SOLUTIONS TO ASSIGNMENT – I FOR ENGINEERING MATHEMATICS – I

\[\color {black} {1.}\ Solve\ the\ following\ equations\ using\ Cramers\ Rule\ \hspace{20cm}\]\[3x + y – z=2,\ 2x – y + 2z = 6\ and\ 2x + y – 2z = -2\ \hspace{10cm}\]
\[\color {black}{Solution:}\ \hspace{20cm}\]
\[3x + y – z=2\ ——————-(1)\ \hspace{6cm}\]
\[2x – y + 2z = 6\ \hspace{15cm}\]
\[2x + y – 2z = -2\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix} 3 & 1 & -1 \\ 2 & -1 & 2 \\ 2 & 1 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta =3\begin{vmatrix} -1 & 2 \\ 1 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 2 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & -1\\ 2 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta =3(2\ -\ 2)\ – 1 (-4\ -\ (4)) – 1(2\ -\ (-2))\ \hspace{9cm}\]
\[\Delta =3(0)\ – 1 (-8) – 1(4)\ \hspace{13cm}\]
\[\Delta =0\ + 8 – 4\ \hspace{14cm}\]
\[\Delta =4\ \hspace{17cm}\]
\[\Delta_x = \begin{vmatrix} 2 & 1 & -1 \\ 6 & -1 & 2 \\ -2 & 1 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =2\begin{vmatrix} -1 & 2 \\ 1 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 6 & 2 \\ -2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 6 & -1\\ -2 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =2(2\ -\ 2) – 1 (-12\ -\ (-4)) – 1(6\ -\ (2))\ \hspace{9cm}\]
\[\Delta_x =2(0)\ – 1 (-8) – 1(4)\ \hspace{13cm}\]
\[\Delta_x = 0\ + 8 – 4\ \hspace{14cm}\]
\[\Delta_x =4\ \hspace{17cm}\]
\[\Delta_y = \begin{vmatrix} 3 & 2 & -1 \\ 2 & 6 & 2 \\ 2 & -2 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =3\begin{vmatrix} 6 & 2 \\ -2 & -2 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 2 \\ 2 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 6\\ 2 & -2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =3(-12\ -\ (-4))\ – 2 (-4\ -\ (4)) – 1(-4\ -\ (12))\ \hspace{9cm}\]
\[\Delta_y =3(- 8)\ – 2 (-8) – 1(-16)\ \hspace{13cm}\]
\[\Delta_y = -24\ + 16\ + 16\ \hspace{14cm}\]
\[\Delta_y = 8\ \hspace{17cm}\]
\[\Delta_z = \begin{vmatrix} 3 & 1 & 2 \\ 2 & -1 & 6 \\ 2 & 1 & -2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =3\begin{vmatrix} -1 & 6 \\ 1 & -2 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 6 \\ 2 & -2 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & – 1\\ 2 & 1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =3(2\ -\ 6)\ – 1 (-4\ -\ (12)) + 2(2\ -\ (-2))\ \hspace{9cm}\]
\[\Delta_z =3(-4)\ – 1 (-16) + 2(4)\ \hspace{13cm}\]
\[\Delta_z =-12\ + 16 + 8\ \hspace{14cm}\]
\[\Delta_z =12\ \hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{4}{4} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{8}{4} =\ 2\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{12}{4} =\ 3\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =1\ y = 2\ z = 3\ in\ equation (1)\ \hspace{18cm}\]
\[LHS = 3(1) + 2 – 3\]\[ = 3 + 2 – 3 = 2\]\[ = RHS\]
