Question 24 MarksFind the value of : $x^3-5 x^2+4 x+8$, if $x=\frac{10}{3-i}$View full question & answer→
Question 44 MarksSolve the following equations for $x, y \in R$ : $2 x+i^9 y(2+i)=x i^7+10 i^{16}$Answer$2 x+i^9 y(2+i)=x i^7+10 i^{16}$ $\therefore 2 x+\left(i^4\right)^2 \cdot i \cdot y(2+i)=x\left(i^2\right)^3 \cdot i+10 \cdot\left(i^4\right)^4$ $\therefore 2 x+(1)^2 \cdot i y(2+i)=x(-1)^3 \cdot i+10(1)^4 \ldots \ldots\left[\because i^2=-1, i^4=1\right]$ $\therefore 2 x+2 y i+y i^2=-x i+10$ $\therefore 2 x+2 y i-y+x i=10$ $\therefore(2 x-y)+(x+2 y) i=10+0 . i$ Equating real and imaginary parts, we get $2 x-y=10 \ldots . .$ $\text { and } x+2 y=0$ $\text { Equation }(i) \times 2$ $5 x=20$ $\therefore x=4 $ Equation (i) $\times 2+$ equation (ii) gives $5 x=20$ $\therefore x=4$ Putting $x=4$ in (i), we get $2(4)-y=10$ $\therefore y=8-10$ $\therefore y=-2 \therefore x=4 and y=-2 $View full question & answer→
Question 54 MarksSolve the following equations for $x, y \in R$ : $\frac{x+i y}{2+3 i}=7- i$Answer$ \frac{x+i y}{2+3 i}=7-i$ $\therefore x+i y=(7-i)(2+3 i)$ $\therefore x+i y=14+21 i-2 i-3 i^2$ $\therefore x+i y=14+19 i-3(-1) \ldots . .\left[\because i^2=-1\right]$ $\therefore x+i y=17+19 i$Equating real and imaginary parts, we get $x=17 \text { and } y=19$View full question & answer→
Question 64 MarksIf $x=a+b, y=\alpha a+\beta b$ and $z=a \beta+b \alpha$, where $\alpha$ and $\beta$ are the complex cube roots of unity, show that $x y z=a^3+b^3$.Answer$x=a+b, y=\alpha a+\beta b, z=a \beta+b \alpha$$\alpha$ and $\beta$ are the complex cube roots of unity.$\begin{aligned}\therefore \quad \alpha & =\frac{-1+i \sqrt{3}}{2} \text { and } \beta=\frac{-1-i \sqrt{3}}{2} \\\therefore \quad \alpha \beta & =\left(\frac{-1+i \sqrt{3}}{2}\right)\left(\frac{-1-i \sqrt{3}}{2}\right) \\& =\frac{(-1)^2-(i \sqrt{3})^2}{4} \\& =\frac{1-(-1)(3)}{4} \\& =\frac{1+3}{4} \quad \ldots\left[\because i^2=-1\right] \\\therefore \quad \alpha \beta & =1\end{aligned}$$\therefore \quad \alpha \beta=1$Also, $\alpha+\beta=\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2}$$\begin{aligned}& =\frac{-1+i \sqrt{3}-1-i \sqrt{3}}{2} \\& =\frac{-2}{2}\end{aligned}$$\therefore \quad \begin{aligned}\alpha+\beta & =-1\\\text { L.H.S } & =x y z=(a+b)(\alpha a+\beta b)(a \beta+b \alpha) \\& =(a+b)\left(\alpha \beta a^2+\alpha^2 a b+\beta^2 a b+\alpha \beta b^2\right) \\& =(a+b)\left[1 .\left(a^2\right)+\left(\alpha^2+\beta^2\right) a b+1 .\left(b^2\right)\right] \\& =(a+b)\left\{a^2+\left[(\alpha+\beta)^2-2 \alpha \beta\right] a b+b^2\right\} \\& =(a+b)\left\{a^2+\left[(-1)^2-2(1)\right] a b+b^2\right\} \\& =(a+b)\left[a^2+(1-2) a b+b^2\right] \\& =(a+b)\left(a^2-a b+b^2\right) \\& =a^3+b^3 \\& =\text { R.H.S. }\end{aligned}$View full question & answer→
