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KC Sinha Mathematics Solution Class 11 Chapter 10 समिश्र संख्याएँ (complex number) Exercise 10.3

Exercise 10.3

Question 1

Find the modulus of the following :
(i) \frac{1-\sqrt{3}i}{2+2 i}
Sol :
\frac{1-\sqrt{3} i}{2+2 i} \times \frac{2-2 i}{2-2 i}

=\frac{2-2 i-2 \sqrt{3} i+2 \sqrt{3} i^{2}}{2^{2}-(2 i)^{2}}

=\frac{2-2 \sqrt{3}-i(2+2 \sqrt{3})}{4+4}

=\frac{(2-2 \sqrt{3})-i(2+2 \sqrt{3})}{8}

=\frac{2(1-\sqrt{3})}{8}-\frac{1 \cdot 2(1+\sqrt{3})}{8}

z=\frac{1-\sqrt{3}}{4}-\frac{i(1+\sqrt{3})}{4}

|z|=\sqrt{\left(\frac{1-\sqrt{3}}{4}\right)^{2}+\left[-\left(\frac{1+\sqrt{3}}{4}\right]\right]^{2}}

=\sqrt{\frac{1^{2}+(\sqrt{3})^{2}-2. 1.\sqrt{3}}{16}+\frac{1^{2}+(\sqrt3)^{2}+2 .1. \sqrt{3}}{16}}

=\sqrt{\frac{1+3-2 \sqrt{3}+1+3+2 \sqrt{3}}{16}}

=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}}


(ii) \frac{2+i}{4 i+(1+i)^{2}}
Sol :
z=\frac{2+i}{4 i+(1+i)^{2}}

z=\frac{2+i}{4 i+1^{2}+i^{2}+2 \cdot 1 \cdot i}

z=\frac{z+i}{4 i+1-1+2 i}

z=\frac{2+i}{6 i} \times \frac{i}{i}=\frac{2 i+i^{2}}{6 i^{2}}

z=\frac{2 i-1}{-6}

z=\frac{-2 i+1}{6}

z=\frac{1}{6}-\frac{2 i}{6}

z=\frac{1}{6}-\frac{1}{3}i

|z|=\sqrt{\left(\frac{1}{6}\right)^{2}+\left(-\frac{1}{3}\right)^{2}}

=\sqrt{\frac{1}{36}+\frac{1}{9}}

=\sqrt{\frac{1+4}{36}}=\frac{\sqrt5}{6}


(iii) \frac{1+i}{1-i}-\frac{1-i}{1+i}
Sol :
z=\frac{1+i}{1-i}-\frac{1-i}{1+i}

z=\frac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)}

z=\frac{\left(1^{2}+i^{2}+2 \cdot 1 \cdot i\right)-\left(1^{2}+1^{2}-2 \cdot 1 \cdot i\right)}{1^{2}-i^{2}}

z=\frac{(1-1+2 i)-(1-1-2 i)}{1+1}

2=\frac{2 i+2 i}{2}

z=\frac{4{i}}{2}

=2i

=0+2i

|z|=\sqrt{0^{2}+2^{2}}

=\sqrt{0+4=\sqrt{4}

=2

Question 2

Find the argument of the following :
(i) -\sqrt{3}-i
Sol :
z=-\sqrt{3}-i

x=-\sqrt{3} , y=-1

\tan \theta=\left|\frac{y}{x}\right|=\left|\frac{-1}{-\sqrt{3}}\right|

\tan \theta=\tan \left(\pi+\frac{\pi}{6}\right)

\arg z=\theta=\pi+\frac{\pi}{6}=\frac{7 \pi}{6}


(ii) \frac{1+i}{1-\sqrt{3} i}
Sol :
z=\frac{1+i}{1-\sqrt{3} i} \times \frac{1+\sqrt{3} i}{1+\sqrt{3}i}