\[2.\ Find\ the\ inverse\ of\ \begin{bmatrix} 1 & 1 & -1\\ 2 & 1 & 0 \\ -1 & 2 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 1 & 1 & -1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =1\begin{vmatrix} 1 & 0 \\ 2 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 0 \\ -1 & 3 \\ \end{vmatrix}\ -\ 1\begin{vmatrix} 2 & 1\\ -1 & 2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =1(3\ -\ 0)\ – 1 (6\ -\ 0) – 1(4\ +\ 1)\ \hspace{9cm}\]
\[ =1(3)\ – 1 (6) – 1(5)\ \hspace{13cm}\]
\[ =\ 3\ -\ 6\ -\ 5\ \hspace{14cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ =\ -\ 8\ \neq\ 0\ \hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix} 1 & 0 \\ 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (3 – 0)\ \hspace{15cm}\]
\[= (1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 3\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 2}\ \begin{vmatrix} 2 & 0 \\ -1 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (6 – 0)\ \hspace{15cm}\]
\[= (-1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 1 =\ -\ 6\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 3}\ \begin{vmatrix} 2 & 1 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (4\ +\ 1)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ -1 =\ 5\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix} 1 & -1 \\ 2 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (3 + 2)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -5\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{2\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ -1 & 3 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3\ -\ 1)\ \hspace{15cm}\]
\[= (1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 =\ 2\ \hspace{15cm}\]
\[cofactor\ of\ 0 = (-1)^{2\ +\ 3}\ \begin{vmatrix} 1 & 1 \\ -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2\ +\ 1)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 0 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 1}\ \begin{vmatrix} 1 & -1 \\ 1 & 0 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (0 + 1)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{3\ +\ 2}\ \begin{vmatrix} 1 & -1 \\ 2 & 0 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (0 + 2)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{3\ +\ 3}\ \begin{vmatrix} 1 & 1 \\ 2 & 1 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (1 – 2)\ \hspace{15cm}\]
\[= (1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix} 3 & -6 & -5 \\ -5 & 2 & -3 \\ 1 & -2 & -1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[adj.\ A\ =\begin{bmatrix} 3 & -5 & 1 \\ -6 & 2 & -2 \\ 5 & -3 & -1 \\ \end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{-1}{8}\ \begin{bmatrix} 3 & -5 & 1 \\ -6 & 2 & -2 \\ 5 & -3 & -1 \\ \end{bmatrix}\ \hspace{2cm}\]
\[3.\ Find\ the\ rank\ of\ the\ matrix\ \begin{bmatrix} 5 & 3 & 14 & 4 \\ 0 & 1 & 2 & 1 \\ 1 & -1 & 2 & 0 \\ \end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix} 5 & 3 & 14 & 4 \\ 0 & 1 & 2 & 1 \\ 1 & -1 & 2 & 0 \\ \end{bmatrix}\ \hspace{15cm}\]
\[Order\ of\ A = 3 × 4\ \hspace{15cm}\]
\[\therefore \ rank\ of\ A=\rho(A)\ \leq\ Min{3,4} =3\ . \hspace{15cm}\]
\[The\ highest\ order\ of\ minors\ of\ A = 3.\ \hspace{15cm}\]
