# ALGEBRA OF COMPLEX NUMBERS (Text)

$\color {royalblue} {Introduction}:\ \hspace{20cm}$
$In\ earlier\ classes,\ we\ have\ studied\ linear\ equations\ in\ one\ and\ two\ variables\ and\ quadratic$
$equations\ in\ one\ variable.\ We\ have\ seen\ that\ the\ equation\ x^2 + 1 = 0\ has\ no\ real solution\ as\ x^2 + 1 = 0$
$gives\ x^2 = – 1\ and square\ of\ every\ real\ number\ is\ non-negative.\ So,\ we\ need\ to\ extend\ the\ real\ number$
$system\ to\ a\ larger\ system\ so\ that\ we\ can\ find\ the\ solution\ of\ the\ equation\ x^2 = – 1.$
$Let\ us\ denote\ \sqrt{-1}\ by\ the\ symbol\ ‘i ‘.\ Then,\ we\ have\ i^2 = -1.\ This\ means\ that$
$‘i ‘\ is\ a\ solution\ of\ the\ equation\ x^2 + 1 = 0.$
$\color {purple}{Example\ 1:}\ \color {red}{Find\ the\ value\ of}\ i^2\ +\ i^3\ +\ i^4\ \hspace{15cm}$
$\color {blue}{Solution:}\ i^2\ +\ i^3\ +\ i^4\ \hspace{19cm}$
$=\ -1\ +\ i^2\ i\ +\ (i^2)^2\ \hspace{15cm}$
$=\ -1\ +\ (-1)\ i\ +\ (-1)^2\ \hspace{15cm}$
$=\ -1\ -\ i\ +\ 1\ \hspace{15cm}$
$=\ -i\ \hspace{15cm}$
$\color {royalblue} {Definition\ of\ complex\ number}:\ \hspace{20cm}$
$A\ number\ which\ is\ of\ the\ form\ a + ib\ where\ a, b ∈ R\ and\ i^2 = -1\ is\ called\ a\ complex$
$number\ and\ it\ is\ denoted\ by\ Z.$
$If\ Z = a + ib ,\ then\ a\ is\ called\ the\ real\ part\ of\ Z\ and\ b\ is\ called\ the\ imaginary\ part\ of\ Z.$
$Re ( Z ) = a\ and\ Im ( Z ) = b$
$Ex:\ If\ Z = 3 + 4i\ then\ Re ( Z ) = 3\ and\ Im ( Z ) = 4$
$\color {royalblue} {Conjugate\ of\ a\ complex\ number}:\ \hspace{20cm}$
$If\ Z = a + ib\ then\ the\ conjugate\ of\ Z\ is\ denoted\ by\ \bar z\ and\ is\ defined\ as\ \bar z\ = a – ib$
$\color {royalblue} {Algebra\ of\ Complex\ numbers}:\ \hspace{20cm}$
$\color {brown} {(i)\ Addition\ of\ two\ Complex\ numbers}:\ \hspace{20cm}$
$Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then$
$Z_1 + Z_2 = a + ib\ +\ c + id$
$=\ a + c + i(b + d )$
$\color {purple}{Example\ 2:}\ If\ Z_1 = 1 + i,\ Z_2 = 3 + 2i,\ \color {red} {find\ Z_1 + Z_2}\ \hspace{18cm}$
$\color {blue}{Solution:}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i\ \hspace{18cm}$
$Z_1 + Z_2 = 1 + i + 3 + 2i\ \hspace{10cm}$
$= 1 + 3 + i (1 +2)\ \hspace{10cm}$
$= 4 + 3i\ \hspace{10cm}$
$\color {purple} {Example\ 3:}\ If\ Z_1 = 1 + 3i,\ Z_2 = 6 – 5i,\ \color {red} {find\ Z_1 + Z_2}\ \hspace{18cm}$
$\color {blue}{Solution:}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i\ \hspace{18cm}$
$Z_1 + Z_2 = 1 + 3i + 6 – 5i\ \hspace{10cm}$
