Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)

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

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(c)

I.

Question 1.
Express the following complex numbers in modulus-amplitude form.
(i) 1 – i
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
r2 (cos2θ + sin2θ) = 1 + 1 = 2
⇒ r2 = 2
⇒ r = √2
tan θ = – 1
⇒ θ = \(-\frac{\pi}{4}\)
∴ 1 – i = √2(cos(\(-\frac{\pi}{4}\)) + i sin(\(-\frac{\pi}{4}\)))

(ii) 1 + i√3
Solution:
1 + i√3 = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1 ………(1)
r sin θ = √3 ……..(2)
θ lies in the I quadrant
Squaring and adding (1) and (2)
r2 (cos2θ + sin2θ) = 1 + 3
⇒ r2 = 4
⇒ r = 2
Dividing (2) by (1)
\(\frac{r \sin \theta}{r \cos \theta}\) = √3
⇒ tan θ = √3
⇒ θ = \(\frac{\pi}{3}\)
∴ 1 + i√3 = 2(cos \(\frac{\pi}{3}\) + i sin \(\frac{\pi}{3}\))

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)

(iii) -√3 + i
Solution:
-√3 + i = r(cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = -√3 …….(1)
r sin θ = 1 …….(2)
⇒ θ lies in II quadrant
Squaring and adding (1), (2)
r2 (cos2θ + sin2θ) = 3 + 1 = 4
⇒ r2 = 4
⇒ r = 2
Dividing (2) by (1)
\(\frac{r \sin \theta}{r \cos \theta}=-\frac{1}{\sqrt{3}}\)
⇒ tan θ = \(-\frac{1}{\sqrt{3}}\) and θ lies in II quadrant
⇒ θ = 180° – 30° = 150° = \(\frac{5 \pi}{6}\)
∴ -√3 + i = 2(cos \(\frac{5 \pi}{6}\) + i sin \(\frac{5 \pi}{6}\))

(iv) -1 – i√3
Solution:
1 – i√3 = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = -1, r sin θ = -√3
Squaring and adding
r2 (cos2θ + sin2θ) = 1 + 3
⇒ r2 = 4
⇒ r = 2
∴ cos θ = \(-\frac{1}{2}\), sin θ = \(-\frac{\sqrt{3}}{2}\)
⇒ θ = -120° = \(-\frac{2 \pi}{3}\)
∴ -1 – i√3 = 2[cos(\(-\frac{2 \pi}{3}\)) + i sin(\(-\frac{2 \pi}{3}\))]

Question 2.
Simplify -2i(3 + i) (2 + 4i) (1 + i) and obtain the modulus of that complex number.
Solution:
-2i(3 + i) (2 + 4i) (1 + i)
= (-6i – 2i2) (2 + 2i + 4i + 4i2)
= (2 – 6i) (-2 + 6i)
= -4 + 12i – 36i2 + 12i
= 32 + 24i
= 8(4 + 3i)
Modulus = |8(4 + 3i)|
= 8\(\sqrt{(4)^{2}+(3)^{2}}\)
= 8(5)
= 40

Question 3.
(i) If z ≠ 0, find Arg z + Arg \((\bar{z})\).
Solution:
If z = x + iy, then Arg(z) = \(\tan ^{-1}\left(\frac{y}{x}\right)\)
and \((\bar{z})\) = x – iy, then Arg (z) = \(\tan ^{-1}\left(\frac{-y}{x}\right)\) = –\(\tan ^{-1}\left(\frac{y}{x}\right)\)
∴ Arg (z) + Arg \((\bar{z})\) = \(\tan ^{-1}\left(\frac{y}{x}\right)\) + (-\(\tan ^{-1}\left(\frac{y}{x}\right)\)) = 0