\[A\ has\ the\ following\ minors\ of\ order 3.\ \hspace{15cm}\]
\[ A_1\ =\begin{vmatrix} 5 & 3 & 14 \\ 0 & 1 & 2 \\ 1 & -1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[=5\begin{vmatrix} 1 & 2 \\ -1 & 2 \\ \end{vmatrix}\ -\ 3\begin{vmatrix} 0 & 2 \\ 1 & 2 \\ \end{vmatrix}\ +\ 14\begin{vmatrix} 0 & 1\\ 1 & -1 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =\ 5(2\ +\ 2)\ – 3 (0\ -\ 2) +\ 14(0\ -\ 1)\ \hspace{9cm}\]
\[ =\ 5(4)\ – 3 (-2) + 14(-1)\ \hspace{13cm}\]
\[ =\ 20\ +\ 6 -\ 14\ =\ 12\ \neq 0\ \hspace{14cm}\]
\[ A_2\ =\begin{vmatrix} 5 & 3 & 4 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \\ \end{vmatrix}\ \hspace{15cm}\]
\[=5\begin{vmatrix} 1 & 1 \\ -1 & 0 \\ \end{vmatrix}\ -\ 2\begin{vmatrix} 2 & 1 \\ 1 & 5 \\ \end{vmatrix}\ +\ 2\begin{vmatrix} 2 & 3\\ 1 & 3 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =5(0\ +\ 1)\ – 3 (0\ -\ 1) +\ 4(0\ -\ 1)\ \hspace{9cm}\]
\[ =5(1)\ – 3 (-1) + 4(-1)\ \hspace{13cm}\]
\[ = 5\ +\ 3 – 4\ =\ 4\ \neq 0\ \hspace{14cm}\]
\[ A_3\ =\begin{bmatrix} 5 & 14 & 4 \\ 0 & 2 & 1 \\ 1 & 2 & 0 \\ \end{bmatrix}\ \hspace{15cm}\]
\[=5\begin{vmatrix} 2 & 1 \\ 2 & 0 \\ \end{vmatrix}\ -\ 14\begin{vmatrix} 0 & 1 \\ 1 & 0 \\ \end{vmatrix}\ +\ 4\begin{vmatrix} 0 & 2\\ 1 & 2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =5(0\ -\ 2)\ – 14 (0\ -\ 1) + 4(0\ -\ 2)\ \hspace{9cm}\]
\[ =5(-2)\ – 14 (-1) + 4(-2)\ \hspace{13cm}\]
\[ = -10\ +\ 14 – 8\ =\ -4\ \neq 0\ \hspace{14cm}\]
\[ A_4\ =\begin{bmatrix} 3 & 14 & 4 \\ 1 & 2 & 1 \\ -1 & 2 & 0 \\ \end{bmatrix}\ \hspace{15cm}\]
\[=3\begin{vmatrix} 2 & 1 \\ 2 & 0 \\ \end{vmatrix}\ -\ 14\begin{vmatrix} 1 & 1 \\ -1 & 0 \\ \end{vmatrix}\ +\ 4\begin{vmatrix} 1 & 2\\ -1 & 2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[ =3(0\ -\ 2)\ – 14 (0\ +\ 1) + 4(2\ +\ 2)\ \hspace{9cm}\]
\[ =3(-2)\ – 14 (1) + 4(4)\ \hspace{13cm}\]
\[ = -6\ – 14 + 16\ =\ -12\ \neq 0\ \hspace{14cm}\]
\[\therefore\ \rho(A)\ =\ 3\ \hspace{15cm}\]
\[4 .\ Find\ the\ middle\ term\ in\ the\ expansion\ of\ (2x\ -\ \frac{x^2}{4})^{16}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ Here\ n\ is\ even\ number,\ so\ there\ is\ one\ middle\ term\ \hspace{18cm}\]
\[(\frac{n+2}{2})^{th}\ term\ =\ (\frac{16+2}{2})^{th}\ term\ =\ (\frac{18}{2})^{th}\ term\ =\ 9^{th}\ term\ \hspace{16cm}\]
\[T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}\]
\[Here\ X\ =\ 2x,\ a\ =\ -\frac{x^2}{4},\ n\ =\ 16,\ r\ =\ 8\ \hspace{14cm}\]
\[T_{8 + 1} = 16C_8\ (2x)^{16-8}\ (\frac{-x^2}{4})^8 \hspace{15cm}\]
\[T_9 = 16C_8\ (2x)^8\ \frac{((-x)^2)^8}{4^8}\ \hspace{15cm}\]
\[ = 16C_8\ \ 2^8\ \ x^8\ x^{16}\ \ 4^{-8}\ \hspace{15cm}\]
\[ = 16C_8\ \ 2^8\ \ 4^{-8}\ x^{8+ 16}\ \hspace{15cm}\]
\[T_9 = 16C_8\ 2^8\ 4^{-8}\ x^{24}\ \hspace{15cm}\]
\[The\ middle term = 16C_8\ 2^8\ 4^{-8}\ x^{24}\ \hspace{15cm}\]
\[5.\ Find\ the\ coefficient\ of\ x^5\ in\ the\ expansion\ of\ (x\ -\ \frac{1}{x})^{11}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}\]
\[Here\ X\ =\ x,\ a\ =\ -\frac{1}{x},\ n\ =\ 11,\ r\ =\ r\ \hspace{14cm}\]
\[T_{r + 1} = 11C_r\ x^{11-r}\ (\frac{-1}{x})^r \hspace{15cm}\]