$= 1 + 6 + i (3 – 5)\ \hspace{10cm}$
$= 7 – 2i\ \hspace{10cm}$
$\color {purple} {Example\ 4:}\ If\ Z_1 = 1 + 4i,\ Z_2 = -3 + 6i,\ \color {red} {find\ 3Z_1 + 2Z_2}\ \hspace{18cm}$
$\color {blue}{Solution:}\ Z_1 = 1 + 4i,\ Z_2 = -3 + 6i\ \hspace{18cm}$
$3Z_1 + 2Z_2 = 3(1 + 4i) + 2( -3 + 6i)\ \hspace{10cm}$
$= 3 + 12i – 6 + 12i\ \hspace{10cm}$
$= 3 – 6 + i (12 + 12)\ \hspace{10cm}$
$= -3 + 24i\ \hspace{10cm}$
$\color {brown} {(ii)\ Difference\ of\ two\ Complex\ numbers}:\ \hspace{20cm}$
$Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then$
$Z_1 – Z_2 = a + ib\ -\ (c + id)$
$= a + ib – c – id$
$=\ a – c + i(b – d )$
$\color {purple} {Example\ 5:}\ If\ Z_1 = 1 + i,\ Z_2 = 3 + 2i,\ \color {red} {find\ Z_1 – Z_2}\ \hspace{18cm}$
$\color {blue}{Solution:}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i\ \hspace{18cm}$
$Z_1 – Z_2 = 1 + i – (3 + 2i)\ \hspace{10cm}$
$Z_1 – Z_2 = 1 + i – 3 – 2i\ \hspace{10cm}$
$Z_1 – Z_2 = 1 – 3 + i – 2i\ \hspace{10cm}$
$Z_1 – Z_2 = – 2 – i\ \hspace{10cm}$
$\color {purple} {Example\ 6:}\ If\ Z_1 = (3 , -1),\ Z_2 = (4 , 2),\ \color {red} {find\ 4Z_1 – Z_2}\ \hspace{18cm}$
$\color {blue}{Solution:}\ Z_1 = 3 – i,\ Z_2 = 4 + 2i\ \hspace{18cm}$
$4Z_1 – Z_2 = 4(3 – i ) – (4 + 2i)\ \hspace{10cm}$
$4Z_1 – Z_2 = 12 – 4i – 4 – 2i\ \hspace{10cm}$
$4Z_1 – Z_2 = 12 – 4 – 4i – 2i\ \hspace{10cm}$
$4Z_1 – Z_2 = 8 – 6i\ \hspace{10cm}$
$\color {brown} {(ii)\ Multiplication\ of\ two\ Complex\ numbers}:\ \hspace{20cm}$
$Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then$
$Z_1 Z_2 = (a + ib) (c + id)$
$= ac + iad + ibc + i^2bd$
$= ac + i(ad + bc) – bd$
$= (ac – bd) + i(ad + bc)$
$\color {purple} {Example\ 7:}\ If\ Z_1 = 2 + 2i,\ Z_2 = 3 + i,\ \color {red} {find\ the\ value\ of\ z_1z_2}\ \hspace{18cm}$
$\color {blue}{Solution:}\ Z_1 = 2 + 2i,\ Z_2 = 3 + i\ \hspace{18cm}$
$Z_1 Z_2 = (2 + 2i) (3 + i)\ \hspace{10cm}$
$Z_1 Z_2 = 6 + 2i + 6i + 2 (-1)\ \hspace{5cm}\ \because i^2\ =\ -1$
$= 6 + 8i – 2\ \hspace{10cm}$
$= 4 + 8i\ \hspace{10cm}$
$\color {brown} {(ii)\ Division\ of\ two\ Complex\ numbers}:\ \hspace{20cm}$
$Let\ Z_1= a + ib,\ Z_2 = c + id\ be\ any\ two\ complex\ numbers.\ Then$
$\frac{z_1}{z_2}\ =\frac{a+ib}{c+id}\ ×\ \frac{c-id}{c-id}$
$= \frac{ac- iad + ibc – i^2bd}{c^2 – i^2d^2}$
$= \frac{ac + i (bc – ad) + bd}{c^2 – i^2d^2}$
$= \frac{ac + bd + i(bc – ad)}{c^2 + d^2}$
$= \frac{ac + bd} {c^2 + d^2}\ +\ i\ \frac{bc – ad}{c^2 + d^2}$
$\color {purple} {Example\ 8:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{2+ 3i}{4+ 5i}\ \hspace{18cm}$