(ii) If z1 = -1 and z2 = -i, then find Arg(z1z2)
Solution:
z1 = -1
⇒ z1 = cos π + i sin π
⇒ Arg z1 = π
z2 = -i
⇒ z2 = \(\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\)
⇒ Arg z2 = \(-\frac{\pi}{2}\)
∴ Arg (z1z2) = Arg z1 + Arg z2
= π – \(\frac{\pi}{2}\)
= \(\frac{\pi}{2}\)

(iii) If z1 = -1, z2 = i then find Arg\(\left(\frac{\mathbf{z}_{1}}{\mathbf{z}_{2}}\right)\)
Solution:
z1 = -1 = cos π + i sin π
⇒ Arg z1 = π
z2 = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
⇒ Arg z2 = \(\frac{\pi}{2}\)
∴ Arg\(\left(\frac{\mathbf{z}_{1}}{\mathbf{z}_{2}}\right)\) = Arg z1 – Arg z2
= π – \(\frac{\pi}{2}\)
= \(\frac{\pi}{2}\)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)

Question 4.
(i) If (cos 2α + i sin 2α) (cos 2β + i sin 2β) = cos θ + i sin θ, then find the value of θ.
Solution:
cos θ + i sin θ
= (cos 2α + i sin 2α) (cos 2β + i sin 2β)
= cos 2α . cos 2β + i sin 2α . cos 2β + i cos 2α sin 2β + i2 sin 2α . sin 2β
= (cos 2α . cos 2β – sin 2α . sin 2β) + i(sin 2α cos 2β + cos 2α sin 2β)
= cos 2(α + β) + i sin 2(α + β)
∴ θ = 2(α + β)

(ii) If √3 + i = r (cos θ + i sin θ), then find the value of θ in radian measure.
Solution:
Given that
√3 + i = r(cos θ + i sin θ)
⇒ r cos θ = √3, r sin θ = 1
⇒ r2 (cos2θ + sin2θ) = 3 + 1
⇒ r2 = 4
⇒ r = 2
∴ cos θ = \(\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{6}\)

(iii) If x + iy = cis α . cis β then find the value of x2 + y2.
Solution:
Given x + iy = cis α . cis β
⇒ x + iy = (cos α + i sin α) (cos β + i sin β)
⇒ x + iy = cos(α + β) + i sin(α + β)
Equating real and imaginary parts
x = cos(α + β) and y = sin(α + β)
∴ x2 + y2 = cos2(α + β) + sin2(α + β) = 1

(iv) If \(\frac{z_{2}}{z_{1}}\), (z1 ≠ 0) is an imaginary number, then find the value of \(\left|\frac{2 z_{1}+z_{2}}{2 z_{1}-z_{2}}\right|\)
Solution:
\(\frac{z_{2}}{z_{1}}\), (z1 ≠ 0) is purely imaginary.
We can suppose that \(\frac{z_{2}}{z_{1}}\) = iy, where y ∈ R – {0}
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) I Q1(iv)

(v) If (√3 + i)100 = 299 (a + ib), then show that a2 + b2 = 4.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) I Q1(v)

Question 5.
(i) If z = x + iy and |z| = 1, find the locus of z.
Solution:
Given z = x + iy
Also |z| = 1
⇒ |x + iy| = 1
⇒ \(\sqrt{x^{2}+y^{2}}=1\)
⇒ x2 + y2 = 1
∴ Locus of z is x2 + y2 = 1

(ii) If the amplitude of (z – 1) is \(\frac{\pi}{2}\), then find the locus of z.
Solution:
Given z = x + iy
z – 1 = x + iy – 1 = (x – 1) + iy
Since Amplitude (z – 1) is \(\frac{\pi}{2}\)
Real part of z – 1 is zero.
∴ x – 1 = 0
∴ Locus of z is x – 1 = 0

