Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Complex Numbers Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2A Complex Numbers Important Questions

Question 1.

If z_{1} = -1, z_{2} = i then find Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) (AP Mar. 17) (TS Mar.’ 16; May ‘11)

Solution:

Z_{1} = -1 = cos π + i sin π

⇒ Arg z_{1} = π

z_{2} = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)

⇒ Arg z_{2} = \(\frac{\pi}{2}\)

⇒ Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) = Arg z_{1} – Arg z_{2} = π – \(\frac{\pi}{2}\).

= \(\frac{\pi}{2}\)

Question 2.

If z = 2 – 3i, show that z^{2} – 4z + 13 = 0. (Mar. ‘08)

Solution:

∴ z = 2 – 3i

⇒ z – 2 = -3i

⇒ (z – 2)^{2} = (-3i)^{2}

⇒ z^{2} – 4z + 4 = 9i^{2}

⇒ z^{2} – 4z + 4 = 9(-1)

⇒ z^{2} – 4z + 13 = 0

Question 3.

Find the multiplicative inverse of 7 + 24i. (TS Mar. 16)

Solution:

Since (x + iy)\(\left[\frac{x-i y}{x^{2}+y^{2}}\right]\) = 1, it follows that the multiplicative inverse of (x + iy) is \(\frac{x-i y}{x^{2}+y^{2}}\)

Hence the multiplicative inverse of 7 + 24i is

Question 4.

Write the following complex numbers in the form A + iB. (2 – 3i) (3 + 4i) (AP Mar. ’17)

Solution:

(2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i^{2}

= 6 – i + 12

= 18 – i = 18 + i(-1)

Question 5.

Write the following complex numbers in the form A + iB. (1 + 2i)^{3} (TS Mar. ’17)

Solution:

(1 + 2i)^{3} = 1 + 3.i^{2}.2i + 3.1. 4i^{2} + 8i^{3}

= 1 + 6i – 12 – 8i

= -11 – 2i = (-11) + i(-2)

Question 6.

Write the conjugate of the following complex number \(\frac{5 i}{7+i}\) (AP Mar. ’15)

Solution:

Question 7.

Find a square root for the complex number 7 + 24i. (Mar. ‘14)

Solution:

7 + 24i

Question 8.

Find a square root for the complex number 3 + 4i (Mar. ’13)

Solution:

Question 9.

Express the following complex numbers in modulus amplitude form. 1 – i (AP Mar. 15)

Solution:

1 – i

Let 1 – i = r (cos θ + i sin θ)

Equating real and imaginary parts

r cos θ = 1

r sin θ = -1

⇒ θ lies in IV quadrant .

Squaring and adding

r^{2} (cos^{2} θ + sin^{2} θ) = 1 + 1 = 2

r^{2} = 2 ⇒ r = \(\sqrt{2}\)

tan θ = -1

⇒ θ = -π/4

Question 10.

Express the complex numbers in modulus — amplitude form 1 + i\(\sqrt{3}\) (TS Mar. ’17)

Solution:

1 + i\(\sqrt{3}\) = r (cos θ + i sin θ)

Equating real and imaginary parts

r cos θ = 1 —– (1)

r sin θ = \(\sqrt{3}\) —– (2)

θ lies in I quadrant

Squaring and adding (1) and (2)

r^{2} (cos^{2} θ – sin^{2} θ) = 1 + 3

r^{2} = 4 ⇒ r = 2

Dividing (2) by (1)

Question 11.

If the Arg \(\overline{\mathbf{z}}_{1}\) and Arg \(z_{2}\) are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, find (Arg z_{1} + Arg z_{2}) (AP Mar. ’16)

Solution:

Arg \(\overline{\mathbf{z}}_{1}\) = \(\frac{\pi}{5}\) ⇒ Arg z_{1} = – Arg z_{1} = – \(\frac{\pi}{5}\)

Arg z_{2} = \(\frac{\pi}{3}\)

∴ Arg z_{1} + Arg z_{2} = – \(\frac{\pi}{5}\) + \(\frac{\pi}{3}\)

= \(\frac{-3 \pi+5 \pi}{15}\) = \(\frac{2 \pi}{15}\)

Question 12.

If |z – 3 + i| = 4 determine the locus of z. (May. ’14)

Solution:

Let z = x + iy

Given |z – 3 + i| = 4

|x + iy – 3 + i| = 4

⇒ (x – 3) + i(y + 1) = 4

⇒ \(\sqrt{(x-3)^{2}+(y+1)^{2}}\) = 4

⇒ (x – 3)^{2} + (y + 1)^{2} = 16

⇒ x^{2} – 6x + 9 + y^{2} + 2y + 1 = 16

⇒ x^{2} + y^{2} – 6x + 2y – 6 = 0

∴ The locus õf z is x^{2} + y^{2} – 6x + 2y – 6 = 0

Question 13.

