Question 13 Marks
Write in $(\text{i}^{25})^3$ polar form.
Answer$(\text{i}^{25} )^3=\text{i}^{75}$ $=\text{i}^4\times 18+3$ $=(\text{ i}^4)^{ 18}.\text{ i}^3$ $=\text{i}^3\ \big[\because \text{i}^4=1\big]$ $=\text{-i}\ \big[\because \text{i}^3=-\text{i}\big]$ Let $\text{z} = 0 - \text{i}$ Then, $\text{|z|}=\sqrt{0^2+(-1)^2}=1$ Let $\theta$ be the argument of z and a be the acute angle given by $\tan \alpha=\frac{|\text{Im}\text{(z)|}}{|\text{Re}\text{(z)}|}$
Then, $\tan\alpha=\frac{1}{0}=\infty$ $\Rightarrow \alpha =\frac{\pi}{2}$ Clearly, z lies in fourth quadrant. so $\text{arg(z)}=-\alpha =-\frac{\pi}{2}$ $\therefore$ the polar from of z is $\text{|z|}(\cos\theta+\text{i }\sin\theta)=\cos(-\frac{\pi}{2})+\text{i }\sin(-\frac{\pi}{2})$ Thus, the polar from of $(\text{i}^{25})^3$ is $\cos(\frac{\pi}{2})-\text{i }\sin(\frac{\pi}{2})$
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Find the conjugates of the following complex numbers: $\frac{1}{3+5\text{i}}$
AnswerLet $\text{z}=\frac{1}{3+5\text{i}}$ $=\frac{1}{3+5\text{i}}\times\frac{(3+5\text{i})}{3+5\text{i}}$ (On rationalising the denominator) $=\frac{3+5\text{i}}{3^2+5^2}$ $\Rightarrow\text{z}=\frac{3-5\text{i}}{9+25}$ So $\bar{\text{z}}=\frac{3-5\text{i}}{34}$ $=\frac{3}{34}+\frac{5}{34}\text{i}$
View full question & answer→Question 33 Marks
Find the values of the following expressions: $\text{i}^{30}+\text{i}^{80}+\text{i}^{120}$
Answer$\text{i}^{30}+\text{i}^{80}+\text{i}^{120}=\text{i}^{4\times7}\times\text{i}^2+\text{i}^{4\times20}+\text{i}^{4\times30}$ $=1\times\text{i}^2+1+1$ $=-1+1+1$ $=1$$\therefore\text{i}^{30}+\text{i}^{80}+\text{i}^{120}=1$
View full question & answer→Question 43 Marks
Find the conjugates of the following complex numbers: $\frac{(1+\text{i})(2+\text{i})}{3+\text{i}}$
AnswerLet $\text{Z}=\frac{(1+\text{i})(2+\text{i})}{3+\text{i}}$ $=\frac{2+\text{i}+\text{i}(2+\text{i})}{3+\text{i}}$ $=\frac{2+\text{i}+2\text{i}+\text{i}}{3+\text{i}}$ $=\frac{1+3\text{i}}{3+\text{i}}$ $=\frac{(1+3\text{i})}{(3+\text{i})}\times\frac{(3-\text{i})}{(3-\text{i})}$ $=\frac{3-\text{i}+3\text{i}(3-\text{i})}{3^2+1^2}$ $=\frac{3-\text{i}+9\text{i}+3}{9+1}$ $=\frac{6+8\text{i}}{10}$ $=\frac{2(3+4\text{i})}{10}$ $\Rightarrow\text{z}=\frac{3+4\text{i}}{5}$ Hence $\bar{\text{z}}=\frac{3-4\text{i}}{5}$ $=\frac{3}{5}-\frac{4}{5}\text{i}$
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$(1+\text{i})^6+(1-\text{i})^3$
Answer$(1+\text{i})^6+(1-\text{i})^3=\Big[(1+\text{i})^2\Big]^3+(1-\text{i})^3$ $=\big(1+\text{i}^2+2\text{i}\big)^3+\big(1-3\text{i}+3\text{i}^2-\text{i}^3\big)$ $=\big(1-1+2\text{i}\big)^3+\big(1-3\text{i}-3+\text{i}\big)$ $=8\text{i}^3-2-2\text{i}$ $=-2-10\text{i}$
View full question & answer→Question 63 Marks
Find the multiplicative inverse of the following complex numbers: $1-\text{i}$
