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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Miscellaneous Exercise on Chapter 5 | Set 1

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  • Last Updated : 01 Apr, 2021
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Question 1. Evaluate \left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3

Solution:

\displaystyle\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3\\ =\left[i^{4\times4+2}+\frac{1}{i^{4\times6+1}}\right]^3\\ =\left[(i^4)^{4}.i^2+\frac{1}{(i^4)^6.i}\right]^3\\ =\left[i^2+\frac{1}{i}\right]^3\ \ \ \ \ \ [i^4=1]\\ =\left[-1+\frac{1}{i}\times\frac{i}{i}\right]^3\ \ \ \ \ \ [i^2=-1]\\ =\left[-1+\frac{i}{i^2}\right]^3

= [-1 – i]3

= (-1)3 [1 + i]3

= -[13 + i3 + 3 × 1 × i (1 + i)]

= -[1 + i3 + 3i + 3i2]

= -[1 – i + 3i – 3]

= -[2 + 2i]

= 2 – 2i 

Question 2. For any two complex numbers z1 and z2, prove that, Re (z1z2) = Re zRe z2 – Im z1 Im z2

Solution:

Let’s assume z1 = x1 + iy1 and z2 = x2 + iy2 as two complex numbers

Product of these complex numbers, z1z2

z1z2 = (x1 + iy1)(x2 + iy2)

= x1(x2 + iy2) + iy1(x2 + iy2)

= x1x2 + ix1y2 + iy1x2 + i2y1y2

= x1x2 + ix1y2 + iy1x2 – y1y2             [i2 = -1]

= (x1x2 – y1y2) + i(x1y2 + y1x2)

Now, 

Re(z1z2) = x1x2 – y1y2

⇒ Re(z1z2) = Rez1Rez2 – Imz1Imz2

Hence, proved.

Question 3. Reduce to the standard form \displaystyle\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)

Solution:

\displaystyle\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)=\left[\frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}\right]\left[\frac{3-4i}{5+i}\right]\\ = \left[\frac{1+i-2+8i}{1+i-4i-4i^2}\right] \left[\frac{3-4i}{5+i}\right]= \left[\frac{-1+9i}{5-3i}\right] \left[\frac{3-4i}{5+i}\right]\\ =  \left[\frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}\right]=\frac{33+31i}{28-10i}=\frac{33+31i}{2(14-5i)}\\ = \frac{(33+31i)}{2(14-5i)}\times\frac{14+5i}{14+5i}

On multiplying numerator and denominator by (14+5i)

\\ =\frac{462+165i+434i+155i^2}{2[(14)^2-(5i)^2]}=\frac{307+599i}{2(196-25i^2)}\\ =\frac{307+599i}{2(221)}=\frac{307+599i}{442}=\frac{307}{442}+\frac{599i}{442}

Hence, this is the required standard form.

Question 4. If x – iy = \sqrt{\frac{a-ib}{c-id}}  prove that (x2 + y2)2=\frac{a^2+b^2}{c^2+d^2}

Solution:

Given:

x – iy = \sqrt{\frac{a-ib}{c-id}}

=\sqrt{\frac{a-ib}{c-id}\times\frac{c+id}{c+id}}

On multiplying numerator and denominator by (c+id)

\\ =\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}

So,

(x – iy)2=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}  

x2 – y2 – 2ixy  =\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}

On comparing real and imaginary parts, we get

x2 – y2 \frac{ac+bd}{c^2+d^2} , -2xy = \frac{ad-bc}{c^2+d^2}        (1)

(x2 + y2)2 = (x2 – y2)2 + 4x2y2

\left( \frac{ac+bd}{c^2+d^2}\right)^2+\left(\frac{ad-bc}{c^2+d^2} \right)^2\ \ \ \ \ [Using (1)\\ =\frac{a^2c^2+b^2d^2+2acbd+a^2d^2+b^2c^2-2adbc}{(c^2+d^2)^2}\\ =\frac{a^2c^2+b^2d^2+a^2d^2+b^2c^2}{(c^2+d^2)^2}\\ = \frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^2}\\ =\frac{(c^2+d^2)(a^2+b^2)}{(c^2+d^2)^2}\\ = \frac{a^2+b^2}{c^2+d^2}

Hence proved

Question 5. Convert the following in the polar form:

(i) \frac{1+7i}{(2-i)^2}

(ii) \frac{1+3i}{1-2i}

Solution:

(i) Here, z = \frac{1+7i}{(2-i)^2}

\frac{1+7i}{(2-i)^2}=\frac{1+7i}{4+i^2-4i}=\frac{1+7i}{4-1-4i}\\ = \frac{1+7i}{(2-i)^2}\times\frac{3+4i}{3+4i}=\frac{3+4i+21i+28i^2}{3^2+4^2}

Multiplying by its conjugate in the numerator and denominator

\\ =\frac{3+4i+21i-28}{3^2+4^2}=\frac{-25+25i}{25}

= -1+i

Let r cos θ = -1 and r sin θ = 1

On squaring and adding, we get

r2 (cos2θ + sin2θ) = 1 + 1 = 2

r2 = 2           [cos2θ + sin2θ = 1]

r = √2

So, 

√2 cosθ = -1 and √2 sinθ = 1

⇒ cosθ = \frac{-1}{\sqrt2}  and sinθ  = \frac{1}{\sqrt2}  

Therefore,

θ = \pi-\frac{\pi}{4}=\frac{3\pi}{4}            [As θ lies in II quadrant]

Expressing as, z = r cos θ + i r sin θ 

\sqrt2cos\frac{3\pi}{4}+i\sqrt2sin\frac{3\pi}{4}=\sqrt2\left(cos\frac{3\pi}{4}+isin\frac{3\pi}{4}\right)

Therefore, this is the required polar form.