The points P, Q denote the complex numbers z_{1}, z_{2} in the Argand diagram. O is the origin. If z_{1}z_{2} + z_{2}z_{1} = 0, show that POQ = 90°. (Mar. ‘07)

Solution:

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

Question 14.

Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\). (TS Mar. 15)

Solution:

Question 15.

Write z = –\(\sqrt{7}\) + i\(\sqrt{21}\) in the polar form. (Mar. ’11)

Solution:

Since the given point lies in the second quadrant we look for a solution of tan θ = –\(\sqrt{3}\) that lies in \(\left[\frac{\pi}{2}, \pi\right]\), we find that θ = \(\frac{2 \pi}{3}\) is such a solution.

Question 16.

z = x + iy and the point P represents z in the Argand plane and \(\left|\frac{z-a}{z+\bar{a}}\right|\) = 1, Re (a) ≠ 0, then find the locus of P. (TS Mar. ’17)

Solution:

Let z = x + iy and a = α + iβ

Locus of P is x = 0 i.e., Y – axis

Question 17.

If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\), show that 4x^{2} – 1 = 0 (AP Mar. ’16, TS Mar. ’17, ’15, ’06 )

Solution:

Equating real parts on both sides, we have

x = \(\frac{1}{2}\)

2x = 1

⇒ 4x^{2} = 1

4x^{2} – 1 = 0

Question 18.

If (\(\sqrt{3}\) + 1)^{100} = 2^{99} (a + ib), then show that a^{2} + b^{2} = 4. (AP Mar. ‘16)

Solution:

Question 19.

Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2\(\sqrt{3}\) + 2\(\sqrt{3}\)i are the vertices of an equilateral triangle. (Mar ‘07)

Solution:

A (2, 2), B (-2, -2), C (-2\(\sqrt{3}\), 2\(\sqrt{3}\)) represents the given complex number in the Argand diagram.

Question 20.

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. (June 04) (TS Mar. ’16; AP Mar.’15 ’05; May ’05)

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

AC^{2} = (-2 – 4)^{2} + (7 + 3)^{2}

= 36 +100 = 136

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

= 25 + 9 = 34

AC ≠ BD

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

Question 21.

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 exists three real numbers p, q, r not all zero, satisfying pz_{1} + qz_{2} + rz_{3} = 0 and p + q + r = 0. (Mar. ‘07)

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

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

⇔ z_{1}, z_{2}, z_{3} are collinear

Question 22.

If the amplitude of \(\left(\frac{z-2}{z-6 i}\right) \frac{\pi}{2}\), find its locus. (Mar. ’06)

Solution:

Let z = (x + iy)

Hence a = 0 and b > 0

∴ x(x – 2) + y(y – 6) = 0

or x^{2} + y^{2} – 2x – 6y = 0.

Question 23.

If x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then show that x^{2} + y^{2} = 4x – 3 (TS Mar ’17)

Solution:

Equating real and imaginary parts on both sides, we have

Question 24.

Express \(\frac{4+2 i}{1-2 i}\) + \(\frac{3+4 i}{2+3 i}\) in the form a + ib, a ∈ R, b ∈ R.

Solution:

Question 25.

Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\) (TS Mar ’15)

Solution:

Question 26.

Express (1 – 3)^{3} (1 + i) in the form of a + ib.

Solution:

(1 – i)^{3} (1 + j) = (1 – j)^{2} (1 – i) (1 + j)

= (1 + i^{2} – 2i) (1^{2} – i^{2})

= (1 – 1 – 2i) (1 + 1)

= 0 – 4i = 0 + i (-4)

Question 27.

Find the multiplicative inverse of 7 + 24i. (TS. Mar. ’16 )

Solution:

Since (x + iy)\(\left[\frac{x-i y}{x^{2}+y^{2}}\right]\) = 1, it follows that the multiplicative inverse of

Question 28.

Determine the locus of z, z ≠ 2i, such that Re\(\left(\frac{z-4}{z-2 i}\right)\) = 0

Solution:

Let z = x + iy

Hence the locus of the given point representing the complex number is the circle with (2, 1) as centre and \(\sqrt{5}\) units as radius, excluding the point (0, 2).

Question 29.