AnswerIf $\text{z}=\text{x}+\text{iy}$ is a complex number, then the multiplicative inverse of z, denoted by $\text{z}^{-1}$ or $\frac{1}{\text{z}}$ is defined as $\text{z}^{-1}=\frac{1}{\text{z}}$ $=\frac{1}{\text{x}+\text{iy}}$ $=\frac{1}{\text{x}+\text{iy}}\times\frac{\text{x}-\text{iy}}{\text{x}-\text{iy}}$ $=\frac{\text{x}-\text{iy}}{\text{x}^2+\text{y}^2}$ $=\frac{\text{x}}{\text{x}^2+\text{y}^2}-\frac{\text{y}}{\text{x}^2+\text{y}^2}\text{i}$ Given $\text{z}=1-\text{i}$ $\therefore \ \text{z}^{-1}=\frac{1}{1^2+1^2}-\frac{(-1)}{1^2+1^2}\times\text{i}$
View full question & answer→Question 73 Marks
Find the conjugates of the following complex numbers: $\frac{1}{1+\text{i}}$
AnswerLet $\text{z}=\frac{1}{1+\text{i}}$ $=\frac{1}{1+\text{i}}\times\frac{(1-\text{i})}{1-\text{i}}$ $=\frac{1+\text{i}}{1^2+1^2}$ $=\frac{1-\text{i}}{2}$ $\therefore\bar{\text{z}}=\frac{1-\text{i}}{2}$ $=\frac{1}{2}+\frac{1}{2}\text{i}$
View full question & answer→Question 83 Marks
Evaluate the following: $\big(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3$
AnswerWe know that $\text{i}=\sqrt{-1}$ $\text{i}^2 = -1$ $\text{i}^3 = -\text{i}$ $\text{i}^4 = 1$ In order to find $i^n$ Where n > 4, we divide n by 4 to get quotient p and remainder q, So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $ Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$ $=\text{i}^{4\text{p}}\times\text{i}^\text{q}$ $=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$ $=\text{i}^{\text{p}}\times\text{i}^\text{q}$ $=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$ Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $ $\big(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3=\big(\text{i}^{4\times9}\times\text{i}^{1}+\text{i}^{4\times17}\times\text{i}^{2}+\text{i}^{4\times21}\times\text{i}^3+\text{i}^{4\times103}\times\text{i}^2\big)^3$ $=\big(1\times\text{i}+1\times\text{i}^2+1\times\text{i}^3+1\times\text{i}^2\big)^3$ $=(\text{i}-1-\text{i}-1)^3$ $=(-2)^3$ $=-8$ $\big(\therefore(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3=-8$
View full question & answer→Question 93 Marks
Find the multiplicative inverse of the following complex numbers: $4-3\text{i}$
AnswerLet $\text{z}=4-3\text{i}$ $\text{z}^{-1}=\frac{4}{4^2+(-3)^2}-\frac{(-3)}{4^2+(-3)^2}$ $=\frac{4}{16+9}+\frac{3}{16+9}\text{i}$ $=\frac{4}{25}+\frac{3}{25}\text{i}$
View full question & answer→Question 103 Marks
Evaluate the following: $\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9$
AnswerWe know that $\text{i}=\sqrt{-1}$ $\text{i}^2 = -1$ $\text{i}^3 = -\text{i}$ $\text{i}^4 = 1$ In order to find $i^n$ Where n > 4, we divide n by 4 to get quotient p and remainder q, So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $ Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$ $=\text{i}^{4\text{p}}\times\text{i}^\text{q}$ $=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$ $=\text{i}^{\text{p}}\times\text{i}^\text{q}$ $=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$ Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $ $\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9=\Bigg(\text{i}^{4\times10}\times\text{i}^1+\frac{1}{\text{i}^{4\times64}\times\text{i}^1}\Bigg)^9$ $=\Big(1\times\text{i}+\frac{1}{1\times\text{i}}\Big)^9$ $=\Big(1+\frac{1}{\text{i}}\Big)^9$ $=\Big(\text{i}+\frac{1}{\text{i}\times\text{i}}\times\text{i}\Big)^9$ $=\Big(\text{i}+\frac{\text{i}}{-1}\Big)^9$ $=(\text{i}-\text{i})^9$ $=0$