=\frac{1+\sqrt{3} i+i+\sqrt{3} i^{2}}{(1)^{2}-(\sqrt{3} i)^{2}}

z=\frac{1+\sqrt{3} i+i-\sqrt{3}}{1-3(-1)}

z=\frac{-\sqrt{3}+1+i(\sqrt{3}+1)}{4}

z=\frac{-(\sqrt{3}-1)}{4}+\frac{i(\sqrt{3}+1)}{4}

x=\frac{-(\sqrt{3}-1)}{4}y=\frac{\sqrt{3}+1}{4}

\tan \theta=\left|\frac{y}{x}\right|

\tan \theta=\left|\frac{\frac{\sqrt{3}+1}{4}}{-\frac{(\sqrt{3}-1)}{4}}\right|

\tan \theta=\left| \frac{\sqrt{3}+1}{-(\sqrt{3}-1 )}\right|

\cot \frac{\pi}{12}=\cot 15^{\circ}=\operatorname{cot{}(45^{\circ}-30^{\circ})}

=\frac{\cot 45^{\circ} \cot 30^{\circ}+1}{\cot30^{\circ}-\cot 45^{\circ}}

=\frac{1 \cdot \sqrt 3+1}{\sqrt{3}+1}

=\frac{\sqrt{3}+1}{\sqrt{3}-1}


\tan \theta=-\cot \frac{\pi}{12}

\tan \theta=-\tan \left[\frac{\pi}{2}-\frac{\pi}{12}\right]

\tan \theta=-\tan \left[\frac{6 \pi-\pi}{12}\right]

\tan \theta=-\tan \frac{5 \pi}{12}

\Rightarrow \theta=\pi-\frac{5 \pi}{12}

=\frac{12 \pi-5 \pi}{12}=\frac{7 \pi}{12}


Question 3

Find the modulus and arguments of the following complex numbers:
(i) -\sqrt{3}+i
Sol:
z=-\sqrt{3}+i

x=-\sqrt{3} , y=1

|z|=\sqrt{(-\sqrt{3})^{2}+1^{2}}

=\sqrt{3+1}=\sqrt{4}

=2

arg \Rightarrow \tan \theta=\left|\frac{y}{2}\right|

\tan \theta=\left|-\frac{1}{\sqrt{3}}\right|


\tan \theta=\tan \left(\pi-\frac{\pi}{6}\right)

\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}


(ii) -1-i \sqrt{3}
Sol :


(iii) \frac{1}{1+i}
Sol :
\frac{1}{1+i} \times \frac{1-i}{1-i}

=\frac{1-i}{1^{2}-i^{2}}=\frac{1-i}{1+1}

z=\frac{1}{2}-\frac{1}{2} i

x=\frac{1}{2} y=-\frac{1}{2}

|z|=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{2}}

=\sqrt{\frac{1}{4}+\frac{1}{4}}

=\sqrt{\frac{1+1}{4}}

=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}

\tan \theta=\left|\frac{y}{x}\right|=\left|\frac{-\frac{1}{2}}{\frac{1}{2}}\right|=|-1|

\tan \theta=\tan \left(-\frac{\pi}{4}\right) \Rightarrow \theta=\frac{-\pi}{4}

\tan \theta=\tan \left(2 \pi-\frac{\pi}{4}\right) \Rightarrow \theta=2 \pi-\frac{\pi}{4} =\frac{7 \pi}{4}


\tan \theta=-\tan \frac{5 \pi}{12}

\tan \theta=\tan \left(\frac{-5 \pi }{12}\right)

\theta=\frac{-5 \pi}{12}


(iv) \frac{1+i}{1-i}
Sol :

Question 5

Sol :
Let z=x+iy

|z|=\sqrt{x^{2}+y^{2}}

|2z-1|=|z-2|

|2(x+iy)-1|=|x+iy-2|

|2x+i.2y-1|=|x-1+iy|

|(2x-1)+i.2y|=|(x+2)+iy|

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

Squaring both sides

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

(2 x)^{2}-2 \cdot 2 x \cdot 1+1^{2}+4 y^{2}=x^{2}-2 \cdot x \cdot 2+2^{2}+y^{2}

4 x^{2}-4 x+1+4 y^{2}=x^{2}-4 x+4+y z

3 x^{2}+3 y^{2}=3

3\left(x^{2}+y^{2}\right)=3

x^{2}+y^{2}=1

\sqrt{x^{2}+y^{2}}=\sqrt{1}=1

|z|=1

Question 6

If (a+ib)(c+id)=x+iy show that
(i) (a+ib)(c-id)=x-iy
Sol :
(a+ib)(c-id)=x-iy..(i)

Note: \overline{z_{1} \cdot z_{2}}=\bar{z}_{1} \cdot \overline{z_{2}}

(a+ib)(c+id)=\overline{x+i y}

(\overline{a+ib})(\overline{c+i d})=\overline{x+i y}

(a-i b)(c-i d)=x-i y..(ii)

Multiplying (i) and (ii)

\left[a^{2}-(i b)^{2}\right]\left[c^{2}-\left(i d)^{2}\right]=x^{2}-(i y)^{2}\right.