If 4x + i (3x – y) = 3 -6i where x and y are real numbers, then find the values of x and y.

Solution:

∵ 4x + i(3x – y) = 3 – 6i

Equating real and imaginary parts, we get 4x = 3 and 3x – y = -6

4x = 3 and 3x – y = -6

⇒ x = 3/4 and 3\(\left(\frac{3}{4}\right)\) – y = -6

\(\frac{9}{4}\) + 6 = y

⇒ y = \(\frac{33}{4}\)

∴ x = \(\frac{3}{4}\) and y = \(\frac{33}{4}\)

Question 30.

If z = 2 – 3i, show that z^{2} – 4z + 13 = 0. (Mar. ’08)

Solution:

∴ z = 2 – 3i

⇒ z – 2 = – 3i

⇒ (z – 2)^{2} = (-3i)^{2}

⇒ z^{2} – 4z + 4 = 9i^{2}

⇒ z^{2} – 4z + 13 = 0

⇒ z^{2} – 4z + 4 = 9

Question 31.

Find the complex conjugate of (3 + 4i) (2 – 3i).

Solution:

The given complex number is

(3 + 4i) (2 – 3i) = 6 + 8i – 9i – 12i^{2}

= 6 – i – 12(-1) = 18 + i

Its complex conjugate is 18 + i

Question 32.

Show that z_{1} = \(\frac{2+11 i}{25}\), z_{2} = \(\frac{-2+i}{(1-2 i)^{2}}\), are conjugate to each other.

Solution:

Since, this complex number is the conjugate of \(\frac{2+11 i}{25}\), the two given complex numbers

are conjugate to each other.

Question 33.

Find the square root of (-5 + 12i).

Solution:

We have \(\sqrt{a+i b}\) =

In this example a = -5, b = 12

Question 34.

Write z = –\(\sqrt{7}\) + i\(\sqrt{21}\) in the polar form. (Mar ’11)

Solution:

Since the given point lies in the second quadrant we look for a solution of

tan θ = – \(\sqrt{3}\) that lies in [\(\frac{\pi}{2}\), π] we find that θ = \(\frac{2 \pi}{3}\) is such a solution.

∴ –\(\sqrt{7}\) + i\(\sqrt{21}\) = 2\(\sqrt{7}\) cis \(\frac{2 \pi}{3}\)

(or) 2\(\sqrt{7}\)(cos \(\frac{2 \pi}{3}\) + i sin \(\frac{2 \pi}{3}\))

Question 35.

Express -1 – i in polar form with principle value of the amplitude.

Solution:

Let -1 – i = r (cos θ + i sin θ), then

-1 = r cos θ, -1 = r sin θ, tan θ = 1 ——— (1)

∴ r^{2} = 2

⇒ r = ±\(\sqrt{2}\)

Since θ is positive, -π < θ < π, the value θ satisfying the equation (1) is

θ = -135° = \(\frac{-3 \pi}{4}\)

Question 36.

If the amplitude of \(\left(\frac{z-2}{z-6 i}\right) \frac{\pi}{2}\), find its locus. (Mar. ’06)

Solution:

By hypothesis, amplitude of a + ib = \(\frac{\pi}{2}\)

So \(\frac{\pi}{2}\) = tan^{-1} \(\frac{b}{a}\)

Hence a = 0 and b > 0

∴ x(x – 2) + y(y – 6) = 0

or x^{2} + y^{2} – 2x – 6y = 0.

Question 37.

Show that the equation of any circle in the complex plane is of the form z\(\overline{\mathbf{z}}\) + b\(\overline{\mathbf{z}}\) + b\(\overline{\mathbf{z}}\) + c = 0, 1(b ∈ C, c ∈ R).

Solution:

Assume the general form of the equation of a circle in cartesian co-ordinates as

x^{2} + y^{2} + 2gx + 2fy + c = 0, (g, f ∈ R) —— (1)

To write this equation in the complex variable form, let (x, y) = z.

Then \(\frac{z+\bar{z}}{2}\) = x, \(\frac{z-\bar{z}}{2 i}\)

= y = \(\frac{-i(z-\bar{z})}{2}\)

∴ x^{2} + y^{2} = |z|^{2} = z\(\overline{\mathbf{z}}\)

Substituting these results in equation (1), we obtain

z\(\overline{\mathbf{z}}\) + g(z + \(\overline{\mathbf{z}}\)) + f(z – \(\overline{\mathbf{z}}\))(-i) + c = 0

i.e., z\(\overline{\mathbf{z}}\) + (g – if)z + (g + if)\(\overline{\mathbf{z}}\) + c = 0 ——-(2)

If (g + if) = b, then equation (2) can be written as z\(\overline{\mathbf{z}}\) + \(\overline{\mathbf{b}}\)z + b\(\overline{\mathbf{z}}\) + c = 0

Question 38.