View full question & answer→Question 113 Marks
If $\text{z}_1=2-\text{i}, \ \text{z}_2=-2+\text{i},$ Find $\text{Im}\Big(\frac{1}{\text{z}_1\bar{\text{z}}_1}\Big)$
Answer$\frac{1}{\text{z}_1\bar{\text{z}}_1}=\frac{1}{|\text{z}|^2}$ $=\frac{1}{|2-\text{i}|^2}$ $=\frac{1}{2^2+(-1)^2}$ $=\frac{1}{4+1}$ $=\frac{1}{5},$ which is purely real $\therefore \ \text{Im}\Big(\frac{1}{\text{z}_1\bar{\text{z}}_1}\Big)=0$
View full question & answer→Question 123 Marks
Find the real values of x and y, if $\frac{(1+\text{i})\text{x}-2\text{i}}{3+\text{i}}+\frac{(2-3\text{i})\text{y}+\text{i}}{3-\text{i}}=\text{i}$
Answer$\frac{(1+\text{i})\text{x}-2\text{i}}{3+\text{i}}+\frac{(2-3\text{i})\text{y}+\text{i}}{3-\text{i}}=\text{i}$ $\Rightarrow\frac{(3-\text{i})((1+\text{i})\text{x}-2\text{i})+(3+\text{i})((2-3\text{i})\text{y}+\text{i})}{(3+\text{i})(3-\text{i})}=\text{i}$ $\Rightarrow\frac{(3-\text{i})(1+\text{i})\text{x}-2\text{i}(3-\text{i})+(3+\text{i})(2-3\text{i})\text{y}+\text{i}(3+\text{i})}{3^2+1^2}=\text{i}$ $\Rightarrow\frac{(3+3\text{i}-\text{i}+1)\text{x}-6\text{i}-2+(6-9\text{i}+2\text{i}+3)\text{y}+3\text{i}}{9+1}=\text{i}$ $\Rightarrow\frac{(4+2\text{i})\text{x}-6\text{i}-2+(9-7\text{i})\text{y}+3\text{i}-1}{10}=1$ $\Rightarrow4\text{x}+2\text{ix}-6\text{i}-2+9\text{y}-7\text{iy}+3\text{i}-1=10\text{i}$ $\Rightarrow4\text{x}+9\text{y}-3+\text{i}(2\text{x}-7\text{y}-3)=10\text{i}$ Equating the real and imaginary parts we get $4\text{x}+9\text{y}-3=0 \ ...(\text{i})$ and $2\text{x}-7\text{y}-3=10$ i.e., $2\text{x}-7\text{y}=13 \ ...(\text{ii})$ Multiplying (i) by 7 and (ii) by 9 and adding we get $28\text{x}+18\text{x}+63\text{y}-63\text{y}=117+21$ $\Rightarrow46\text{x}=117+21$ $\Rightarrow46\text{x}=138$ $\Rightarrow\text{x}=\frac{138}{46}$ $=3$ Substituting the value of x = 3 in (i), we get $4\times3+9\text{y}=3$ $\Rightarrow9\text{y}=-9$ $\Rightarrow\text{y}=\frac{-9}{9}$ $\Rightarrow\text{y}=-1$ Hence $\text{x}=3, \ \text{y}=-1$
View full question & answer→Question 133 Marks
Find the real values of x and y, if $(1+\text{i})(\text{x}+\text{iy})=2-5\text{i}$
Answer$(1+\text{i})(\text{x}+\text{iy})=2-5\text{i}$ $\Rightarrow1(\text{x}+\text{iy})+\text{i}(\text{x}+\text{iy})=2-5\text{i}$ $\Rightarrow\text{x}+\text{iy}+\text{i}\text{x}-\text{y}=2-5\text{i}$ $\Rightarrow\text{x}-\text{y}+\text{i}(\text{x}+\text{y})=2-5\text{i}$ Equating the real and imaginary parts we get $\text{x}-\text{y}=2 \ ...(\text{i})$ $\text{x}+\text{y}=-5 \ ...(\text{ii})$ Adding (i) and (ii) we get $2\text{x}=2-5$ $\Rightarrow2\text{x}=-3$ $\Rightarrow\text{x}=\frac{-3}{2}$ Substituting the value of x in (i), we get $\frac{-3}{2}-\text{y}=2$ $\Rightarrow\frac{-3}{2}-2=\text{y}$ $\Rightarrow\text{y}=\frac{-3-4}{2}$ $\Rightarrow\text{y}=\frac{-7}{2}$ Hence $\text{x}=\frac{-3}{2},\text{y}=\frac{-7}{2}$
