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=0Sol :
$\frac{a-i b}{a+i b}=\frac{1+i}{1-i}$
$\frac{a-i b}{a+i b} \times \frac{a-i b}{a-i b}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$
$\frac{a^{2}-a bi-a b i+i^{2} b^{2}}{a^{2}-(i b)^{2}}=\frac{1+i+i+i^{2}}{1^{2}-i^{2}}$
$\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}=\frac{1+2 i-1}{1}$
$\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2 a b}{a^{2}+b^{2}} i=\frac{2 i}{2}$
$\frac{a^{2}-b^{2}}{a^{2}+b^{2}}-\frac{2 a b}{a^{2}+b^{2}} i=0+i$
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|<9Sol :
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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In hindi
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