Exercise 5(A)
Question 1
(i)
$\begin{aligned} =&(i)^{48} \\ = & \left(i^2\right)^{24} \\ = & (-1)^{24} \\ = & +1\end{aligned}$
(ii)
$\begin{aligned}= & (-i)^{12}-(i)^{12} \\ = & +(i)^{12}-(i)^{12} \\ = & 0\end{aligned}$
(iii)
$\begin{aligned} =& (i)^{61} \\ = & \left(i^2\right)^{30} \times i \\ = & (-1)^{30} \times i \\ = & 1 \times i \Rightarrow i\end{aligned}$
(iv)
$\begin{aligned} & = i^{-101} \\ & =\frac{1}{i^{101}} \\ & =\frac{1}{i^{100} \times i} \\ & =\frac{1}{\left(i^2\right)^{50} \times i} \\ & =\frac{1}{(-1)^{50} \times i} \\ & =\frac{1}{1 \times i} \\& \Rightarrow \frac{1}{i} \times \frac{i}{i} \\ & =\frac{i}{i^2} \\ & =\frac{i}{-1} \\ & =-i\end{aligned}$
Question 2
$ \begin{aligned} &= i^9+i^{19} \\ & =i^8 \times i+i^8 \times i \\& \begin{array}{l} =\left(i^2\right)^4 \times i+\left(i^2\right)^9 \times i \end{array} \\ & =(-1)^4 \times i+(-1)^9 \times i \\ & =1 \times i+(-1) \times i \\ & =i-i \\ & =0 \end{aligned} $
(ii)
$\begin{aligned} & =i^{-39} \\ & =\frac{1}{i^{39}} \\ & =\frac{1}{i^{38} \times i} \\ & =\frac{1}{\left(i^2\right)^{19} \times i} \\ & =\frac{1}{(-1)^{19} \times i} \\ & =\frac{1}{-1 \times i} \\ & =\frac{-1}{i} \\& =\frac{-1}{i} \times \frac{i}{i} \\ & =\frac{-i}{i^2} \\ & =\frac{+i}{71} \\ & =i\end{aligned}$
Question 3
(i) (2,5)
$ =2+5 i$
(ii) $\begin{aligned} & -4,-2 \\ = & -4-2 i\end{aligned}$
(iii) $\begin{aligned} & 1,0 \\ = & 1+0 \times i \\ = & 1\end{aligned}$
(iv) $\begin{aligned} & 0,1 \\ = & 0+1 \times i \\ = & i\end{aligned}$
(v) $\begin{aligned} & 3,-4 \\ = & 3-4 i\end{aligned}$
(viii) $\begin{aligned} & \frac{1}{\sqrt{3}}, \frac{\sqrt{3}}{2} \\ = & \frac{1}{\sqrt{3}}+\frac{\sqrt{3}}{2} i\end{aligned}$
Question 4
(a) (5-5i) & (-1-8i)
$\begin{aligned}& =5-5 i+(-1)-8 i \\ & =5-5 i-1-8i \\ & =4-13 i\end{aligned}$
(b) $(-2 a+3 b i)+(3 a+5 b i)+(8 a-6 b i)$
$\begin{aligned} & =(-2 a+3 a+8 a)+i(3 b+5 b-6 b) \\ & =9 a+i(2 b) \\ & =9 a+2 b i\end{aligned}$
(c) $\begin{aligned} & (\sqrt{12} a-\sqrt{3} b i)-(2 \sqrt{3} a+2 \sqrt{3} b i) \\ = & 2 \sqrt{3} a-\sqrt{3} b i-2 \sqrt{3} a-2 \sqrt{3} b i \\ =&-3 \sqrt{3} b i\end{aligned}$
(d) $\begin{aligned} & (x-\sqrt{-y}) \times(x+\sqrt{-y}) \\ = & (x)^2-(\sqrt{-y})^2 \\ = & x^2-(-y) \\ = & x^2+y\end{aligned}$
(e) $\begin{aligned} & (4+5 i) \times(5-4 i) \\ = & 20-16 i+25 i-20 i^2 \\ = & 20+9 i-20 \times(-1) \\ = & 20+9 i+20 \\ = & 40+9 i\end{aligned}$
Question 5
