Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(d) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Complex Numbers Solutions 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 in the Argand diagram.

Solution:

A(7, 7); B(7, -7) represents the given complex numbers in the Argand diagram.

O is the mid-point of AB.

Co-ordinates of O are \(\left(\frac{7+7}{2}, \frac{7-7}{2}\right)\) = (7, 0)

Slope of \(\overleftrightarrow{\mathrm{AB}}=\frac{7+7}{7-7}=\frac{14}{0}\) = ∞

AB is parallel to Y-axis

PQ is perpendicular to AB

PQ is parallel to X-axis

Slope of PQ = 0

Equation of PQ is y – 0 = 0(x – 7)

i.e., y = 0

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

Solution:

Given points are -9 + 6i, 11 – 4i

Let A = (-9, 6); B = (11, -4)

Equation of the straight line AB is y – 6 = \(\frac{-4-6}{11+9}\) (x + 9)

⇒ y – 6 = \(\frac{-1}{2}\) (x + 9)

⇒ 2y – 12 = -x – 9

⇒ x + 2y – 3 = 0

Question 2.

If Z = x + iy and if the point P in the Argand plane represents Z, then describe geometrically the locus of z satisfying the equation.

(i) |z – 2 – 3i| = 5

(ii) 2|z – 2| = |z – 1|

(iii) Img z^{2} = 4

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

Solution:

(i) z = x + iy and |z – 2 – 3i| = 5

|z – 2 – 3i| = 5

⇒ |x + iy – 2 – 3i| = 5

⇒ |(x – 2) + i(y – 3)| = 5

⇒ \(\sqrt{(x-2)^{2}+(y-3)^{2}}\) = 5

⇒ (x – 2)^{2} + (y – 3)^{2} = 25

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

∴ Locus of P is x^{2} + y^{2} – 4x – 6y – 12 = 0

(ii) 2|z – 2| = |z – 1|

2|x + iy – 2| = |x + iy – 1|

⇒ 2|(x – 2) + iy| = |(x – 1) + iy|

⇒ \(2 \sqrt{(x-2)^{2}+y^{2}}=\sqrt{(x-1)^{2}+y^{2}}\)

Squaring both sides

⇒ 4[(x – 2)^{2} + y^{2}] = (x – 1)^{2} + y^{2}

⇒ 4(x^{2} – 4x + 4 + y^{2}) = x^{2} – 2x + 1 + y^{2}

⇒ 4x^{2} + 4y^{2} – 16x + 16 = x^{2} + y^{2} – 2x + 1

∴ Locus of P is 3x^{2} + 3y^{2} – 14x + 15 = 0

(iii) Img z^{2} = 4

∵ z = x + iy

⇒ z^{2} = (x + iy)^{2}

⇒ z^{2} = x^{2} + i^{2}y^{2} + 2ixy

⇒ z^{2} = (x^{2} – y^{2}) + i(2xy)

∴ Img (z^{2}) = 2xy = 4

∴ The locus of P is xy = 2

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

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

∴ \(\frac{2 y}{x^{2}+y^{2}-1}=\tan \frac{\pi}{4}\)

⇒ 2y = x^{2} + y^{2} – 1

⇒ x^{2} + y^{2} – 2y – 1 = 0

∴ The locus of z is x^{2} + y^{2} – 2y – 1 = 0.

Question 3.

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

Solution:

A(2, 2), B(-2, -2), C(-2√3, 2√3) represent the given complex number in the Argand diagram.

AB^{2} = (2 + 2)^{2} + (2 + 2)^{2}

= 16 + 16

= 32

BC^{2} = (-2 + 2√3)^{2} + (-2 – 2√3)^{2}

= 4 + 12 – 8√3 + 4 + 12 + 8√3

= 32

AC^{2} = (-2√3 – 2)^{2} + (2√3 – 2)^{2}

= 12 + 4 + 8√3 + 12 + 4 – 8√3

= 32

AB^{2} = BC^{2} = AC^{2}

⇒ AB = BC = CA

∴ ∆ABC is an Equilateral triangle.

Question 4.

Find the eccentricity of the ellipse whose equation 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:

Let z_{1} = x_{1} + iy_{1}; z_{2} = x_{2} + iy_{2}; z_{3} = x_{3} + iy_{3}

Given \(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\) is a real number.

Imaginary part = 0

⇒ (y_{3} – y_{1}) (x_{2} – x_{1}) – (x_{3} – x_{1}) (y_{2} – y_{1}) = 0

⇒ (y_{3} – y_{1}) (x_{2} – x_{1}) = (x_{3} – x_{1}) (y_{2} – y_{1})

⇒ \(\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

The points A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) represents the complex numbers z_{1}, z_{2}, z_{3} respectively.

Slope of \(\stackrel{\leftrightarrow}{\mathrm{AC}}\) = Slope of \(\stackrel{\leftrightarrow}{\mathrm{AB}}\)

∴ A, B, C are collinear.

Question 2.

