Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.6

Question 1. If \( z = x + iy \) is a complex number such that \( \left|\frac{z-4 i}{z+4 i}\right| = 1 \) show that the locus of \( z \) is real axis.
Answer: Let the complex number be \( z = x + iy \).
We are given the equation: \( \left|\frac{z-4 i}{z+4 i}\right| = 1 \)
This can be written as \( |z-4i| = |z+4i| \).
Now, substitute \( z = x + iy \) into the equation:
\( |x + iy - 4i| = |x + iy + 4i| \)
Combine the imaginary parts:
\( |x + i(y-4)| = |x + i(y+4)| \)
To find the magnitude of a complex number \( a+ib \), we use \( \sqrt{a^2+b^2} \).
So, \( \sqrt{x^2+(y-4)^2} = \sqrt{x^2 + (y+4)^2} \)
Now, square both sides to remove the square roots:
\( x^2 + (y-4)^2 = x^2 + (y+4)^2 \)
Expand the terms:
\( x^2 + y^2 - 8y + 16 = x^2 + y^2 + 8y + 16 \)
Subtract \( x^2 + y^2 + 16 \) from both sides:
\( -8y = 8y \)
Bring \( 8y \) to the left side:
\( -8y - 8y = 0 \)
\( -16y = 0 \)
So, \( y = 0 \).
This means the imaginary part of \( z \) is zero. The locus of \( z \) is therefore the real axis, which is a straight line. The distance of \( z \) from \( 4i \) is the same as its distance from \( -4i \).
In simple words: We started with an equation about complex numbers and replaced \( z \) with \( x + iy \). After solving, we found that the imaginary part, \( y \), must be zero. This means all possible values of \( z \) lie on the x-axis, which is called the real axis.

🎯 Exam Tip: When dealing with magnitudes of complex numbers, substituting \( z = x + iy \) and squaring both sides is a common and effective method to simplify the equation and find the Cartesian form of the locus.

Question 2. If \( z = x + iy \) is a complex number such that \( Im \left(\frac{2 z+1}{i z+1}\right) = 0 \), show that the locus of \( z \) is \( 2x^2 + 2y^2 - 2y + x = 0 \).
Answer: Let \( z = x + iy \).
First, we need to simplify the expression \( \frac{2z+1}{iz+1} \).
Substitute \( z = x + iy \):
\( \frac{2(x + iy) + 1}{i(x + iy) + 1} = \frac{2x + 2iy + 1}{ix + i^2y + 1} = \frac{(2x+1) + 2iy}{(1-y) + ix} \)
To find the imaginary part, we multiply the numerator and denominator by the conjugate of the denominator, which is \( (1-y) - ix \).
\( \frac{(2x+1) + 2iy}{(1-y) + ix} \times \frac{(1-y) - ix}{(1-y) - ix} \)
Numerator:
\( ((2x+1) + 2iy)((1-y) - ix) \)
\( = (2x+1)(1-y) - i(2x+1)x + 2iy(1-y) - i^2(2xy) \)
\( = (2x+1)(1-y) - ix(2x+1) + 2iy(1-y) + 2xy \)
\( = [(2x+1)(1-y) + 2xy] + i[2y(1-y) - x(2x+1)] \)
Denominator:
\( ((1-y) + ix)((1-y) - ix) = (1-y)^2 + x^2 \)
So, \( \frac{2z+1}{iz+1} = \frac{[(2x+1)(1-y) + 2xy]}{(1-y)^2 + x^2} + i\frac{[2y(1-y) - x(2x+1)]}{(1-y)^2 + x^2} \)
We are given that the imaginary part of this expression is 0.
Therefore,
\( \frac{2y(1-y) - x(2x+1)}{(1-y)^2 + x^2} = 0 \)
Since the denominator cannot be zero for the expression to be defined, the numerator must be zero.
\( 2y(1-y) - x(2x+1) = 0 \)
\( 2y - 2y^2 - 2x^2 - x = 0 \)
Multiply by -1 to make the leading terms positive:
\( 2y^2 - 2y + 2x^2 + x = 0 \)
Rearrange the terms:
\( 2x^2 + 2y^2 - 2y + x = 0 \)
This is the required locus of \( z \). This equation represents a circle. The process of finding a locus helps to visualize all possible points that satisfy a complex number condition.
In simple words: We took the complex fraction and split it into its real and imaginary parts. Since the problem said the imaginary part was zero, we set that part of the equation to zero. After some algebra, we ended up with the equation \( 2x^2 + 2y^2 - 2y + x = 0 \), which describes all the points \( z \) that fit the original rule.