(i) $\begin{aligned} & 3(7+i 7)+i(7+i 7) \\ = & 21+21 i+7 i+7 i^2 \\ = & 21+28 i+7(-1) \\ = & 21+28 i-7 \\ = & 14+28 i\end{aligned}$
(ii) $\begin{aligned} & (-i)-(-1+6 i) \\ = & 1-i+1-6 i \\ = & 2-7 i\end{aligned}$
$\begin{aligned} & \text { (iii) }\left[\left(\frac{1}{3}+\frac{7}{3} i\right)+\left(4+\frac{1}{3} i\right)\right]-\left(-\frac{4}{3}+i\right) \\ & =\left[\frac{1}{3}+4+i\left(\frac{7}{3}+\frac{1}{3}\right)\right]+\frac{4}{3}-i \\ & =\left[\left(\frac{13}{3}\right)+i\left(\frac{8}{3}\right)\right]+\frac{4}{3}-i \\ & =\frac{13}{3}+\frac{4}{3}+i\left(\frac{8}{3}-1\right) \\ & =\frac{17}{3}+\frac{i 5}{3}\end{aligned}$
$ \begin{aligned}& \text { (iv) } \frac{5+\sqrt{2} i}{1-\sqrt{2} i} \\ & =\frac{5+\sqrt{2} i}{1-\sqrt{2} i} \times \frac{1+\sqrt{2} i}{1+\sqrt{2} i} \\ & =\frac{5+5 \sqrt{2} i+\sqrt{2} i+2(i)}{(1)^2-(\sqrt{2} i)^2} \\ & =\frac{5+6 \sqrt{2} i-2}{1-2 i^2} \\ & =\frac{3+6 \sqrt{2} i}{1+2} \\ & =\frac{3+6 \sqrt{2} i}{3} \\ & =\frac{3(1+2 \sqrt{2} i)}{3} \\ &=1+2 \sqrt{2} i \end{aligned}$
(v) $\begin{aligned} & 5 i\left(-\frac{3}{5} i\right) \\ = & -\frac{15}{5}\left(i^2\right) \\ = & -3 \times(-1) \\ = & 3\end{aligned}$
$ \begin{aligned} & \text { (vi) }\left(\frac{1}{5}+\frac{2}{5} i\right)-\left(4+\frac{5}{2} i\right) \\ & =\frac{1}{5}+\frac{2}{5} i-4-\frac{5}{2} i \\ & =\frac{1}{5}-4+i\left(\frac{2}{5}-\frac{5}{2}\right) \\ & =\frac{-19}{5}+i\left(\frac{4-25}{10}\right) \\ & =-\frac{19}{5}+\left(-\frac{21}{10}\right) i \\ & =-\frac{19}{5}-\frac{21}{10} i \end{aligned} $
Question 6
(i)
$\begin{aligned} & 5(i)^{11}+6(i)^7-9(i)^9 \\ = & 5(i)^{10} \times i+6(i)^6 \times i-9(i)^8 \times i \\ = & 5\left(i^2\right)^5 \times i+6\left(i^2\right)^3 \times i-9\left(i^2\right)^4 \times i \\ = & 5(-1)^5 i+6(-1)^3 i-9(-1)^4 \times i \\ = & -5 i-6 i-9 i \\ = & -20 i\end{aligned}$
Question 7
$\begin{aligned} & \frac{1}{(1-i)^2}-\frac{1}{(1+i)^2} \\ = & \frac{1}{1+i^2-2 i}-\frac{1}{1+i^2+2 i} \\= &\frac{1}{1-1-2 i}-\frac{1}{1-1+2 i} \\ & =-\frac{1}{2 i}-\frac{1}{2 i} \\ & =\frac{-2}{2 i} \\ &=-\frac{1}{i} \times \frac{i}{i} \\ & =\frac{-i}{i^2} \\ & =\frac{-i}{-1} \\ & =i\end{aligned}$
Question 8
$ \begin{aligned} & \text { (8) }\left(\frac{1}{3}+\frac{1}{5} i\right)+\left(\frac{1}{5}-\frac{1}{3} i\right) \\ & =\left(\frac{1}{3}\right)+\frac{1}{5} i+\left(\frac{1}{5}\right)-\frac{1}{3} i \\ & =\frac{1}{3}+\frac{1}{5}+i\left(\frac{1}{5}-\frac{1}{3}\right) \\ & =\frac{5+3}{15}+i\left(\frac{3-5}{15}\right) \\ & =\frac{8}{15}+i\left(-\frac{2}{15}\right) \\ & =\frac{8}{15}-\frac{2}{15} i \end{aligned} $
