = (x - 2) + i(y - 3) \)
Now, the inequality becomes:
\( 3 < |(x - 2) + i(y - 3)| < 4 \)
Calculate the magnitude:
\( 3 < \sqrt{(x - 2)^2 + (y - 3)^2} < 4 \)
Square all parts of the inequality:
\( 3^2 < (x - 2)^2 + (y - 3)^2 < 4^2 \)
\( 9 < (x - 2)^2 + (y - 3)^2 < 16 \)
This inequality describes the region between two concentric circles. Both circles have their center at \( (2, 3) \). The inner circle has a radius of \( \sqrt{9} = 3 \) units, and the outer circle has a radius of \( \sqrt{16} = 4 \) units. The region represents an annulus or a circular ring, excluding the boundaries of both circles.

In simple words: This problem asks for the region of complex numbers \( z \) that are further than 3 units but closer than 4 units from the point \( 2 + 3i \). This creates a ring shape in the Argand plane. The center of this ring is at \( (2, 3) \), and the ring lies between a circle with radius 3 and a larger circle with radius 4. Neither the inner nor the outer edge of the ring is included in the solution.

🎯 Exam Tip: An inequality of the form \( r_1 < |z - z_0| < r_2 \) always represents an annulus (a ring-shaped region) centered at \( z_0 \), with the inner radius \( r_1 \) and outer radius \( r_2 \). If the inequalities are strict (using <), the boundaries are not included; if they are non-strict (using ≤), the boundaries are included.