🎯 Exam Tip: Remember to always multiply by the conjugate of the denominator to simplify complex fractions and separate the real and imaginary parts. This is a crucial step for locus problems involving quotients.

Question 3. Obtain the Cartesian form of the locus of \( z = x + iy \) in each of the following cases:
(i) \( [Re (iz)]^2 = 3 \)
(ii) \( Im[(1 – i)z + 1] = 0 \)
(iii) \( |z + i| = |z – 1| \)
Answer:
(i) Given \( [Re (iz)]^2 = 3 \)
Let \( z = x + iy \).
First, find \( iz \):
\( iz = i(x + iy) = ix + i^2y = ix - y = -y + ix \)
The real part of \( iz \) is \( Re(iz) = -y \).
Now substitute this into the given equation:
\( (-y)^2 = 3 \)
\( y^2 = 3 \)
So, \( y = \pm\sqrt{3} \). This represents two horizontal straight lines. The Cartesian form helps us plot these complex number solutions on a standard x-y plane.
(ii) Given \( Im[(1 – i)z + 1] = 0 \)
Let \( z = x + iy \).
First, find \( (1-i)z + 1 \):
\( (1-i)(x + iy) + 1 = x + iy - ix - i^2y + 1 \)
\( = x + iy - ix + y + 1 \)
Group the real and imaginary parts:
\( = (x + y + 1) + i(y - x) \)
The imaginary part of this expression is \( Im[(1-i)z + 1] = y - x \).
We are given that the imaginary part is 0.
\( y - x = 0 \)
So, \( y = x \). This represents a straight line passing through the origin with a slope of 1.
(iii) Given \( |z + i| = |z – 1| \)
Let \( z = x + iy \).
Substitute \( z = x + iy \) into the equation:
\( |x + iy + i| = |x + iy - 1| \)
Combine the imaginary parts on the left and real parts on the right:
\( |x + i(y+1)| = |(x-1) + iy| \)
To find the magnitude of a complex number \( a+ib \), we use \( \sqrt{a^2+b^2} \).
\( \sqrt{x^2 + (y+1)^2} = \sqrt{(x-1)^2 + y^2} \)
Square both sides to remove the square roots:
\( x^2 + (y+1)^2 = (x-1)^2 + y^2 \)
Expand the terms:
\( x^2 + y^2 + 2y + 1 = x^2 - 2x + 1 + y^2 \)
Subtract \( x^2 + y^2 + 1 \) from both sides:
\( 2y = -2x \)
Divide by 2:
\( y = -x \)
So, \( x + y = 0 \). This represents a straight line. The equation shows that any point on this line is equidistant from \( -i \) and \( 1 \).
In simple words: For each case, we changed the complex number \( z \) into its \( x+iy \) form. Then, we used the rules for real parts, imaginary parts, or magnitudes of complex numbers to turn the complex equation into a simpler equation with just \( x \) and \( y \), which shows where all the points lie on a graph.

🎯 Exam Tip: Always remember to substitute \( z = x + iy \) and then separate the real and imaginary components carefully. Mistakes often happen when handling \( i^2 = -1 \).