\[ = 15C_r\ x^{11- r}\ (-1)^r\ x^{-r}\ \hspace{15cm}\]
\[ = 11C_r\ (-1)^r\ x^{11 – r – r} \hspace{15cm}\]
\[T_{r + 1} = 11C_r\ (-1)^r\ x^{11 – 2r}\ ———————— (1)\ \hspace{15cm}\]
\[We\ assume\ that\ x^5\ occurs\ in\ 11C_r\ (-1)^r\ x^{11 – 2r}\ \hspace{15cm}\]
\[\therefore\ x^{11- 2r}\ =\ x^5\ \hspace{15cm}\]
\[Equating\ the\ indices\ 11 – 2r\ =\ 5\ \hspace{15cm}\]
\[-2r\ =\ 5\ -\ 11\ \hspace{15cm}\]
\[-2r\ = -\ 6\ \hspace{15cm}\]
\[r\ = 3\ \hspace{15cm}\]
\[Put\ r\ =\ 3\ in\ (1)\ \hspace{15cm}\]
\[T_{3 + 1} = 11C_3\ (-1)^3\ x^{11\ -\ 6}\ \hspace{15cm}\]
\[T_4 = -\ 11C_3\ x^5\ \hspace{15cm}\]
\[The\ coefficient\ of\ x^5\ =\ -\ 11\ C_3\ \hspace{15cm}\]
\[6.\ Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of\ (4x^3\ +\ \frac{3}{x^2})^{20}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}\]
\[Here\ X\ =\ 4x^3,\ a\ =\ \frac{3}{x^2},\ n\ =\ 20,\ r\ =\ r\ \hspace{14cm}\]
\[T_{r + 1} = 20C_r\ ((4x)^3)^{20-r}\ (\frac{3}{x^2})^r \hspace{15cm}\]
\[ = 20C_r\ 4^{20\ -\ r}\ (x^3)^{20 – r}\ \frac{3^r}{x^{2r}} \hspace{15cm}\]
\[ = 20C_r\ 4^{20\ -\ r}\ x^{60\ -\ 3r}\ 3^r\ x^{-2r}\ \hspace{15cm}\]
\[ = 20C_r\ 4^{20\ -\ r}\ 3^r\ x^{60 – 3r – 2r} \hspace{15cm}\]
\[T_{r + 1} = 20C_r\ 4^{20\ -\ r}\ 3^r\ x^{60 – 5r}\ ———————— (1)\ \hspace{15cm}\]
\[To\ find\ the\ independent\ term\ of\ x,\ find\ the\ coefficient\ of\ x^0.\ \hspace{15cm}\]
\[\therefore\ x^{60\ -\ 5r}\ =\ x^0\ \hspace{15cm}\]
\[60 – 5r\ =\ 0\ \hspace{15cm}\]
\[-5r\ = -\ 60\ \hspace{15cm}\]
\[r\ = 12\ \hspace{15cm}\]
\[Put\ r\ =\ 12\ in\ (1)\ \hspace{15cm}\]
\[\therefore\ independent\ term\ of\ x =\ 20C_{12}\ 4^{20\ -\ 12}\ 3^{12}\ \hspace{15cm}\]
\[=\ 20C_{12}\ 4^8\ 3^{12}\ \hspace{15cm}\]
\[7.\ Express\ \frac{(1+ i)(1 + 2i)}{1+ 3i}\ in\ the\ form\ of\ a\ +\ ib\ \hspace{18cm}\]
\[\color {black}{Solution:} \hspace{20cm}\]
\[Let\ z = \frac{(1+ i)(1 + 2i)}{1+ 3i}\ \hspace{18cm}\]
\[ = \frac{1+ 2i + i + 2\ i^2}{1+ 3i}\ \hspace{15cm}\]
\[ = \frac{1+ 3i – 2}{1+ 3i}\ \hspace{15cm}\]
\[ = \frac{-1+ 3i}{1+ 3i}\ \hspace{15cm}\]
\[= \frac{-1+3i}{1+ 3i}\ ×\ \frac{1 – 3i}{1 – 3i}\ \hspace{15cm}\]
\[ = \frac{-1 + 3i + 3i – 9\ i^2}{(1)^2 + (3^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{-1 + 6i + 9 }{1+ 9}\ \hspace{4cm}\ \because [i^2\ =\ -1]\]
\[ = \frac{8 +\ 6i}{10}\ \hspace{10cm}\]
\[ Z = \frac{8}{10}\ +\ \frac{6i}{10}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{4}{5};\ Im(Z)\ =\ \frac{3}{5}\ \hspace{15cm}\]
\[8.\ Find\ the\ modulus\ and\ amplitude\ of\ \frac{-\ 3\ +\ i}{-\ 1\ +\ i}\ \hspace{18cm}\]
\[\color {black}{Solution:} \hspace{20cm}\]
\[Let\ z = \frac{-\ 3\ +\ i}{-\ 1\ +\ i}\ \hspace{18cm}\]
\[= \frac{-\ 3\ +\ i}{-\ 1\ +\ i}\ ×\ \frac{-\ 1\ -\ i}{-\ 1\ -\ i}\ \hspace{15cm}\]
\[ = \frac{3\ +\ 3i\ -\ i -\ i^2}{(-\ 1)^2 + (1)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 +\ (b)^2]\]
\[ = \frac{3\ +\ 2i\ +\ 1 }{1+ 1}\ \hspace{4cm}\ \because [i^2\ =\ -1]\]
\[ = \frac{4\ +\ 2i}{2}\ \hspace{10cm}\]
\[ Z = \frac{4}{2}\ +\ \frac{2i}{2}\ \hspace{15cm}\]