(ii) Let, z = \frac{1+3i}{1-2i}\\ =\frac{1+3i}{1-2i}\times\frac{1+2i}{1+2i}\\ =\frac{1+2i+3i-6}{1+4}\\ =\frac{-5+5i}{5}

= -1 + i

Now, Let r cosθ = -1 and r sin θ = 1

On squaring and adding, we get

r2(cos2θ + sin2θ) = 1 + 1

r(cos2θ + sin2θ) = 2

r2 = 2          [cos2θ + sin2θ = 1]

= r = √2     [Conventionally, r > 0]

Therefore,

√2 cosθ = -1 and √2 sinθ = 1

cosθ = \frac{-1}{\sqrt2}  and sinθ = \frac{1}{\sqrt2}

Therefore,

θ = \pi-\frac{\pi}{4}=\frac{3\pi}{4}       [As θ lies in II quadrant]

Expressing as, z = r cosθ + i r sinθ 

z = \sqrt2cos\frac{3\pi}{4}+i\sqrt2\frac{3\pi}{4}=\sqrt2\left(cos\frac{3\pi}{4}+isin\frac{3\pi}{4}\right)

Therefore, this is the required polar form.

Solve each of the equation in Exercises 6 to 9.

Question 6. 3x2 – 4x + 20/3 = 0

Solution:

Given quadratic equation, 3x2 – 4x + 20/3 = 0

It can be re-written as: 9x2 – 12x + 20 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 9, b = –12, and c = 20

So, the discriminant of the given equation will be

D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576

Hence, the required solutions are

X = \frac{-b\pm\sqrt{D}}{2a}=\frac{-(-12)\pm\sqrt{-576}}{2\times9}=\frac{12\pm\sqrt{576}i}{18}\\ =\frac{12\pm24i}{18}=\frac{6(2\pm4i)}{18}=\frac{2\pm4i}{3}=\frac{2}{3}\pm\frac{4}{3}i

Question 7. x2 – 2x + 3/2 = 0

Solution:

Given:

Quadratic equation, x2 – 2x + \frac{3}{2} = 0

After re-written 2x2 – 4x + 3 = 0

On comparing it with ax2 + bx + c = 0, 

We get

a = 2, b = –4, and c = 3

So, the discriminant of the given equation will be

D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8

Hence, the required solutions are

x = \frac{-b\pm\sqrt{D}}{2a}=\frac{-(-4)\pm\sqrt{-8}}{2\times2}=\frac{4\pm2\sqrt2i}{4}\ \ \ \ [\sqrt{-1}=i]\\ =\frac{2\pm\sqrt2i}{2}=1\pm\frac{\sqrt2}{2}i

Question 8. 27x2 – 10x + 1 = 0

Solution:

Given:

Quadratic equation, 27x2 – 10x + 1 = 0

On comparing it with ax2 + bx + c = 0, 

We get

a = 27, b = –10, and c = 1

So, the discriminant of the given equation will be

D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8

Hence, the required solutions are

=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-10)\pm\sqrt{-8}}{2\times27}=\frac{10\pm2\sqrt2i}{54}\\ =\frac{5\pm\sqrt2i}{27}=\frac{5}{27}\pm\frac{\sqrt2}{27}i

Question 9. 21x2 – 28x + 10 = 0

Solution:

Given:

Quadratic equation, 21x2 – 28x + 10 = 0

On comparing it with ax2 + bx = 0, 

We have

a = 21, b = –28, and c = 10

So, the discriminant of the given equation will be

D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56

Hence, the required solutions are

=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-28)\pm\sqrt{-56}}{2\times21}=\frac{28\pm\sqrt{56}i}{42}\\ =\frac{28\pm2\sqrt{14}i}{42}=\frac{28}{42}\pm\frac{2\sqrt{14}}{42}i=\frac{2}{3}\pm\frac{\sqrt{14}}{21}i

Question 10. If z1 = 2 – i, z2 = 1 + i, find \left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|

Solution:

Given, z1 = 2 – i, z2 = 1 + i

\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|\\ =\left|\frac{4}{2-2i}\right|=\left|\frac{4}{2(1-i)}\right|\\ =\left|\frac{2}{1-i}\times\frac{1+i}{1+i}\right|=\left|\frac{2(1+i)}{1^2-i^2}\right|\\ =\left|\frac{2(1+i)}{1+1}\right|\ \ \ \ \ [i^2=-1]\\ =\left|\frac{2(1+i)}{2}\right|\\ =|1+i|=\sqrt{1^2+1^2}=\sqrt2

Hence, the value of \left|\frac{z_1+z_2+1}{z_1-z_2+1}\right| is √2


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