Class 11 NCERT Solutions Chapter 5 Complex Numbers And Quadratic Equations Miscellaneous Exercise on Chapter 5 | Set 1 (original) (raw)

Last Updated : 6 Sep, 2024

Complex numbers are fundamental in higher mathematics and have a wide range of applications in engineering, physics, and applied sciences. Chapter 5 of the Class 11 NCERT textbook introduces the concept of complex numbers and their properties. This chapter also covers quadratic equations including the methods to solve them and their relation to complex numbers.

Complex Numbers and Quadratic Equations

The Complex numbers extend the idea of real numbers by incorporating the imaginary unit i where i2=−1. This extension allows for the solutions to all polynomial equations including those with no real roots. The Quadratic equations are a key area where complex numbers are utilized especially when the discriminant is negative leading to the complex roots. This chapter provides a detailed exploration of these concepts including operations with the complex numbers and methods for solving quadratic equations using the various techniques.

**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 z 1 and z 2 , prove that, Re (z 1 z 2 ) = Re z 1 **Re z 2 – Im z 1 Im z 2

**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 (x 2 + y 2 ) 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. 3x 2 – 4x + 20/3 = 0

**Solution:

Given quadratic equation, **3x 2 – 4x + 20/3 = 0

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

On comparing it with __ax_2 + _bx + _c = 0, we get

_a = 9, _b = –12, and _c = 20

So, the discriminant of the given equation will be

D = _b_2 – 4__ac = (–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. x 2 – 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 __ax_2 + _bx + _c = 0,

We get

_a = 2, _b = –4, and _c = 3

So, the discriminant of the given equation will be

D = _b_2 – 4__ac = (–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. 27x 2 – 10x + 1 = 0

**Solution:

Given:

Quadratic equation, 27__x_2 – 10__x_ + 1 = 0

On comparing it with __ax_2 + _bx + _c = 0,

We get

_a = 27, _b = –10, and _c = 1

So, the discriminant of the given equation will be

D = _b_2 – 4__ac = (–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. 21x 2 – 28x + 10 = 0

**Solution:

Given:

Quadratic equation, 21__x_2 – 28__x_ + 10 = 0

On comparing it with __ax_2 + _bx + _c = 0,

We have

_a = 21, _b = –28, and _c = 10

So, the discriminant of the given equation will be

D = _b_2 – 4__ac = (–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 z 1 = 2 – **i**, z** 2 = 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

Conclusion

Understanding complex numbers and quadratic equations is crucial for the solving problems in the various fields of science and engineering. Mastery of these concepts not only helps in academic pursuits but also in the practical applications where complex solutions are necessary. By practicing exercises from the Chapter 5 students can enhance their problem-solving skills and gain a deeper appreciation for the role of the complex numbers in mathematics.