(iii) If the Arg \(\overline{\mathbf{z}}_{1}\) and Arg z2 are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, find (Arg z1 + Arg z2)
Solution:
Arg \(\overline{\mathbf{z}}_{1}\) = \(\frac{\pi}{5}\)
⇒ Arg \(\overline{\mathbf{z}}_{1}\) = -Arg z1 = \(-\frac{\pi}{5}\)
Arg z2 = \(\frac{\pi}{3}\)
∴ Arg z1 + Arg z2 = \(-\frac{\pi}{5}+\frac{\pi}{3}\) = \(\frac{2 \pi}{15}\)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)

(iv) If z = \(\frac{1+2 i}{1-(1-i)^{2}}\) then find Arg(z).
Solution:
Given z = \(\frac{1+2 i}{1-(1-i)^{2}}\)
= \(\frac{1+2 i}{1-(1-2 i+i)^{2}}\)
= \(\frac{1+2 i}{1+2 i}\)
= 1
= cos 0° + i sin 0°
∴ Arg (z) = 0°

II.

Question 1.
Simplify the following complex numbers and find their modulus.
(i) \(\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\)
(ii) \(\frac{(1+i)^{3}}{(2+i)(1+2 i)}\)
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) II Q1(i)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) II Q1(ii)

Question 2.
(i) If (1 – i) (2 – i) ( 3 – i) …… (1 – ni) = x – iy then prove that 2 . 5 . 10 …….. (1 + n2) = x2 + y2.
Solution:
Given (1 – i) (2 – i) (3 – i) (1 – ni) = x – iy
|1 – i| |2 – i| |3 – i| …… |1 – ni| = |x – iy|
\(\sqrt{1+1} \sqrt{4+1} \sqrt{9+1} \ldots \sqrt{1+n^{2}}=\sqrt{x^{2}+y^{2}}\)
∴ 2 . 5 . 10 ……. (1 + n2) = x2 + y2

(ii) If the real part of \(\frac{z+1}{z+i}\) is 1, then find the locus of z.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) II Q2(ii)

(iii) If |z – 3 + i | = 4 determine the locus of z.
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
⇒ x2 – 6x + 9 + y2 + 2y + 1 = 16
⇒ x2 + y2 – 6x + 2y – 6 = 0
∴ The locus of z is x2 + y2 – 6x + 2y – 6 = 0

(iv) If |z + ai| = |z – ai| then find the locus of z.
Solution:
Let z = x + iy
Given |z + ai | = |z – ai|
⇒ |x + iy + ai| = |x + iy – ai|
⇒ |x + i(y + a)| = |x + i(y – a)|
⇒ \(\sqrt{x^{2}+(y+a)^{2}}=\sqrt{x^{2}+(y-a)^{2}}\)
⇒ x2 + (y + a)2 = x2 + (y – a)2
⇒ (y + a)2 – (y – a)2 = 0
⇒ 4ay = 0
⇒ y = 0
∴ The locus of z is y = 0

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)

Question 3.
If z = (x + iy) and if the point P in the Argand plane represents z, then describe geometrically the locus of P satisfying the equation.
(i) |2z – 3| = 7
Solution:
|2z – 3| = 7
Let z = x + iy
Given |2z – 3| = 7
⇒ |2(x + iy) – 3| = 7
⇒ |(2x – 3) + i(2y)| = 7
⇒ \(\sqrt{(2 x-3)^{2}+(2 y)^{2}}\) = 7
⇒ 4x2 – 12x + 9 + 4y2 = 49
⇒ 4x2 + 4y2 – 12x – 40 = 0
⇒ x2 + y2 – 3x – 10 = 0
∴ The locus of z is x2 + y2 – 3x – 10 = 0
This equation represents a circle.
Centre = (\(\frac{3}{2}\), 0)
Radius = \(\sqrt{\frac{9}{4}-(-10)}\) = \(\frac{7}{2}\) units

(ii) |z|2 = 4 Re(z + 2)
Solution:
|z|2 = 4 Re (z + 2)
Given |z|2 = 4 Re (z + 2)
⇒ |x + iy|2 = 4 Re (x + iy + z)
⇒ (\(\sqrt{x^{2}+y^{2}}\)) = 4(x + 2)
⇒ x2 + y2 = 4x + 8
∴ The locus of z is x2 + y2 – 4x – 8 = 0
This equation represents a circle. Center = (2, 0)
Radius = \(\sqrt{4-(-8)}\) = 2√3 units.