Question 9
$\left(\frac{1+i}{1-i}\right)^m=1, m=9$
$\begin{aligned} & \left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^m=1 \\ & \left(\frac{(1+i)^2}{(1)^2-(i)^2}\right)^m=1 \\ & \left(\frac{1+i^2+2 i}{1-(i)^2}\right)^m=1\end{aligned}$
$\begin{aligned}\left(\frac{x-x+2 i}{1+1}\right)^m&=1 \\ \left(\frac{2 i}{2}\right)^m&=1 \\ i^m&=i^4 \\ m&=4\end{aligned}$
Question 10
(a) $\begin{aligned} & \frac{(1+i)^2+(1-i)^2}{(1+i)^2-(1-i)^2} \\ = & \frac{1+i^2+2 i+1+i^2-2 i}{1+i^2+2 i-1-i^2+2 i} \\ = & \frac{2+2 i^2}{4 i} \\ = & \frac{2+2(-1)}{4 i} \\ = & \frac{2-2}{4 i} \\ = & \frac{0}{4 i} \Rightarrow 0\end{aligned}$
(b) $\begin{aligned} & \frac{1+i}{1-i}-\frac{1-i}{1+i} \\ = & \frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)} \\ = & \frac{1+i^2+2 i-1-i^2+2 i}{1^2-i^2} \\ = & \frac{4 i}{1+1} \\ = & \frac{4 i}{2} \\ = & 2 i\end{aligned}$
(c) $\begin{aligned} & (1-i)(1+i) \\ = & (1)^2-(i)^2 \\ = & 1-i^2 \\ = & 1-(-1) \\ = & 1+1 \\ = & 2\end{aligned}$
(d) $\begin{aligned} & (\sqrt{3}+i \sqrt{12})+(\sqrt{12}-\sqrt{3} i) \\ = & \sqrt{3}+2 \sqrt{3} i+2 \sqrt{3}-\sqrt{3} i \\ = & \sqrt{3}+2 \sqrt{3}+2 \sqrt{3} i-\sqrt{3} i \\ = & 3 \sqrt{3}+\sqrt{3} i\end{aligned}$
Question 11
$\begin{aligned} =& (1-i)^4\\ = & \left\{(1-i)^2\right\}^2 \\ = & \left\{1+i^2-2 i\right\}^2 \\ & =\{1-1-2 i\}^2 \\ & =\{-2 i\}^2 \\ & =4 i^2 \\ & =4(-1) \\ & =-4\end{aligned}$
Question 12
(a)
$\begin{aligned} &= \frac{1}{3+4 i} \\ & =\frac{1}{3+4 i} \times \frac{3-4 i}{3-4 i} \\ & =\frac{3-4 i}{(3)^2-(4 i)^2} \\ & =\frac{3-4 i}{9-16 i^2} \\ & =\frac{3-4 i}{9-16} \\& =\frac{3-4 i}{25} \\ & =\frac{3}{25}-\frac{4}{25} i\end{aligned}$
(b)
$\begin{aligned} & \frac{3+2 i}{5-3 i} \times \frac{5+3 i}{5+3 i} \\ = & \frac{(3+2 i)(5+3 i)}{(5)^2-(3 i)^2} \\ = & \frac{15+9 i+10 i+6 i^2}{25-9 i^2} \\ & =\frac{15-6+19 i}{25+9} \\ & =\frac{9+19 i}{34} \\ & =\frac{9}{34}+\frac{19}{34} i\end{aligned}$
(c)
$\begin{aligned} & \frac{(2+3 i)^2}{2-i} \times \frac{2+i}{2+i} \\ = & \frac{\left(4+9 i^2+12 i\right)(2+i)}{(2-i)(2+i)} \\ = & \frac{(4-9+12 i)(2+i)}{(2)^2-(i)^2} \\ = & \frac{(-5+12 i)(2+i)}{4+1} \\ = & \frac{-10-5 i+24 i+12 i^2}{5} \\ & =\frac{-10-12+19 i}{5} \\ & =\frac{-22+19 i}{5} \\ & =\frac{-22}{5}+\frac{19}{5} i\end{aligned}$
(d)