Question 4. Show that the following equations represent a circle, and, find its centre and radius.
(i) \( |z − 2 − i| = 3 \)
(ii) \( |2(x + iy) + 2 − 4i| = 2 \)
(iii) \( |3(x + iy) – 6 + 12i| = 8 \)
Answer:
(i) Given \( |z − 2 − i| = 3 \)
Let \( z = x + iy \).
Substitute \( z = x + iy \):
\( |x + iy − 2 − i| = 3 \)
Group the real and imaginary parts:
\( |(x − 2) + i(y − 1)| = 3 \)
The magnitude of a complex number \( a+ib \) is \( \sqrt{a^2+b^2} \).
So, \( \sqrt{(x-2)^2 + (y-1)^2} = 3 \)
Square both sides:
\( (x-2)^2 + (y-1)^2 = 3^2 \)
\( (x-2)^2 + (y-1)^2 = 9 \)
This equation is in the standard form of a circle \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h,k) \) is the centre and \( r \) is the radius.
Comparing, we find:
Centre \( (h,k) = (2, 1) \). In complex form, this is \( 2+i \).
Radius \( r = 3 \).
We can also expand the equation:
\( x^2 - 4x + 4 + y^2 - 2y + 1 = 9 \)
\( x^2 + y^2 - 4x - 2y + 5 = 9 \)
\( x^2 + y^2 - 4x - 2y - 4 = 0 \)
This also confirms it represents a circle. Every point on a circle is equidistant from its center.
(ii) Given \( |2(x + iy) + 2 − 4i| = 2 \)
Let \( z = x + iy \). The given equation is \( |2z + 2 − 4i| = 2 \).
First, factor out 2 from the term inside the magnitude:
\( |2(z + 1 − 2i)| = 2 \)
Using the property \( |ka| = |k||a| \):
\( 2|z + 1 − 2i| = 2 \)
Divide by 2:
\( |z + 1 − 2i| = 1 \)
Rewrite the complex constant term: \( +1 - 2i = -( -1 + 2i ) \)
\( |z - (-1 + 2i)| = 1 \)
This is in the standard form of a circle \( |z - z_0| = r \).
Comparing, we find:
Centre \( z_0 = -1 + 2i \).
Radius \( r = 1 \).
To show this in Cartesian form, substitute \( z = x + iy \):
\( |x + iy + 1 - 2i| = 1 \)
\( |(x+1) + i(y-2)| = 1 \)
Square both sides:
\( (x+1)^2 + (y-2)^2 = 1^2 \)
\( (x+1)^2 + (y-2)^2 = 1 \)
Expanding:
\( x^2 + 2x + 1 + y^2 - 4y + 4 = 1 \)
\( x^2 + y^2 + 2x - 4y + 5 = 1 \)
\( x^2 + y^2 + 2x - 4y + 4 = 0 \)
This is the Cartesian form of the circle. This method makes it easy to spot the center and radius.
(iii) Given \( |3(x + iy) – 6 + 12i| = 8 \)
Let \( z = x + iy \). The given equation is \( |3z – 6 + 12i| = 8 \).
First, factor out 3 from the terms inside the magnitude:
\( |3(z – 2 + 4i)| = 8 \)
Using the property \( |ka| = |k||a| \):
\( 3|z – 2 + 4i| = 8 \)
Divide by 3:
\( |z – 2 + 4i| = \frac{8}{3} \)
Rewrite the complex constant term: \( -2 + 4i = -(2 - 4i) \)
\( |z - (2 - 4i)| = \frac{8}{3} \)
This is in the standard form of a circle \( |z - z_0| = r \).
Comparing, we find:
Centre \( z_0 = 2 - 4i \).
Radius \( r = \frac{8}{3} \).
To show this in Cartesian form, substitute \( z = x + iy \):
\( |x + iy – 2 + 4i| = \frac{8}{3} \)
\( |(x-2) + i(y+4)| = \frac{8}{3} \)
Square both sides:
\( (x-2)^2 + (y+4)^2 = \left(\frac{8}{3}\right)^2 \)
\( (x-2)^2 + (y+4)^2 = \frac{64}{9} \)
Expanding:
\( x^2 - 4x + 4 + y^2 + 8y + 16 = \frac{64}{9} \)
\( x^2 + y^2 - 4x + 8y + 20 = \frac{64}{9} \)
\( x^2 + y^2 - 4x + 8y + 20 - \frac{64}{9} = 0 \)
\( x^2 + y^2 - 4x + 8y + \frac{180-64}{9} = 0 \)
\( x^2 + y^2 - 4x + 8y + \frac{116}{9} = 0 \)
This is the Cartesian form of the circle. This form allows for easier identification of key properties like the center and radius.
In simple words: For each complex equation, we transformed it into the standard form of a circle, which looks like \( |z - \text{center}| = \text{radius} \). From this, we could easily read off the center (as a complex number) and the radius. Then, we expanded it into \( x \) and \( y \) terms to show the full Cartesian equation.