View full question & answer→Question 143 Marks
If $\frac{\text{z}-1}{\text{z}+1}$ is purely imaginary number $(\text{z}\neq-1),$ find the value of $|\text{z}|.$
AnswerLet $\text{z}=\text{x}+\text{iy},$ $\frac{\text{z}-1}{\text{z}+1}$ $=\frac{\text{x+iy}-1}{\text{x+iy}+1}$ $=\frac{\text{x-1+iy}}{\text{x+1+iy}}$ $=\frac{(\text{x-1+iy)}(\text{x+1-iy)}}{(\text{x+1+iy)}(\text{x+1-iy)}}$ [Rationalizing the denominator] $=\frac{(\text{x-1+iy)}(\text{x+1-iy)}}{{(\text{x}+1)}^2-{(\text{iy)}^2}}$ $=\frac{\text{x}^2+\text{x}-\text{ixy}-\text{x}-1+\text{iy}+\text{ixy}+\text{iy}+\text{y}^2}{\text{x}^2+2\text{x}+1+\text{y}^2}$ $=\frac{\text{x}^2-1+2\text{iy}+\text{y}^2}{\text{x}^2+2\text{x}+1\text{y}^2}$ $=\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+2\text{x}+1+\text{y}^2}+\text{i}\frac{2\text{y}}{\text{x}^2+2\text{x}+1+\text{y}^2}$ $\because$ It is a purely imaginary no. therefore real part = 0 $=\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+2\text{x}+1+\text{y}^2}=0$ $\Rightarrow\text{x}^2+\text{y}^2-1=0$ $\Rightarrow\text{x}^2+\text{y}^2=1$ $\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=1$ $\Rightarrow|\text{z}|=1$
View full question & answer→Question 153 Marks
Find the values of the following expressions: $\text{i}^{5}+\text{i}^{10}+\text{i}^{15}$
Answer$\text{i}^{5}+\text{i}^{10}+\text{i}^{15}=\text{i}^{4\times1}\times\text{i}^1+\text{i}^{4\times2}\times\text{i}^2+\text{i}^{4\times3}\times\text{i}^3$ $=1\times\text{i}+1\times\text{i}^2+1\times\text{i}^3$ $=\text{i}-1-\text{i}$ $=-1$ $\therefore\text{i}^{5}+\text{i}^{10}+\text{i}^{15}=-1$
View full question & answer→Question 163 Marks
Find the values of the following expressions: $\text{i}+\text{i}^{2}+\text{i}^{3}+\text{i}^{4}$
Answer$\text{i}+\text{i}^{2}+\text{i}^{3}+\text{i}^{4}=1+(-1)+(-\text{i})+1$ $=0$$\therefore\text{i}+\text{i}^{2}+\text{i}^{3}+\text{i}^{4}=0$
View full question & answer→Question 173 Marks
Find the modulus of $\frac{1+\text{i}}{1-\text{i}}-\frac{1-\text{i}}{1+\text{i}}$
AnswerLet $\text{z}=\frac{1+\text{i}}{1-\text{i}}-\frac{1-\text{i}}{1+\text{i}}$ $=\frac{(1+\text{i})^2-(1-\text{i})^2}{(1-\text{i})(1+\text{i})}$ $=\frac{1^2+\text{i}^2+2\times1\times\text{i}-(1^2+\text{i}^2-2\times1\times\text{i})}{1^2+1^2}$ $=\frac{1-1+2\text{i}-(1-1-2\text{i})}{2}$ $=\frac{2\text{i}+2\text{i}}{2}$ $=\frac{4\text{i}}{2}$ $\Rightarrow\text{z}=2\text{i}$ $\therefore \ |\text{z}|=|2\text{i}|$ $=2|\text{i}| \ \Big(\because \ |\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|\Big)$ $=2\times\text{i} \ \Big(\because \ |\text{i}|=1\Big)$ $=2$
View full question & answer→Question 183 Marks
Find the square root of the following complex numbers: $1+4\sqrt{-3}$
AnswerLet $\text{z}=1+4\sqrt{-3}$ $=1+4\sqrt{3}\times\sqrt{-1} \ (\therefore \ \sqrt{-3}=\sqrt{3}\times\sqrt{-1})$ $\Rightarrow\text{z}=1+4\sqrt{ 3\text{i}}$ $\therefore|\text{z}|=\sqrt{(1)^2+(4\sqrt{3})^2}$ $=\sqrt{1+48}$ $=\sqrt{49}$ $=7$ Hence $\therefore\sqrt{1+4\sqrt{-3}}=\pm\Bigg\{\sqrt{\frac{7+1}{2}}+\text{i}\sqrt{\frac{7-1}{2}}\Bigg\} \ (\because\text{y}>0)$ $=\pm\Bigg\{\sqrt{\frac{8}{2}}-\text{i}\sqrt{\frac{6}{2}}\Bigg\}$ $=\pm\{\sqrt{4}+\text{i}\sqrt{3}\}$ $=\pm\{2+\text{i}\sqrt{3}\}$