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

Solution:

A(2, 1), B(4, 3) C(2, 5), D(0, 3) represent the given complex number in the Argand plane

AB^{2} = (2 – 4)^{2} + (1 – 3)^{2} = 4 + 4 = 8

BC^{2} = (4 – 2)^{2} + (3 – 5)^{2} = 4 + 4 = 8

CD^{2} = (2 – 0)^{2} + (5 – 3)^{2} = 4 + 4 = 8

DA^{2} = (0 – 2)^{2} + (3 – 1)^{2} = 4 + 4 = 8

AB^{2} = BC^{2} = CD^{2} = DA^{2}

⇒ AB = BC = CD = DA ………(1)

AC^{2} = (2 – 2)^{2} + (1 – 5)^{2} = 0 + 16 = 16

BD^{2} = (4 – 0)^{2} + (3 – 3)^{2} = 16 + 0 = 16

AC^{2} = BD^{2}

⇒ AC = BD …….(2)

By (1), (2)

A, B, C, D are the vertices of a square.

Question 3.

Show that the 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:

A(-2, 7), B(\(-\frac{3}{2}\), \(\frac{1}{2}\)), C(4, -3), D(\(\frac{7}{2}\), \(\frac{7}{2}\)) represents the given complex numbers in the Argand diagram.

∴ AB^{2} = BC^{2} = CD^{2} = DA^{2}

⇒ AB = BC = CD = DA ……….(1)

AC^{2} = (-2 – 4)^{2} + (7 + 3)^{2}

= 36 + 100

= 136

BD^{2} = \(\left(-\frac{3}{2}-\frac{7}{2}\right)^{2}+\left(\frac{1}{2}-\frac{7}{2}\right)^{2}\)

= 25 + 9

= 34

AC ≠ BD ……..(2)

∴ A, B, C, D are the vertices of a Rhombus.

Question 4.

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

Solution:

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

⇔ rz_{3} = -pz_{1} – qz_{2}

⇔ z_{3} = \(\frac{-p z_{1}-q z_{2}}{r}\) [∵ r ≠ 0]

∵ p + q + r = 0

⇔ r = -p – q

⇔ z_{3} = \(-\frac{\left(p z_{1}+q z_{2}\right)}{-(p+q)}\)

⇔ z_{3} = \(\frac{p z_{1}+q z_{2}}{p+q}\)

⇔ z_{3} divides the line segment joining z_{1}, z_{2} in the ratio q : p

⇔ z_{1}, z_{2}, z_{3} are collinear

Question 5.

The points P, Q denotes the complex numbers z_{1}, z_{2} in the Argand diagram. O is the origin. If \(\bar{z}_{1} \bar{z}_{2}+\bar{z}_{2} \bar{z}_{1}=0\), show that POQ = 90°.

Solution:

Let z_{1} = x_{1} + iy_{1} and z_{2} = x_{2} + iy_{2}

Then P(x_{1}, y_{1}), Q(x_{2}, y_{2}), O(0, 0)

\(\overline{\mathbf{z}}_{1}\) = x1 – iy1, \(\overline{\mathbf{z}}_{2}\) = x_{2} – iy_{2}

\(\bar{z}_{1} \bar{z}_{2}+\bar{z}_{2} \bar{z}_{1}\) = (x_{1} + iy_{1}) (x_{2} – iy_{2}) + (x_{2} + iy_{2}) (x_{1} – iy_{1}) = 0

⇒ x_{1}x_{2} + y_{1}y_{2} – ix_{1}y_{2} + ix_{2}y_{1} + x_{1}x_{2} + y_{1}y_{2} – ix_{2}y_{1} + ix_{1}y_{2} = 0

⇒ 2(x_{1}x_{2} + y_{1}y_{2}) = 0

⇒ x_{1}x_{2} + y_{1}y_{2} = 0

⇒ x_{1}x_{2} = -y_{1}y_{2}

⇒ \(\left(\frac{-x_{1}}{y_{1}}\right)\left(-\frac{x_{2}}{y_{2}}\right)=-1\)

Slope of OP × Slope of OQ = -1

∴ OP, OQ are perpendicular

⇒ ∠POQ = 90°

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 – \(\frac{6}{z}\)) = i

Solution:

Let z = cos θ + i sin θ

Given |z – 3i| = 3.

⇒ |(cos θ + i sin θ) – 3i| = 3

⇒ |cos θ + i(sin θ – 3)| = 3

⇒ \(\sqrt{\cos ^{2} \theta+(\sin \theta-3)^{2}}\) = 3

⇒ cos^{2}θ + sin^{2}θ – 6 sin θ + 9 = 9

⇒ 1 – 6 sin θ = 0

⇒ 6 sin θ = 1

⇒ sin θ = \(\frac{1}{6}\)

since 0 < θ < \(\frac{\pi}{2}\)

∴ cos θ = \(\frac{\sqrt{35}}{6}\)

cot θ = √35

∴ cot θ – \(\frac{6}{z}\) = cot θ – \(\frac{6}{\cos \theta+i \sin \theta}\)

= cot θ – 6(cos θ – i sin θ)

= √35 – 6\(\left[\frac{\sqrt{35}}{6}-i \cdot \frac{1}{6}\right]\)

= √35 – √35 + i

= i

Hence cot θ – \(\frac{6}{z}\) = i