(iii) |z + i|2 – |z – i|2 = 2
Solution:
Given |z + i|2 – |z –  i|2 = 2
⇒ |x + iy + i|2 – |x + iy – i|2 = 2
⇒ |x + i(y + 1)|2 – |x + i(y – 1)|2 = 2
⇒ \(\left[\sqrt{x^{2}+(y+1)^{2}}\right]^{2}\) – \(\left[\sqrt{x^{2}+(y-1)^{2}}\right]^{2}\) = 2
⇒ x2 + (y + 1 )2 – [x2 + (y – 1)2] = 2
⇒ x2 + (y + 1)2 – x2 – (y – 1)2 = 2
⇒ 4y = 2
⇒ 2y – 1 = 0
The locus of P is 2y – 1 = 0
This equation represents a straight line parallel to the x-axis.

(iv) |z + 4i| + |z – 4i| = 10
Solution:
Given |z + 4i| + |z – 4i| = 10
⇒ |x + iy + 4i| + |x + iy – 4i| = 10
⇒ |x + i(y + 4)| + |x + i(y – 4)| = 10
⇒ \(\sqrt{x^{2}+(y+4)^{2}}+\sqrt{x^{2}+(y-4)^{2}}\)
⇒ \(\sqrt{x^{2}+(y+4)^{2}}\) = 10 – \(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ x2 + (y + 4)2 = 100 + x2 + (y – 4)2 – 20 \(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ 16y – 100 = – 20\(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ 4y – 25 = -5\(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ (4y – 25)2 = 25[x2 + (y – 4)2]
⇒ 16y2 – 200y + 625 = 25x2 + 25y2 – 200y + 400
⇒ 25x2 + 9y2 = 225
∴ The locus of P is 25x2 + 9y2 = 225
⇒ \(\frac{x^{2}}{9}+\frac{y^{2}}{25}\) = 1
This equation represents an ellipse.
Center = (0, 0)
a2 = 9, b2 = 25
We know a2 – b2(1 – e2)
⇒ 9 = 25(1 – e2)
⇒ 1 – e2 = \(\frac{9}{25}\)
⇒ e = 1 – \(\frac{9}{25}\) = \(\frac{16}{25}\)
∴ Eccentricity = \(\frac{4}{5}\) and major axis parallel to y-axis.

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c)

Question 4.
(i) If z1, z2 are two non-zero complex numbers satisfying |z1 + z2| = |z1| + |z2|, show that Arg z1 – Arg z2 = 0.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) II Q4(i)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) II Q4(i).1
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(c) II Q4(i).2

(ii) 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.
Solution:
Let z = x + iy and a = α + iβ
\(\frac{z-a}{z+\bar{a}}=\frac{(x+i y)-(\alpha+i \beta)}{(x+i y)+(\alpha-i \beta)}\)
\(\left|\frac{z-a}{z+\bar{a}}\right|\) = 1
⇒ \(\frac{|z-a|}{|z+\bar{a}|}\) = 1
⇒ |z – a| = |z + \(\overline{\mathrm{a}}\)|
⇒ |(x – α)| + i(y – β)| = |(x + α) + i(y – β)|
⇒ \(\sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}=\sqrt{(x+\alpha)^{2}+(y-\beta)^{2}}\)
⇒ (x – α)2 + (y – β)2 = (x + α)2 + (y – β)2
⇒ (x – α)2 = (x + α)2
⇒ (x + α)2 – (x – α)2 = 0
⇒ 4αx = 0
Re(a) ≠ 0
⇒ a ≠ 0
⇒ x = 0
Locus of P is x = 0 i.e., Y-axis

Leave a Comment