View full question & answer→Question 193 Marks
Find the real values of x and y, if $(3\text{x}-2\text{iy})(2+\text{i})^2=10(1+\text{i})$
Answer$(3\text{x}-2\text{iy})(2+\text{i})^2=10(1+\text{i})$ $\Rightarrow(3\text{x}-2\text{iy})(2^2+\text{i}^2+2\times2\times\text{i})=10+10\text{i}$ $\Rightarrow(3\text{x}-2\text{iy})(4-1\times4\text{i})=10+10\text{i}$ $\Rightarrow3\text{x}(3+4\text{i})-2\text{iy}(3+4\text{i})=10+10\text{i}$ $\Rightarrow9\text{x}+12\text{xi}-6\text{yi}+8\text{y}=10+10\text{i}$ $\Rightarrow9\text{x}+8\text{y}+\text{i}(12\text{x}-6\text{y})=10+10\text{i}$ Equating the real and imaginary parts we get $9\text{x}+8\text{y}=10 \ ...(\text{i})$ $-12\text{x}-6\text{y}=10 \ ...(\text{ii})$ Multiplying (i) by 6 and (ii) by 8 and adding $54\text{x}+96\text{x}+48\text{y}-48\text{y}=60+80$ $\Rightarrow150\text{x}=140$ $\Rightarrow\text{x}=\frac{140}{150}$ $\Rightarrow\text{x}=\frac{14}{15}$ Substituting the value of x in (i), we get $9\times\frac{14}{13}+8\text{y}=10$ $\Rightarrow\frac{42}{5}+8\text{y}=10$ $\Rightarrow8\text{y}=10-\frac{42}{5}$ $\Rightarrow8\text{y}=\frac{50-42}{5}$ $\Rightarrow8\text{y}=\frac{8}{5}$ $\Rightarrow\text{y}=\frac{1}{5}$ Hence $\text{x}=\frac{14}{5}$ and $\text{y}=\frac{1}{5}$
View full question & answer→Question 203 Marks
$1+\text{i}^2+\text{i}^4+\text{i}^6+\text{i}^8+ ...+\text{i}^{20}$
Answer$1+\text{i}^2+\text{i}^4+\text{i}^6+\text{i}^8+ ...+\text{i}^{20}$ $=1+\text{i}^2+\text{i}^4+\text{i}^{4\times1}\times\text{i}^2+\text{i}^{4\times2}+\text{i}^{4\times2}\times\text{i}^{2}+\text{i}^{4\times3}+\text{i}^{4\times3}\times\text{i}^{2}+\text{i}^{4\times4}+\text{i}^{4\times4}\times\text{i}^2+\text{i}^{4\times5}$ $=1-1+1+1\times\text{i}^2+1+1\times\text{i}^2+1+1\times\text{i}^2+1+1\times\text{i}^2+1$ $=1-1+1-1+1-1+1-1+1-1+1$ $=1$
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Solve the system of equations $\text{Re}(\text{z}^2)=0,|\text{z}|=2.$
Answer$\text{Re}(\text{z}^2)=0,|\text{z}|=2$ Let $\text{z}=\text{x}+\text{iy}$ $\text{z}^2=0$ $\Rightarrow(\text{x}+\text{iy})^2=0$ $\Rightarrow\text{x}^2+2\text{ixy}-\text{y}^2=0$ $\Rightarrow\text{x}^2-\text{y}^2=0 \ ...(\text{i}),$ which is the real part of $(\text{x}+\text{iy})^2.$ $|\text{z}|=2$ $\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=2$ $\Rightarrow\text{x}^2+\text{y}^2=4 \ ...(\text{ii})$ Adding (i) and (ii), we get $2\text{x}^2=4$ $\Rightarrow\text{x}^2=2$ $\Rightarrow\text{x}=\pm\sqrt{2},\text{y}=\pm\sqrt{2}$ $\text{x}+\text{iy}=\sqrt{2}+\text{i}\sqrt{2}$ $=\sqrt{2}-\text{i}\sqrt{2}$ $=\sqrt{2}+\text{i}\sqrt{2}$
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Find the square root of the following complex numbers: $-7+24\text{i}$