$\begin{aligned} & \frac{4+\sqrt{-3}}{4-\sqrt{-3}} \\ = & \frac{4+\sqrt{3} \times \sqrt{-1}}{4-\sqrt{3} \times \sqrt{-1}} \\ = & \frac{4+\sqrt{3} i}{4-\sqrt{3} i} \\ = & \frac{4+\sqrt{3} i}{4-\sqrt{3} i} \times \frac{4+\sqrt{3} i}{4+\sqrt{3} i} \\ = & \frac{(4+\sqrt{3} i)^2}{(4)^2-(\sqrt{3} i)^2} \\& =\frac{(4)^2+(\sqrt{3} i)^2+8 \sqrt{3} i}{16-3 i} \\ & =\frac{16+3 i^2+8 \sqrt{3} i}{16+3} \\ & =\frac{16-3+8 \sqrt{3} i}{19} \\ & =\frac{13+8 \sqrt{3} i}{19} \\ & =\frac{13}{19}+\frac{8 \sqrt{3}}{19} i\end{aligned}$
(g)
$\begin{aligned} & \left(\frac{1}{3}+3 i\right)^3 \\ = & \left(\frac{1}{3}\right)^3+(3 i)^3+3 \times \frac{1}{3} \times 3 i\left(\frac{1}{3}+3 i\right) \\ = & \frac{1}{27}+27 i^3+3 i\left(\frac{1}{3}+3 i\right) \\ = & \frac{1}{27}+27 i^2 \cdot i+3 i \times \frac{1}{3}+9 i^2 \\ = & \frac{1}{27}-27 i+i-9 \\ = & \frac{1}{27}-9-26 i \\ = & -\frac{242}{9}-26 i\end{aligned}$
Question 13
$\begin{aligned} & \text { (13) } 2-3 i \\ & \text { संयुग्मी } \Rightarrow 2+3 i \\ & (2-3 i)(2+3 i) \\ & =(2)^2-(3 i)^2 \\ & =4-9 i^2 \\ & =4+9 \\ & =13\end{aligned}$
Question 14
$\begin{aligned} & Z_1=2-i, \quad \bar{Z}_1=2+i \\ & Z_2=-2+i, \quad \bar{Z}_2=-2-i\end{aligned}$
(i) $R_e\left(\frac{z_1 z_2}{z_1}\right)$
$\begin{aligned} & =\frac{(2-i)(-2+i)}{\bar{z}_1} \\ & =\frac{-(2-i)(2-i)}{2+i} \\ & =\frac{-(2-i)^2}{2+i} \\ & =\frac{-\left[4+i^2-4 i\right]}{2+i} \\ & =\frac{-[4-1-4 i]}{2+i} \\ & =\frac{-[3-4 i]}{2+i} \times \frac{(2-i)}{(2-i)} \\ & =\frac{-\left[6-3 i-8 i+4 i^2\right]}{(2)^2-(i)^2} \\ & =\frac{-[6-4-11 i]}{4-i^2} \\ & =\frac{-[2-11 i]}{4+1} \\ & =\frac{-2}{5}+\frac{11}{5} i\end{aligned}$
$\operatorname{Re}\left(\frac{-2}{8}+\frac{11}{5} i\right)=\frac{-2}{5}$
(ii) $Z_1=2-i \quad \& Z_2=-2+i$
$\begin{aligned} \operatorname{Im}\left(\frac{1}{z_1 \bar{z}_1}\right) & =\frac{1}{(2-i)(2+i)} \\ & =\frac{1}{(2)^2-(i)^2} \\ & =\frac{1}{4-i^2} \\ & =\frac{1}{4+1} \\ & =\frac{1}{5}\end{aligned}$
Question 15
$ \begin{aligned} & \text { (i) } 2 i=x+i y \\ & 0+2 i=x+i y \end{aligned} $
तुलना करने पर
x=0
y=2
(b) $2+yi=x-3i$
तुलना करने पर
x=2
y=-3
$ \begin{aligned} & \text { (c) }(x+y i)-(7-4 i)=3-5 i \\ & n+y i-7+4 i=3-5 i \\ & (x-7)+i(y+4)=3-5 i \end{aligned} $
$\begin{aligned} x-7 & =3 \\ x & =3+7 \\ x & =10\end{aligned}$
$\begin{aligned} y+4 & =-5 \\ y & =-5-4 \\ y & =-9\end{aligned}$
(d)
$(2 y-6)+(4x-20) i=0+0 i$
तुलना करने पर
$\begin{aligned} 2 y-6 & =0 \\ 2 y & =6 \\ y & =\frac{6}{2} =3 \\ y & =3\end{aligned}$
$\begin{aligned} 4 x-20 & =0 \\ 4 x & =20 \\ x & =\frac{20}{4}=5 \\ x & =5\end{aligned}$
Question 16