$\color {blue}{Solution:}\ z = \frac{2+ 3i}{4+ 5i}\ ×\ \frac{4- 5i}{4- 5i}\ \hspace{18cm}$
$= \frac{8 – 10i + 12i + 15}{(4)^2 + (5)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{8 – 10i + 12i + 15}{16+ 25}\ \hspace{15cm}$
$= \frac{23 + 2i}{41}\ \hspace{15cm}$
$= \frac{23}{41}\ +\ i\ \frac{2}{41}\ \hspace{15cm}$
$Re(Z)\ =\ \frac{23}{41};\ Im(Z)\ =\ \frac{23}{41}\ \hspace{15cm}$
$\color {purple} {Example\ 9:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1}{5- 2i} \hspace{18cm}$
$\color {blue}{Solution:}\ z = \frac{1}{5- 2i}\ ×\ \frac{5 + 2i}{5 + 2i}\ \hspace{18cm}$
$= \frac{5 + 2i}{(5)^2 + (2)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{5 + 2i}{25+ 4}\ \hspace{15cm}$
$= \frac{5 + 2i}{29}\ \hspace{15cm}$
$= \frac{5}{29}\ +\ i\ \frac{2}{29}\ \hspace{15cm}$
$Re(Z)\ =\ \frac{5}{29};\ Im(Z)\ =\ \frac{2}{29}\ \hspace{15cm}$
$\color {purple} {Example\ 10:}\ \color {red} {Find\ the\ conjugate\ of}\ \frac{5}{3+ 2i} \hspace{18cm}$
$\color {blue}{Solution:}\ z = \frac{5}{3+ 2i}\ ×\ \frac{3 – 2i}{3 – 2i}\ \hspace{18cm}$
$= \frac{15 – 10i}{(3)^2 + (2)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{15 – 10i}{9+ 4}\ \hspace{15cm}$
$= \frac{15 – 10i}{13}\ \hspace{15cm}$
$Z = \frac{15}{13}\ -\ i\ \frac{10}{13}\ \hspace{15cm}$
$conjugate\ of\ Z\ is\ \bar z\ = \frac{15}{13}\ +\ i\ \frac{10}{13}\ \hspace{10cm}$
$\color {purple} {Example\ 11:}\ \color {red} {Express}\ \frac{2}{4+ 3i}\ +\ \frac{i}{3- 4i}\ \color {red} {in\ a+ib\ form}\ \hspace{18cm}$
$\color {blue}{Solution:} \hspace{20cm}$
$Let\ z = \frac{2}{4+ 3i}\ +\ \frac{i}{3- 4i}\ \hspace{18cm}$
$= \frac{2}{4+ 3i}\ ×\ \frac{4 – 3i}{4 – 3i}\ +\ \frac{i}{3- 4i}\ ×\ \frac{3 + 4i}{3 + 4i}\ \hspace{12cm}$
$= \frac{8 – 6i}{(4)^2 + (3)^2}\ +\ \frac{3i + 4i^2}{(3)^2 + (4)^2}\ \hspace{2cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{8 – 6i}{16 + 9}\ +\ \frac{3i + 4(-1)}{9 + 16}\ \hspace{2cm}\ \because [i^2\ =\ -1]$
$= \frac{8 – 6i}{25}\ +\ \frac{3i – 4}{25}\ \hspace{10cm}$
$= \frac{8 – 6i + 3i – 4}{25}\ \hspace{10cm}$
$= \frac{4 – 3i}{25}\ \hspace{10cm}$
$Z = \frac{4}{25}\ -\ \frac{3i}{25}\ a+ib\ form\ \hspace{15cm}$
$\color {purple} {Example\ 12:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{(1+ i)(2 + i)}{1+ 4i} \hspace{18cm}$
$\color {blue}{Solution:} \hspace{20cm}$
$Let\ z = \frac{(1+ i)(2 + i)}{1+ 4i}\ \hspace{18cm}$
$= \frac{2+ i + 2i + i^2}{1+ 4i}\ \hspace{15cm}$
$= \frac{2+ 3i – 1}{1+ 4i}\ \hspace{15cm}$