AnswerLet $\text{z}=-7+24\text{i}$ $\Rightarrow|\text{z}|=\sqrt{(-7)^2+24^2}$ $=\sqrt{49+576}$ $=\sqrt{625}$ $=25$ $\therefore\sqrt{-7-24\text{i}}=\pm\Bigg\{\sqrt{\frac{25-7}{2}}+\text{i}\sqrt{\frac{25+7}{2}}\Bigg\} \ (\because\text{y}<0)$ $=\pm\Bigg\{\sqrt{\frac{18}{2}}+\text{i}\sqrt{\frac{32}{2}}\Bigg\}$ $=\pm\{\sqrt{9}+\text{i}\sqrt{16}\}$ $=\pm\{3-4\text{i}\}$
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Find the conjugates of the following complex numbers: $4-5\text{i}$
AnswerIf $\text{z}=\text{x}+\text{iy}$ is a complex number, then the conjugate of z denoted by $\bar{\text{z}}$ is defined as $\bar{\text{z}}=\text{x}-\text{iy}$ Let $\text{z}=4-5\text{i}$ $\Rightarrow\bar{\text{z}}=4+5\text{i}$
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Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: $\frac{1+2\text{i}}{1-3\text{i}}$
AnswerThe polar form of a complex number $\text{z}=\text{x}+\text{iy},$ is given by $\text{z}=|\text{z}|(\cos\theta+\text{i}\sin\theta)$ Where, $|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$ and $\text{arg(z)}=\theta=\tan^{-1}\Big(\frac{\text{b}}{\text{a}}\Big)$ Let $\text{z}=\frac{1+2\text{i}}{1-3\text{i}}$ $=\frac{1+2\text{i}}{1-3\text{i}}\times\frac{1+3\text{i}}{1+3\text{i}}$ $=\frac{1(1+3\text{i})+2\text{i}(1+3\text{i})}{1^2+3^2}$ $=\frac{1+3\text{i}+2\text{i}-6}{1+9}$ $=\frac{-5+5\text{i}}{10}$ $=\frac{-5}{10}+\frac{5}{10}\text{i}$ $=\frac{-1}{2}+\frac{1}{2}\text{i}$ $\therefore \ |\text{z}|=\sqrt{\Big(\frac{-1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}$ $=\sqrt{\frac{1}{4}+\frac{1}{4}}$ $=\sqrt{\frac{2}{4}}$ $=\frac{1}{\sqrt{2}}$ Here $\text{x}=\frac{-1}{2}<0$ & $\text{y}=\frac{1}{2}>0, \ \therefore\theta$ lies i quadrantII $\theta=\text{arg(z)}=\tan^{-1}\frac{\frac{1}{2}}{\frac{-1}{2}}$ $=\tan^{-1}(-1)$ $=\tan^{-1}\Big(-\tan\frac{\pi}{4}\Big)$ $=\tan^{-1}\Big(\tan\big(\pi-\frac{\pi}{4}\big)\Big) \ \big(\therefore\tan(\pi-\theta)=-\tan\theta\big)$ $=\pi-\frac{\pi}{4}$ $=\frac{3\pi}{4}$ The polar form is given by $\text{z}=\frac{1}{\sqrt{2}}\Big(\cos\frac{3\pi}{4}+\text{i}\sin\frac{3\pi}{4}\Big)$
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Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: $\frac{1-\text{i}}{1+\text{i}}$
Answer$\frac{1-\text{i}}{1+\text{i}}=\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}=\frac{(1-\text{i})^2}{1^2-\text{i}^2}=\frac{1-2\text{i}-1}{1+1}=\frac{-2\text{i}}{2}=\text{-i}$ Modulus, $\Big|\frac{1-\text{i}}{1+\text{i}}\Big|=|\text{-i}|=1$ Argument, $\tan^{-1}\Big(\frac{-1}{0}\Big)=-\frac{\pi}{2}$ Polar form, $\text{z}=\text{r}(\cos\theta+\text{i}\sin\theta)$ $\text{z}=\Big(\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}\Big)$
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If $(1+\text{i})\text{z}=(1-\text{i})\bar{\text{z}},$ then show that $\text{z}=-\text{i}\bar{\text{z}}.$
Answer$(1+\text{i})\text{z}=(1-\text{i})\bar{\text{z}}$ $\Rightarrow\text{z}=\frac{(1-\text{i})}{(1+\text{i})}\bar{\text{z}}$ $\Rightarrow\text{z}=\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\bar{\text{z}}$ [Rationalizing the denominator] $\Rightarrow\text{z}=\frac{(1-2\text{i}-1)}{(1+1)}\bar{\text{z}}$ $\Rightarrow\text{z}=\frac{-2\text{i}}{2}\bar{\text{z}}$ $\Rightarrow\text{z}=-\text{i}\bar{\text{z}}$