$\begin{aligned} & \text { (a) } 1+\sqrt{-1} \\ & =1+1i\end{aligned}$
क्रमित युग्म =(1,1)
(b) $x+3 y i$
क्रमित युग्म $=(n, 3 y)$
(c) $4-\sqrt{7} i$
क्रामित युग्न $=(4,-\sqrt{7})$
(d) $-5 i \Rightarrow 0-5 i$
क्रामित युग्न=(0,-5)
Question 17
(i) 4-3i
4-3i का गुथात्मक प्रलिलोम
$\begin{aligned} & =\frac{1}{4-3 i} \times \frac{4+3 i}{4+3 i} \\ & =\frac{4+3 i}{(4)^2-(3 i)^2} \\ & =\frac{4+3 i}{16-9 i^2} \\ & =\frac{4+3 i}{16+9} \\ & =\frac{4+3 i}{25} \\ & =\frac{4}{25}+\frac{3}{95} i\end{aligned}$
(ii) $\sqrt{5}+3i$ का गु० प्रतिलोम
$\begin{aligned} & =\frac{1}{\sqrt{5}+3 i} \times \frac{\sqrt{5}-3 i}{\sqrt{5}-3 i} \\ & =\frac{\sqrt{5}-3 i}{(\sqrt{5})^2-(3 i)^2} \\ & =\frac{\sqrt{5}-3 i}{5-9 i^2} \\ & =\frac{\sqrt{5}-3 i}{5+9} \\ & =\frac{\sqrt{5}-3 i}{14} \\ &= \frac{\sqrt{5}}{14}-\frac{3}{14} i\end{aligned}$
(iii) -i का गु० प्रतिलोम
$\begin{aligned} & =\frac{-1}{i} \times \frac{i}{i} \\ & =\frac{-i}{i^2} \\ & =\frac{-i}{-1} \\ & =i\end{aligned}$
Question 18
$\left(\frac{1+i}{\sqrt{2}}\right)^8+\left(\frac{11-i}{\sqrt{2}}\right)^8=2$
Sol :
$\begin{aligned} & =\left(\left(\frac{1+i}{\sqrt{2}}\right)^2\right]^4+\left[\left(\frac{1-i}{\sqrt{2}}\right)^2\right]^4 \\ & =\left[\left(\frac{1+i^2+2 i}{2}\right)^4+\left(\frac{1+i^2-2 i}{2}\right)^4\right] \\ & =\left[\left(\frac{1-1+2 i}{2}\right)^4+\left(\frac{1-1-2 i}{2}\right)^4\right]\end{aligned}$
$=\left[\left(\frac{2 i}{x}\right)^4+\left(\frac{-2 i}{2}\right)^4\right]$
$\begin{aligned} & =i^4+i^4 \\ & =1+1 \\ & =2\end{aligned}$
R.H.S
L.H.S=R.H.S
Question 19
$\frac{1-i n}{1+i n}=a-i b$....(i)
संयुगमी लिखने पर
$\frac{1+i n}{1-i n}=a+i b$....(ii)
समी (i) ,(ii)
$\frac{1-i n}{1+i n} \times \frac{1+i n}{1-i n}=(a-i b)(a+i b)$
$\begin{aligned} & 1=(a)^2-(i b)^2 \\ & 1=a^2-b^2 i^2 \\ & 1=a^2+b^2\end{aligned}$
Question 20
$(\cos 2 p \pi+i \sin 2 p \pi)(\cos 2 q \pi+i \sin 2 q \pi)=1$
$=\cos 2 p \pi \cdot \cos 2 q \pi +i \sin 2 q \pi \cdot \cos 2 p \pi +\operatorname{} i^2 \sin 2 p \pi \cdot \sin 2 q \pi$
$=\cos 2 p \pi \cdot \cos 2 q \pi-\sin 2 p \pi \cdot \sin 2 q \pi+i(\sin 2 p \pi \cdot \cos 2 q \pi+ \sin 2 q \pi \cdot \cos 2 p \pi)$
$\begin{aligned} & =\cos (2 p \pi+2 q \pi)+i \sin (2 p \pi+2 q \pi) \\ & =\cos 2 \pi(p+q)+i \sin 2 \pi(p+q)\end{aligned}$
$\begin{aligned} & =(\cos \pi+i \sin \pi)^{2(p+q)} \\ & =(-1+0)^{2(p+q)}\end{aligned}$
$\begin{aligned} & =(-1)^{2(p+q)} \\ & =1\end{aligned}$
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