\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=x^{2}+y^{2}


(ii) \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=x^{2}+y^{2}
Sol :


Question 7

Sol :
\frac{1-i x}{1+i x}=m+i n..(i)

Note: \overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\overline{z_{1}}}{z_{2}}

\overline{\left(\frac{1-i x}{1+x x}\right)}=\overline{m+i n}

\frac{\overline{1-i x}}{\overline{1+i x}}=\overline{m+in}

\frac{1+i x}{1-i x}=m-i n..(ii)

Multiplying (i) with (ii)

\frac{1-i x}{1+ix} \times \frac{1+i x}{1-i x}=(m+i n)(m-i n)

1=m^{2}-(i n)^{2}

i=m^{2}+n^{2}


Question 8

Sol :
\frac{1}{m+i n}-\frac{x-i y}{x+iy}=0

\frac{1}{m+i n}=\frac{x-i y}{x+i y}..(i)

\overline{\frac{1}{m+i n}}=\overline{\frac{x-i y}{x+i y}}

\frac{1}{\overline{m+1}}=\frac{\overline{x-i y}}{\overline{x+i y}}

\frac{1}{m-i n}=\frac{x+i y}{x-i y}..(ii)

Multiplying (i) with (ii)

\frac{1}{m+in } \times \frac{1}{m-in}=\frac{x-iy}{x+iy} \times \frac{x+i y}{x-i y}

\frac{1}{m^{2}-(i n)^{2}}=1

\frac{1}{m^{2}+n^{2}}=1

1=m^{2}+n^{2}


(iii) 
Sol :
a+i b=\frac{(x+i)^{2}}{2 x^{2}+1}..(i)

\overline{a+i b}=\overline{\left(\frac{(x+i)^{2}}{2 x^{2}+1}\right)}

a-i b=\frac{(x-i)^{2}}{2 x^{2}+1}..(ii)

Multiplying (i) with (ii)

(a+i b)(a-i b)=\frac{(x+i)^{2}(x-i)^{2}}{\left(2 x^{2}+1\right)^{2}}

a^{2}-(i b)^{2}=\frac{[(x+i)(x-i)]^{2}}{\left(2 x^{2}+1\right)^{2}}

a^{2}+b^{2}=\frac{\left[x^{2}-i^{2}\right]^{2}}{\left(2 x^{2}+1\right)^{2}}

a^{2}+b^{2}=\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}


Question 9

If \left(1+i \frac{x}{a}\right)\left(1+i \frac{x}{b}\right)\left(1+i \frac{x}{c}\right) \ldots=A+i B..(i) then prove that \left(1+\frac{x^{2}}{a^{2}}\right)\left(1+\frac{x^{2}}{b^{2}}\right)\left(1+\frac{x^{2}}{c^{2}}\right) \ldots=A^{2}+B^{2}
Sol :
\left(1+i \frac{x}{a}\right)\left(1+i \frac{x}{b}\right)\left(1+i \frac{x}{c}\right) \ldots=A+i B..(i)

\overline{\left(1+i \frac{x}{a}\right)\left(1+i \frac{x}{b}\right)\left(1+i \frac{x}{c}\right) \ldots}=\overline{A+i B}

\left(\overline{1+i \frac{x}{a}}\right)\left(\overline{1+i \frac{x}{b}}\right)\left(\overline{1+i \frac{x}{c}}\right) \overline{\ldots}=\overline{A+i B}


\left(1-i \frac{x}{a}\right)\left(1-i \frac{x}{b}\right)\left(1-i \frac{x}{c}\right) \ldots-=A-i B..(ii)

Multiplying (i) with (ii)

\left[1^{2}-\left(i\frac{x}{a}\right)^{2}\right]\left[1^{2}-\left(i\frac{x}{b}\right)^{2}\right]\left[1^{2}-\left(i \frac{x}{c}\right)^{2}\right] \ldots=A^{2}-(i B)^{2}

\left(1+\frac{x^{2}}{a^{2}}\right)\left(1+\frac{x^{2}}{b^{2}}\right)\left(1+\frac{x^{2}}{c^{2}}\right)-\ldots=A^{2}+B^{2}

Question 10

If \frac{a-i b}{a+i b}=\frac{1+i}{1-i} then show that a+b=0
Sol :
\frac{a-i b}{a+i b}=\frac{1+i}{1-i}