Question 74 MarksFind the square root of the following complex numbers:2(1 – √3i)AnswerLet $\sqrt{2(1-\sqrt{3} i)}=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$Squaring on both sides, we get$\begin{aligned}& 2(1-\sqrt{ } 3 i)=(a+b i)^2 \\& 2(1-\sqrt{ } 3 i)=a^2+b^2 i^2+2 a b i \\& 2-2 \sqrt{ } 3 i=\left(a^2-b^2\right)+2 a b i \quad \ldots . .\left[\because i^2=-1\right]\end{aligned}$Equating real and imaginary parts, we get$\begin{array}{ll} & a^2-b^2=2 \text { and } 2 a b=-2 \sqrt{3} \\\therefore & a^2-b^2=2 \text { and } b=-\frac{\sqrt{3}}{a} \\\therefore & a^2-\left(-\frac{\sqrt{3}}{a}\right)^2=2 \\& \text { } \\\therefore \quad & a^2-\frac{3}{a^2}=2 \\\therefore & a^4-3=2 a^2 \\\therefore & a^4-2 a^2-3=0 \\\therefore & \left(a^2-3\right)\left(a^2+1\right)=0 \\\therefore & a^2=3 \text { or } a^2=-1 \\& B a \in R\end{array}$$\begin{array}{ll}\therefore & a^2-\frac{3}{a^2}=2 \\\therefore & a^4-3=2 a^2 \\\therefore & a^4-2 a^2-3=0 \\\therefore & \left(a^2-3\right)\left(a^2+1\right)=0 \\\therefore & a^2=3 \text { or } a^2=-1\end{array}$But $\mathrm{a} \in \mathrm{R}$$\begin{array}{ll}\therefore & \mathrm{a}^2 \neq-1 \\\therefore & \mathrm{a}^2=3 \\\therefore & \mathrm{a}= \pm \sqrt{3}\end{array}$When $\mathrm{a}=\sqrt{3}, \mathrm{~b}=\frac{-\sqrt{3}}{\sqrt{3}}=-1$When $\mathrm{a}=-\sqrt{3}, \mathrm{~b}=\frac{-\sqrt{3}}{-\sqrt{3}}=1$$\therefore \quad \sqrt{2(1-\sqrt{3} \mathrm{i})}= \pm(\sqrt{3}-\mathrm{i})$View full question & answer→
Question 84 MarksFind the square root of the following complex numbers:3 + 2√10iAnswerLet $\sqrt{3+2 \sqrt{10}} i=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get$\begin{aligned}& 3+2 \sqrt{ } 10 \mathrm{i}=(\mathrm{a}+\mathrm{bi})^2 \\& 3+2 \sqrt{ } 10 \mathrm{i}=\mathrm{a}^2+\mathrm{b}^2 \mathrm{i}^2+2 \mathrm{abi} \\& 3+2 \sqrt{ } 10 \mathrm{i}=\left(\mathrm{a}^2-\mathrm{b}^2\right)+2 \mathrm{abi} \ldots . . .\left[\mathrm{i}^2=-1\right]\end{aligned}$Equating real and imaginary parts, we get$\begin{array}{ll}a^2-b^2=3 \text { and } 2 a b=2 \sqrt{ } 10 \\a^2-b^2=3 \text { and } b=\frac{\sqrt{10}}{a} \\\therefore & a^2-\left(\frac{\sqrt{10}}{a}\right)^2=3 \\\therefore & a^2-\frac{10}{a^2}=3 \\\therefore & a^4-10=3 a^2 \\\therefore & a^4-3 a^2-10=0 \\\therefore & \left(a^2-5\right)\left(a^2+2\right)=0 \\\therefore & a^2=5 \text { or } a^2=-2\end{array}$But $\mathrm{a} \in \mathrm{R}$$\begin{array}{ll}\therefore & a^2 \neq-2 \\\therefore & a^2=5 \text { } \\\therefore & a= \pm \sqrt{5}\end{array}$When $a=\sqrt{5}, \mathrm{~b}=\frac{\sqrt{10}}{\sqrt{5}}=\sqrt{2}$When $\mathrm{a}=-\sqrt{5}, \mathrm{~b}=\frac{\sqrt{10}}{-\sqrt{5}}=-\sqrt{2}$$\therefore \quad \sqrt{3+2 \sqrt{10}} \mathrm{i}= \pm(\sqrt{5}+\sqrt{2} \mathrm{i})$View full question & answer→