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Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: $\frac{1}{1+\text{i}}$
Answer$\frac{1}{1+\text{i}}$ Rationalising the denominator: $\frac{1}{1+\text{i}}\times \frac{1-\text{i}}{1-\text{i}}$ $\Rightarrow \frac{1-\text{i}}{1-\text{i}^2}$ $\Rightarrow \frac{1-\text{i}}{2}\ (\because\ \text{i}^2=-1)$ $\Rightarrow \frac{1}{2}-\frac{\text{i}}{2}$ $\text{r}=|\text{z}|$ $= \sqrt{\frac{1}{4}+\frac{1}{4}}$ $=\frac{1}{\sqrt{2}}$ Let $\tan\alpha=\bigg|\frac{\text{Im}\text{(z)}}{\text{Re}\text{(z)}}\bigg|$ $\therefore\tan\alpha=\Bigg|\frac{\frac{1}{2}}{\frac{-1}{2}}\Bigg|$ $=1$ $\Rightarrow \alpha =\frac{\pi}{4}$ Since ponit $\Big(\frac{1}{2},-\frac{1}{2}\Big)$ lies in the fourth quadrant, the argurment is given by $\theta=-\alpha =\frac{\pi}{4}$ Polar form $=\text{r}(\cos \theta + \text{i } \sin\theta )$ $=\frac{1}{\sqrt{2}}\big\{\cos \big(\frac{-\pi}{4}\big)+\text{i }\sin\big(\frac{-\pi }{4}\big)\big\}$ $=\frac{1}{\sqrt{2}}\big(\cos \frac{\pi}{4}-\text{i }\sin\frac{\pi }{4}\big)$
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Show that $1+\text{i}^{10}+\text{i}^{20}+\text{i}^{30}$ is a real number.
Answer$1+\text{i}^{10}+\text{i}^{20}+\text{i}^{30}=1+\text{i}^{4\times2}\times\text{i}^2+\text{i}^{4\times5}+\text{i}^{4\times7}\times\text{i}^2$ $=1+1\times\text{i}^2+1+1\times\text{i}^2$ $=1-1+1-1$$=0,$ which is real number.
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Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: $\sin120^\circ-\text{i}\cos120^\circ$
Answer$\sin120^\circ-\text{i}\cos120^\circ$ $\frac{\sqrt{3}}{2}+\frac{\text{i}}{2}$ $\text{r}=|\text{z}|$ $=\sqrt{\frac{3}{4}+\frac{1}{4}}$ $=1$ Let $\tan\alpha=\bigg|\frac{\text{Im}\text{(z)}}{\text{Re}\text{(z)}}\bigg|$ Then, $\tan\alpha=\Bigg|\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg|$ $=\frac{1}{\sqrt{3}}$ $\Rightarrow \alpha =\frac{\pi}{6}$ Since point $\Big(\frac{\sqrt{3}}{2},\frac{1}{2}\Big)$ lies in the fourth quadrant, the argurment s given by $\theta=\alpha =\frac{\pi}{6}$ Polar form $=\text{r}(\cos \theta + \text{i } \sin\theta )$ $=\cos\frac{\pi}{6}+\text{i }\sin\frac{\pi}{6}$
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Find the square root of the following complex numbers: $-11-60\sqrt{-1}$
AnswerLet $\text{z}=11-60\sqrt{-1}$ $\Rightarrow\text{z}=-11-60\text{i} \ (\therefore \ \sqrt{-1}=\text{i})$ Then, $|\text{z}|=\sqrt{(-11)^2+(-60)^2}$ $=\sqrt{121+3600}$ $=\sqrt{3721}$ $=61$ $\therefore\sqrt{-11-60\text{i}}=\pm\Bigg\{\sqrt{\frac{61-11}{2}}-\text{i}\sqrt{\frac{61+11}{2}}\Bigg\} \ (\because\text{y}<0)$ $=\pm\Bigg\{\sqrt{\frac{50}{2}}-\text{i}\sqrt{\frac{72}{2}}\Bigg\}$ $=\pm\{\sqrt{5}-6\text{i}\}$
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Find the values of the following expressions: $\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}$