\[ z\ =\ 2\ +\ i\ = a\ + ib\ \hspace{14cm}\]
\[a = 2,\ b\ = \ 1\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(2)^2 + {1}^2}\ \hspace{12cm}\]
\[ = \sqrt{4\ +\ 1}\ =\ \sqrt{5}\ \hspace{12cm}\]
\[|z| = \sqrt{5}\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} \frac{2}{1}\ =\ tan^{-1} {2}\]
\[9.\ Show\ that\ the\ complex\ numbers\ 3\ +\ 2i,\ 5\ +\ 4i,\ 3\ +\ 6i\ and\ 1\ +\ 4i\ form\ a\ square\ \hspace{10cm}\]
\[\color {black}{Solution:} \hspace{20cm}\]
\[Let\ A = 3\ +\ 2i\ = (3, 2)\ \hspace{15cm}\]
\[ B = 5\ +\ 4i\ = (5, 4)\ \hspace{13cm}\]
\[ C = 3\ +\ 6i\ = (3, 6)\ \hspace{13cm}\]
\[ D = 1\ +\ 4i\ = (1, 4)\ \hspace{13cm}\]
\[AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (3\ -\ 5)^2 +\ (2\ -\ 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((-2)^2 + (-2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[AB = \sqrt {8}\ \hspace{8cm}\]
\[BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (5\ -\ 3)^2 +\ (4\ -\ 6)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((2)^2 + (-2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[BC = \sqrt{8}\ \hspace{8cm}\]
\[CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (3\ -\ 1)^2 +\ (6\ -\ 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((2)^2 + (2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[CD = \sqrt{8}\ \hspace{8cm}\]
\[DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (1\ -\ 3)^2 +\ (4\ -\ 2)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((-2)^2 + (2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[DA = \sqrt{8}\ \hspace{8cm}\]
\[∴\ AB = BC = CD = DA\]
\[AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (3\ -\ 3)^2 + (2\ -\ 6)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((0)^2 + (-4)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (0 + 16 )}\ \hspace{8cm}\]
\[AC = \sqrt{16}\ =\ 4\ \hspace{8cm}\]
\[BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (5\ -\ 1)^2 + (4\ – 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((4)^2 + (0)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (0 + 16 )}\ \hspace{8cm}\]
\[BD = \sqrt{16}\ =\ 4\ \hspace{8cm}\]
\[∴\ AC = BD\]
\[∴\ The\ given\ complex\ numbers\ form\ a\ square\]
\[10.\ Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^4\ (cos⁡\ 4θ + i sin⁡\ 4θ)^2} {(cos⁡\ 2θ + i sin⁡\ 2θ)^5\ (cos⁡\ 5θ + i sin⁡\ 5θ)^3}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ \frac{(cos⁡\ 3θ + i sin⁡\ 3θ)^4\ (cos⁡\ 4θ + i sin⁡\ 4θ)^2} {(cos⁡\ 2θ + i sin⁡\ 2θ)^5\ (cos⁡\ 5θ + i sin⁡\ 5θ)^3}\ \hspace{18cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^{3 \times 4}\ (cos⁡\ θ + i sin⁡\ θ)^{4 \times 2}} {(cos⁡\ θ + i sin⁡\ θ)^{2 \times 5}\ (cos⁡\ θ + i sin⁡\ θ)^{5 \times 3 }}\ \hspace{8cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^{12}\ (cos⁡\ θ + i sin⁡\ θ)^{8}} {(cos⁡\ θ + i sin⁡\ θ)^{10}\ (cos⁡\ θ + i sin⁡\ θ)^ {15}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{12 + 8 – 10 – 15 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{-5}\ \hspace{10cm}\]
\[= cos\ 5θ – i sin⁡\ 5θ\ \hspace{10cm}\]