🎯 Exam Tip: Remember the general equation of a circle in Cartesian form: \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The center is \( (-g, -f) \) and radius is \( \sqrt{g^2+f^2-c} \). This can be used to cross-check your results when expanding the squared terms.

Question 5. Obtain the Cartesian equation for the locus of \( z = x + iy \) in each of the following cases:
(i) \( |z − 4| = 16 \)
(ii) \( |z – 4|^2 – |z – 1|^2 = 16 \)
Answer:
(i) Given \( |z − 4| = 16 \)
Let \( z = x + iy \).
Substitute \( z = x + iy \):
\( |x + iy − 4| = 16 \)
Group the real and imaginary parts:
\( |(x − 4) + iy| = 16 \)
The magnitude of a complex number \( a+ib \) is \( \sqrt{a^2+b^2} \).
\( \sqrt{(x-4)^2 + y^2} = 16 \)
Square both sides:
\( (x-4)^2 + y^2 = 16^2 \)
\( (x-4)^2 + y^2 = 256 \)
Expand the terms:
\( x^2 - 8x + 16 + y^2 = 256 \)
Rearrange into the Cartesian form:
\( x^2 + y^2 - 8x + 16 - 256 = 0 \)
\( x^2 + y^2 - 8x - 240 = 0 \)
This equation represents a circle with center \( (4,0) \) and radius \( \sqrt{256} = 16 \). Finding the Cartesian form helps describe the shape of the complex locus geometrically.
(ii) Given \( |z – 4|^2 – |z – 1|^2 = 16 \)
Let \( z = x + iy \).
Substitute \( z = x + iy \):
\( |x + iy – 4|^2 – |x + iy – 1|^2 = 16 \)
Group the real and imaginary parts for each term:
\( |(x – 4) + iy|^2 – |(x – 1) + iy|^2 = 16 \)
The square of the magnitude of a complex number \( a+ib \) is \( a^2+b^2 \).
So, \( ((x-4)^2 + y^2) - ((x-1)^2 + y^2) = 16 \)
Expand the squared terms:
\( (x^2 - 8x + 16 + y^2) - (x^2 - 2x + 1 + y^2) = 16 \)
Remove the parentheses. Be careful with the minus sign:
\( x^2 - 8x + 16 + y^2 - x^2 + 2x - 1 - y^2 = 16 \)
Combine like terms. The \( x^2 \) and \( y^2 \) terms cancel out:
\( (-8x + 2x) + (16 - 1) = 16 \)
\( -6x + 15 = 16 \)
Subtract 15 from both sides:
\( -6x = 16 - 15 \)
\( -6x = 1 \)
Divide by -6:
\( x = -\frac{1}{6} \)
This equation represents a straight line parallel to the y-axis. The locus of \( z \) is a vertical line at \( x = -1/6 \).
In simple words: For the first part, we substituted \( z = x + iy \) and squared both sides to get rid of the magnitude. This gave us a circle's equation. For the second part, we used the square of the magnitude and expanded everything. Most terms canceled out, leaving us with a simple equation for \( x \), which is a straight line.

🎯 Exam Tip: When dealing with \( |z-a|^2 - |z-b|^2 = k \), always expect the \( x^2 \) and \( y^2 \) terms to cancel out, resulting in a linear equation (a straight line). This is a helpful check to ensure your algebraic manipulation is correct.