Question 94 MarksFind the square root of the following complex numbers:1 + 4√3iAnswerLet $\sqrt{1}+4 \sqrt{3} i=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get$\begin{aligned}& 1+4 \sqrt{ } 3 i=(a+b i)^2 \\& 1+4 \sqrt{ } 3 i=a^2+b^2 i^2+2 a b i \\& 1+4 \sqrt{ } 3 i=\left(a^2-b^2\right)+2 a b i \quad \ldots . . .\left[\because i^2=-1\right]\end{aligned}$Equating real and imaginary parts, we get$\begin{array}{ll} & a^2-b^2=1 \text { and } 2 a b=4 \sqrt{3} \\\therefore \quad & a^2-b^2=1 \text { and } b=\frac{2 \sqrt{3}}{a} \\\therefore \quad & a^2-\left(\frac{2 \sqrt{3}}{a}\right)^2=1 \\\therefore \quad & a^2-\frac{12}{a^2}=1 \\\therefore \quad & a^4-12=a^2 \\\therefore \quad & a^4-a^2-12=0 \\\therefore \quad & \left(a^2-4\right)\left(a^2+3\right)=0 \\\therefore \quad & a^2=4 \text { or } a^2=-3 \\\therefore \quad & \text { But } a \in R \\\therefore \quad & a^2 \neq-3=4 \text { } \\\therefore \quad & a= \pm 2 \\& \text { When } a=2, b=\frac{2 \sqrt{3}}{2}=\sqrt{3} \\& \text { When } a=-2, b=\frac{2 \sqrt{3}}{-2}=-\sqrt{3} \\\therefore \quad & \sqrt{1+4 \sqrt{3} i}= \pm(2+\sqrt{3} i)\end{array}$View full question & answer→
Question 104 MarksFind the square root of the following complex numbers: 7 + 24iAnswerLet $\sqrt{7+24 i}=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get$\begin{aligned}& 7+24 i=(a+b i)^2 \\& 7+24 i=a^2+b^2 i^2+2 a b i \\& 7+24 i=\left(a^2-b^2\right)+2 a b i \quad \ldots . .\left[\because i^2=-1\right]\end{aligned}$Equating real and imaginary parts, we get$\begin{array}{ll} & a^2-b^2=7 \text { and } 2 a b=24 \\\therefore & a^2-b^2=7 \text { and } b=\frac{12}{a} \\\therefore & a^2-\left(\frac{12}{a}\right)^2=7 \\\therefore & a^2-\frac{144}{a^2}=7 \\\therefore & a^4-144=7 a^2 \\\therefore & a^4-7 a^2-144=0 \\\therefore & \left(a^2-16\right)\left(a^2+9\right)=0 \\\therefore & a^2=16 \text { or } a^2=-9 \\\therefore & \text { But } a \in R \\\therefore & a^2 \neq-9 \\\therefore & a^2=16 \\& a= \pm 4 \text { } \\& \text { When } a=4, b=\frac{12}{4}=3 \\& \text { When } a=-4, b=\frac{12}{-4}=-3 \\\therefore & \sqrt{7+24 i}= \pm(4+3 i)\end{array}$View full question & answer→
Question 114 MarksFind the square root of the following complex numbers:-8 – 6iAnswerLet $\sqrt{-8-6 i}=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get$\begin{aligned}-8-6 i & =(a+b i)^2 \\-8-6 i & =a^2+b^2 i^2+2 a b i \\-8-6 i & =\left(a^2-b^2\right)+2 a b i \quad . . . .\left[\because i^2=-1\right]\end{aligned}$Equating real and imaginary parts, we get$\begin{array}{ll} & a^2-b^2=-8 \text { and } 2 a b=-6 \\\therefore & a^2-b^2=-8 \text { and } b=\frac{-3}{a} \\\therefore & a^2-\left(-\frac{3}{a}\right)^2=-8 \\\therefore & a^2-\frac{9}{a^2}=-8 \\\therefore & a^4-9=-8 a^2 \\\therefore & a^4+8 a^2-9=0 \\\therefore & \left(a^2+9\right)\left(a^2-1\right)=0 \\\therefore & a^2=-9 \text { or } a^2=1\end{array}$But $a \in R$$\begin{array}{ll}\therefore & \mathrm{a}^2 \neq-9 \\\therefore & \mathrm{a}^2=1 \text { } \\\therefore & \mathrm{a}= \pm 1\end{array}$When $a=1, b=\frac{-3}{1}=-3$When $\mathrm{a}=-1, \mathrm{~b}=\frac{-3}{-1}=3$$\therefore \quad \sqrt{-8-6 i}= \pm(1-3 i)$View full question & answer→
Question 124 MarksFind the value of : $2 x^3-11 x^2+44 x+27$, if $x=\frac{25}{3-4 i}$View full question & answer→