\frac{a-i b}{a+i b} \times \frac{a-i b}{a-i b}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}

\frac{a^{2}-a bi-a b i+i^{2} b^{2}}{a^{2}-(i b)^{2}}=\frac{1+i+i+i^{2}}{1^{2}-i^{2}}

\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}=\frac{1+2 i-1}{1}

\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2 a b}{a^{2}+b^{2}} i=\frac{2 i}{2}

\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2 a b}{a^{2}+b^{2}} i=0+i

By equating real part

\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=0

a^{2}-b^{2}=0

(a-b)(a+b)=0

a+b=0


Question 11

If \frac{3}{2+\cos \theta+i \sin \theta}=a+i b then prove that a^{2}+b^{2}=4 a-3
Sol :
\frac{3}{2+\cos \theta+i\sin\theta}=\frac{a+i b}{1}

\frac{2+\cos \theta+i\sin \theta}{3}=\frac{1}{a+i b} \times \frac{a-ib}{a-i b}

\frac{2+\cos }{3}+\frac{i\sin \theta}{3}=\frac{a-i b}{a^{2}-(i b)^{2}}

\frac{2+\cos \theta}{3}+\frac{i \sin \theta}{3}=\frac{a-i b}{a^{2}+b^{2}}

\frac{2+\cos \theta}{3}+\frac{i \cdot \sin \theta}{3}=\frac{a}{a^{2}+b^{2}}-i \frac{b}{a^{2}+b^{2}}

By equating real and imaginary part ,

\frac{2+\cos \theta}{3}=\frac{a}{a^{2}+b^{2}}

2+\cos \theta=\frac{3 a}{a^{2}+b^{2}}

\cos \theta=\frac{3 a}{a^{2}+b^{2}}-2..(i)


or \frac{\sin \theta}{3}=\frac{-b}{a^{2}+b^{2}}

\sin \theta=\frac{-3 b}{a^{2}+b^{2}}..(ii)

Squaring and adding (i) and (ii)

\cos ^{2} \theta+\sin ^{2} \theta=\left(\frac{3 a}{a^{2}+b^{2}}-2\right)^{2}+\left(\frac{-3 b}{a^{2}+b^{2}}\right)^{2}

1=\left(\frac{3 a}{a^{2}+b^{2}}\right)^{2}+2^{2}-2 \cdot \frac{3 a}{a^{2}+b^{2}} \cdot 2+\frac{9 b^{2}}{\left(a^{2}+b^{2}\right)^{2}}

1=\frac{9 a^{2}}{\left(a^{2}+b^{2}\right)^{2}}+\frac{9 b^{2}}{\left(a^{2}+b^{2}\right)^{2}}-\frac{12a}{a^{2}+b^{2}}+4

1-4=\frac{9 a^{2}+9 b^{2}}{\left(a^{2}+b^{2}\right)^{2}}-\frac{12 a}{a^{2}+b^{2}}

-3=\frac{9\left(a^{2}+b^{2}\right)}{\left(a^{2}+b^{2}\right)^{2}}-\frac{12 a}{a^{2}+b^{2}}

-3=\frac{9}{a^{2}+b^{2}}-\frac{12 a}{a^{2}+b^{2}}

-3=\frac{9-12 a}{a^{2}+b^{2}}

-3\left(a^{2}+b^{2}\right)=9-12 a

-3\left(a^{2}+b^{2}\right)=-12 a+9

3\left(a^{2}+b^{2}\right)=3(4 a-3)

a^{2}+b^{2}=4 a-3

Question 12

If a^{2}+b^{2}+c^{2}=1 , b+ic=(1+a)z , then prove that \frac{a+i b}{1+c}=\frac{1+i z}{1-i z}
Sol :
a^{2}+b^{2}+c^{2}=1 \quad \Rightarrow \quad a^{2}+b^{2}=1-c^{2}

a^{2}-(i b)^{2}=1^{2}-c^{2}

(a-ib)(a+ib)=(1-c)(1+c)

\frac{a+i b}{1+c}=\frac{1-c}{a-i b}..(i)


R.H.S \frac{1+i z}{1-i z}=\frac{1+i\left(\frac{b+i c}{1+a}\right)}{1-i\left(\frac{b+i c}{1+a}\right)}

