Students must practice this TS Intermediate Maths 2A Solutions Chapter 1 Functions Ex 1(d) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(d)

I.

Question 1.

i) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i.

ii) Find the equation of the straight line joining the points – 9 + 6i, 11 – 4i in the Argand plane.

Solution:

i) z_{1} = 7 + 7i

z_{2} = 7 – 71

A (7, 7) B (7, – 7)

M(7, 0)

Slope of AB = \(\frac{7+7}{7-7}\) → ∞

Line ⊥ to AB slope is zero,

y = 0 is line.

ii) A (- 9, 6) B (11, – 4)

Slope of AB = \(\frac{6+4}{-9-11}\)

= \(\frac{10}{-20}=\frac{-1}{2}\)

Equation of line AB,

y – 6 = \(\frac{- 1}{2}\) (x + 9)

2y – 12 = – x – 9

x + 2y = 3.

Question 2.

If z = x + ily and if the point Pin the Argand plane represents z, then describe geometrically the locus of z satisfying the equations

i) |z – 2 – 3i| = 5

ii) 2|z – 2| = |z – 1|

iii) im z^{2} = 4

iv) Arg \(\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)

Solution:

i) |z – 2 – 3i| = 5

|(x – 2) + (y – 3)i| = 5

(x – 2)^{2} + (y – 3)^{2} = 25

x^{2} + y^{2} – 4x – 6y + 4 + 9 – 25 = 0

x^{2} + y^{2} – 4x – 6y – 12 = 0

ii) 2|z – 2| = |z – 1|

4(z – 2) (- 2) = (z – 1) (\(\overline{\mathbf{z}}\) – 1)

4z\(\overline{\mathbf{z}}\) – 8z – 8\(\overline{\mathbf{z}}\) + 16 = z\(\overline{\mathbf{z}}\) – z – \(\overline{\mathbf{z}}\) + 1

3z\(\overline{\mathbf{z}}\) – 7z – 7\(\overline{\mathbf{z}}\) + 15 = 0

3(x^{2} + y^{2}) – 7(2x) + 15 = 0.

iii) Im (z^{2}) = 4

Im (z^{2}) = 4

z = x + iy

z^{2} = (x + iy)^{2}

z^{2} = x^{2} – y^{2} + 2xyi

Im(z^{2}) = 2xy

2xy = 4

xy = 2 rectangualr hyperbola

iv) Arg \(\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)

Question 3.

Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, – 2 – 2i, – 2√3 + 2√3i are the vertices of an equilateral triangle..

Solution:

A(2, 2), B(- 2, – 2), C (- 2√3 + 2√3)

AB = \(\sqrt{(2+2)^2+(2+2)^2}\) = 4√2

BC = \(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\)

BC = \(\sqrt{4+12-8 \sqrt{3}+4+12+8 \sqrt{3}}\)= 4√2

AC = \(\sqrt{\left(2+2 \sqrt{3}^2\right)+(2-2 \sqrt{3})^2}\) = 4√2

AB = AC = BC

∆ ABC is equilateral.

Question 4.

Find the eccentricity of the ellipse whose equtaion is | z – 4 | + |z – \(\frac{12}{5}\)| = 10

Solution:

SP + S’P = 2a

S (4, 0) S’(\(\frac{12}{5}\), 0)

2a = 10

a = 5

SS’ = 2ae

4 – \(\frac{12}{5}\) = 2 × 5e

\(\frac{8}{5}\) = 10e

e = \(\frac{4}{25}\).

II.

Question 1.

If \(\frac{z_3-z_1}{z_2-z_1}\) is a real number, show that the points represented by the complex numbers z_{1}, z_{2}, z_{3} are collinear.

Solution:

Arg \(\left(\frac{z_1-z_3}{z_1-z_2}\right)\) = 0 then \(\frac{z_1-z_3}{z_1-z_2}\) is real θ = 0.

∴ z_{1}, z_{2}, z_{3} are collinear.

Question 2.

Show that the four points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are vertices of a square.

Solution:

A (2, 1), B (4, 3), C (2, 5), D (0, 3)

AB = \(\sqrt{(4-2)^2+(3-1)^2}\) = 2√2

BC = \(\sqrt{(4-2)^2+(3-5)^2}\) = 2√2

CD = \(\sqrt{(2-0)^2+(5-3)^2}\) = 2√2

AD = \(\sqrt{(2-0)^2+(1-3)^2}\) = 2√2

Slope of AB = \(\frac{5-3}{2-4}\)= 1

Slope of BC = \(\frac{3-1}{4-2}\) = 1

AB ⊥ BC

BC ⊥ CD

⇒ ABCD is a square.

Question 3.

Show that die points in the Argand plane represented by the complex numbers – 2 + 7i, – \(\frac{-3}{2}\) + \(\frac{1}{2}\)i, 4 – 3i, \(\frac{7}{2}\) (1 + i) are the vertices of a rhombus.

Solution:

AC ⊥ BD

∴ ABCD is rhombus.

Question 4.

Show that the points in the Argand diagram represented by the complex numbers z_{1}, z_{2}, z_{3} are collhitear if and only if there exists three real numbers p, q, r not all zero satisfying p + qz_{2} + rz_{3} = 0 and p + q + r = 0.

Solution:

pz_{1} + qz_{2} + rz_{3} = 0

pz_{1} + qz_{2} = – rz_{3}

\(\left(\frac{p z_1+q z_2}{p+q}\right)\) (p + q) = – rz_{3}

Now p + q = – r

\(\left(\frac{p z_1+q z_2}{p+q}\right)\) = z_{3}

⇒ z_{3} divides z_{1} and z_{2} is q : p ratio.

∴ z_{1}, z_{2}, z_{3} are collinear.

Question 5.

The points P, Q denote the complex numbers z_{1}, z_{2} in the Argand diagram. O is

origin. If z_{1}\(\overline{\mathbf{z}}_2\) + \(\overline{\mathbf{z}}_1\)z_{2} = 0 tlien show that ∠POQ = 90°.

Solution:

z_{1}\(\overline{\mathbf{z}}_2\) + \(\overline{\mathbf{z}}_1\)z_{2} = 0

\(\frac{\mathbf{z}_1 \overline{\mathbf{z}}_2+\overline{\mathbf{z}}_1 \mathbf{z}_2}{\mathbf{z}_2 \overline{\mathrm{z}}_2}\) = 0

⇒ Real of \(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) = 0

\(\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}+\frac{\overline{\mathrm{z}}_1}{\mathrm{z}_2}\right)\)

Imaginary part of (\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\)) is k.

\(\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)+\left(\frac{\overline{\mathrm{z}}_1}{\mathrm{z}_2}\right)\) = 0 or

\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) is purely imaginary.

\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) = ki

⇒ Arg\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) = \(\frac{\pi}{2}\).

Question 6.

The complex number z has argument θ 0 < θ < \(\frac{\pi}{2}\) and satisfy the equation |z – 3i| = 3. Then prove that (cot θ – \(\frac{6}{z}\)) = 1.

Solution:

(x^{2}) + (y – 3)^{2} = 9

x + y = 0

x + y = 6y