Show that the complex numbers z satisfying z^{2} + \((\overline{\mathbf{z}})^{2}\) = 2 constitute a hyperbola.

Solution:

Substituting z = x + iy in the given equation

z^{2} + (\(\overline{\mathbf{z}}\))^{2} = 2, we obtain the cartesian form of the given equation.

∴ (x + iy)^{2} + (x – iy)^{2} = 2

i.e., x^{2} – y^{2} + 2ixy + x^{2} – y^{2} – 2ixy = 2

i.e., x^{2} – y^{2} = 1.

Since, this equation denotes a hyperbola, all the complex numbers satisfying

lie on the hyperbola x^{2} – y^{2} = 1.

Question 39.

Show that the points in the Argand diagram represented by the complex numbers 1 + 3i, 4 – 3i, 5 – 5i are collinear.

Solution:

Let the three complex numbers be represented in the Argand plane by the points P, Q, R respectively. Then P = (1, 3), Q = (4, -3), R = (5, -5). The slope of the line segment joining P,Q is \(\frac{3+3}{1-4}\) = \(\frac{6}{-3}\) = -2.

Similarly the slope of the line segment joining Q, R is \(\frac{-3+5}{4-5}\) = \(\frac{2}{-1}\) = -2.

Since the slope of PQ is the slope of QR, the points P, Q and R are collinear.

Question 40.

Find the equation of the straight line joining the points represented by (- 4 + 3i), (2 – 3i) in the Argand plane.

Solution:

Take the given points as

A = -4 + 3i = (-4, 3)

B = 2 – 3i = (2, -3)

Then equation of the straight line \(\overleftrightarrow{\mathrm{AB}}\) is

y – 3 = \(\frac{3+3}{-4-2}\)(x + 4)

i.e., x + y + 1 = 0.

Question 41.

z = x + iy represents a point in the Argand plane, find the locus of z. Such that |z| = 2.

Solution:

|z| = 2, z = x + iy

if \(\sqrt{x^{2}+y^{2}}\) = 2

if \(\sqrt{x^{2}+y^{2}}\) = 2

if and only if x^{2} + y^{2} = 4

The equation x^{2} + y^{2} = 4 represents the circle with centre at the origin (0, 0) and radius 2 units.

∴ The locus of |z| = 2 is the circle

x^{2} + y^{2} = 4

Question 42.

The point P represents a complex number z in the Argand plane. If the amplitude of z is \(\frac{\pi}{4}\), determine the locus of P.

Solution:

Let z = x + iy.

By hypothesis, amplitude of z = \(\frac{\pi}{4}\)

Hence tan^{-1} \(\left(\frac{y}{x}\right)\) = \(\frac{\pi}{4}\) and \(\frac{y}{x}\) = tan \(\frac{\pi}{4}\)

Hence x = y

∴ The locus of P is x = y.

Question 43.

If the point P denotes the complex number z = x + iy in the Argand plane and if \(\frac{z-i}{z-1}\) is a purely imaginary number, find the locus of P.

Solution:

We note that the quotient \(\frac{z-i}{z-1}\) is not defined if z = 1.

∴ The locus of P is the circle

x^{2} + y^{2} – x – y = 0

excluding the point (1, 0).

Question 44.

Describe geometrically the following subsets of C.

i) {z ∈ C| |z – 1 + i| = 1}

ii) {z ∈ C| |z + i| ≤ 3|

Solution:

i) Let S = {z ∈ C| z – 1 + i| = 1}

If we write z = (x, y), then

S = {(x, y) ∈ R^{2}||x + iy – 1 + i| = 1}

= {x, y) ∈ R^{2} || x + i(y – 1)| ≤ 3}

= {(x, y) ∈ R^{2} || (x – 1)^{2} + (y + 1)^{2} = i}

Hence S is a circle with centre (1, -1) and radius 1 unit.

ii) Let S’ = {z ∈ C || z + i| ≤ 3}

Then S = {(x, y ∈ R^{2} || x + iy + i| ≤ 3}

= {(x, y) ∈ R^{2} || x^{2} + i(y + 1) ≤ 3}

= {(x, y) ∈ R^{2} || x^{2} + (y + 1)^{2} ≤ 9}

Hence S’ is the closed circular disc with centre at (0, -1) and radius 3 units.