=\dfrac{\frac{1+a+i b-c}{1+a}}{\frac{1+a-i b+c}{1+a}}

=\frac{1+a+i b-c}{1+a-i b+c}

=\dfrac{a+i b+\frac{(a-i b)(a+i b)}{1+c}}{1+c+\frac{(1-c)(1+c)}{a+i b}}

=\frac{(a+i b)\left[1+\frac{a-i b}{1+c}\right]}{(1+c)\left[1+\frac{1-c}{a+i b}\right]}

=\dfrac{(a+i b)}{(1+c)} \cdot \dfrac{\left[\frac{1+c+a-i b}{1+c}\right]}{\left[\frac{a+i b+1-c}{a+ib}\right]}

=\dfrac{(a+i b)}{1+c} \cdot \frac{(a+i b)(1+c+a-i b)}{(1+c)(a+i b+1-c)}

=\frac{a+1 b}{1+c} \cdot\left(\frac{a+a c+a^{2}-i a b+ib+i b c+i a b+b^2}{a+i b+1-c+a c+i b c+c-c^{2}}\right)^{2}

=\frac{(a+i b)}{1+c} \frac{\left(a+a c+1-c^{2}+i b+i b c\right)}{\left(a+a c+1-c^{2}+i b+i b c\right)}

\frac{1+i z}{1-i z}=\frac{a+i b}{1+c}


Question 13

If |z|<4 then prove that |iz+3-4i|<9
Sol :
L.H.S
|i z+3-4 i| \leq|i z|+|3-4 i|

Note:
|z|=|i z |
\left|z_1+z_{2}|\leq| z_{1}|+| z_{2} |\right.

<4+\sqrt{3^{2}+(-4)^{2}}

<4+\sqrt{9+16}

<4+5

|iz+3-4i|<9


Question 15

\left|1-\bar{z}_{1} z_{2}\right|^{2}-| z_{1}-\left.z_{2}\right|^{2}=\left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^{2}\right)
Sol :

Note: |z|^{2}=z \cdot \bar{z}

L.H.S
\left|1-\bar{z}_{1} z_{2}\right|^{2}-\left|z_{1}-z_{2}\right|^{2}

\left(1-\bar{z}_{1} z_{2}\right) \cdot(\overline{1-\bar{z}_{1} z_{2}}]-\left(z_{1}-z_{2}\right) \cdot(\overline{z_{1}-z_{2}})

=\left(1-\bar{z}_{1} z_{2}\right)\left(1-z_{1} \bar{z}_{1}\right)-\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)

=\left(1-z_{1} \cdot \bar{z}_{2}-\bar{z}_{1} z_{2}+z_{1} \cdot \bar{z}_{1} \cdot z_{2} \cdot \bar{z}_{2}\right)-\left(z_{1} \bar{z}_{1}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+z_2.\bar{z_2}\right)

=1-z_{1} \cdot \bar{z}-\bar{z}_{1} \cdot z_{2}+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-\left|z_{1}\right|^{2}+z_{1} \cdot \bar{z}_{2}+z_{2} \cdot \bar{z}_{1}-|z_2|^2

=1-\left|z_{1}|^{2}-\left|z_{2}\right|^{2}+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}\right.

=1\left(1-| z_{1}|^{2}\right)-\left|z_{2}\right|^{2}\left(1-\left|z_{1}\right|^{2}\right)

=\left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^{2}\right)


Question 16

If a=\frac{1+i}{\sqrt{2}}, a^{6}+a^{4}+a^{2}+1 Find the value of a^{6}+a^{4}+a^{2}+1
Sol :
a=\frac{1+i}{\sqrt{2}}

\sqrt 2 a=1+i

Squaring both sides

(\sqrt{2} a)^{2}=(1+i)^{2}

2 a^{2}=1^{2}+i^{2}+2 \cdot 1 i

2 a^{2}=1-1+2 i

Squaring both sides

\left(a^{2}\right)^{2}=i^{2}

a^{4}=-1 \Rightarrow a^{4}+1=0

[]


Question 17

If x=\sqrt{-2}-1 find the value of x^{4}+4 x^{3}+6 x^{2}+4 x+9
Sol :
x=\sqrt{-2}-1

x+1=\sqrt{2}i

Squaring both sides

(x+1)^{2}=(\sqrt{2} i)^{2}

x^{2}+2 \cdot x \cdot 1+1^{2}=-2

x^{2}+2 x+1=-2

x^{2}+2 x+3=0


Question 21

WORKING....





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