Answer$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}$ $=\frac{\text{i}^{4\times48}+\text{i}^{147}\times\text{i}^{2}+\text{i}^{4\times147}+\text{i}^{4\times146}\times\text{i}^2+\text{i}^{4\times146}}{\text{i}^{4\times145}+\text{i}^{2}\times\text{i}^{4\times145}+\text{i}^{4\times144}+\text{i}^{4\times144} +\text{i}^{4\times143}\times\text{i}^2}$ $=\frac{1+1\times\text{i}^2+1+1\times\text{i}^2+1}{1\times\text{i}^2+1+1\times\text{i}^2+1+1\times\text{i}^2}$ $=\frac{1-1+1-1+1}{-1+1-1+1-1}$ $=\frac{1}{-1}$ $=-1$ $\therefore \ \frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}=-1$
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Find the values of the following expressions: $\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}$
Answer$\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}=\text{i}^{4\times12}\times\text{i}^1+\text{i}^{4\times17}+\text{i}^{4\times22}\times\text{i}^1+\text{i}^{4\times7}\times\text{i}^2$ $=1\times\text{i}+1+1\times\text{i}+1\times\text{i}^2$ $=\text{i}+1+\text{i}-1$ $=2\text{i}$$\therefore\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}=2\text{i}$
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Find the real values of $\theta$ for which the complex number $\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ is purely real.
AnswerLet $\text{z}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ $=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}\times\frac{1+2\text{i}\cos\theta}{1+2\text{i}\cos\theta}$ $=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta(1+2\text{i}\cos\theta)}{1^2+(2\cos\theta)^2}$ $=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta-2\cos^2\theta)}{1+4\cos^2\theta}$ $=\frac{1-2\cos^2\theta+3\text{i}\cos\theta}{1+4\cos^2\theta}$ $=\frac{1-2\cos^2\theta}{1+4\cos^2\theta}+\frac{3\cos\theta}{1+4\cos^2\theta}\text{i}$ We know that z is purely real if and on;y if Im z=0 $\therefore \ \frac{3\cos\theta}{1+4\cos^2\theta}=0$ $\big(\because$ z is given to be purely real $\big)$ $\Rightarrow3\cos\theta=0$ $\Rightarrow\cos\theta=0$ $\Rightarrow\cos\theta=\cos\frac{\pi}{2}$ $\therefore$ The general solution is given by $\theta=2\text{n}\pi\pm\frac{\pi}{2},\text{n}\in\text{z}$
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$\text{x}^6+\text{x}^4+\text{x}^2+1,$ when $\text{x}=\frac{1+\text{i}}{\sqrt{2}}.$
AnswerWe have $\text{x}=\frac{1+\text{i}}{\sqrt{2}}$ $\Rightarrow\sqrt{2}\text{x}=1+\text{i}$ $\Rightarrow\sqrt({2}\text{x})^2=(1+\text{i})^2$ (squaring both sides) $\Rightarrow2\text{x}^2=1^2+(\text{i})^2+2\times1\times\text{i}$ $=1-1+2\text{i}$ $\Rightarrow2\text{x}^2=2\text{i}$ $\Rightarrow\text{x}^2=\text{i}$ $\Rightarrow(\text{x}^2)^2=(\text{i})^2$ (squaring both sides) $\Rightarrow\text{x}^4=-1$ $\Rightarrow\text{x}^4+1=0 \ ...(1)$ Now $\text{x}^6+\text{x}^4+\text{x}^2+1$ $=\text{x}^6+\text{x}^4+\text{x}^2+1$ $=\text{x}^2(\text{x}^4+1)+1(\text{x}^4+1)$ $=\text{x}^2\times0+1\times0$ (using(i)) $=0$
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