$= \frac{1+ 3i}{1+ 4i}\ \hspace{15cm}$
$= \frac{1+3i}{1+ 4i}\ ×\ \frac{1 – 4i}{1 – 4i}\ \hspace{15cm}$
$= \frac{1 – 4i + 3i – 12i^2}{(1)^2 + (4)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{1 – i + 12 }{1+ 16}\ \hspace{4cm}\ \because [i^2\ =\ -1]$
$= \frac{13 – i}{17}\ \hspace{10cm}$
$Z = \frac{13}{17}\ -\ i\ \frac{1}{17}\ \hspace{15cm}$
$Re(Z)\ =\ \frac{13}{17};\ Im(Z)\ =\ \frac{-1}{17}\ \hspace{15cm}$
$\color {royalblue} {Polar\ form\ of\ a\ Complex\ number}:\ \hspace{20cm}$
$Z = r cosθ + i r sin θ = r ( cosθ + i sin θ )$
$\color {royalblue} {Modulus\ and\ Amplitude\ (or)\ Argument\ of\ a\ Complex\ number}:\ \hspace{20cm}$
$If\ Z = a + ib\ then\ Modulus is |z| = \sqrt{a^2 + b^2}\ and\ Amplitude\ is\ θ = tan^{-1} (\frac{b}{a})$
$\color {purple} {Example\ 13:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ 1 + \sqrt{3}i\ \hspace{18cm}$
$\color {blue}{Solution:} \hspace{20cm}$
$Let\ z = 1 + \sqrt{3}i\ = a\ + ib\ \hspace{15cm}$
$a = 1,\ b\ = \sqrt{3}\ \hspace{15cm}$
$\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}$
$|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}$
$= \sqrt{(1)^2 + (\sqrt{3})^2}\ \hspace{12cm}$
$= \sqrt{1 + 3}\ =\ \sqrt{4}\ \hspace{12cm}$
$|z| = 2\ \hspace{15cm}$
$\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}$
$θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} \frac{\sqrt{3}}{1}\ =\ tan^{-1} \sqrt{3}$
$θ = 60^0\ \hspace{15cm}$
$\color {purple} {Example\ 14:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ \frac{1}{2} + i\ \frac{\sqrt{3}}{2}\ \hspace{18cm}$
$\color {bllue}{Solution:} \hspace{20cm}$
$a =\ \frac{1}{2} ,\ b\ = \frac{\sqrt{3}}{2}\ \hspace{15cm}$
$Let\ z = \frac{1}{2} + i\ \frac{\sqrt{3}}{2}\ = a\ + ib\ \hspace{15cm}$
$\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}$
$|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}$
$= \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}\ \hspace{12cm}$
$= \sqrt{\frac{1}{4} +\ \frac{3}{4}}\ \hspace{12cm}$
$= \sqrt{\frac{1\ +\ 3}{4}}\ \hspace{12cm}$
$= \sqrt{\frac{4}{4}}\ \hspace{12cm}$
$= \sqrt{1}\ \hspace{12cm}$
$|z| = 1\ \hspace{15cm}$
$\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}$
$θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} (\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}})\ =\ tan^{-1} (\sqrt{3})$
$θ = 60^0\ \hspace{15cm}$
$\color {purple} {Example\ 15:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ \frac{1\ -\ i}{1\ +\ i}\ \hspace{18cm}$
$\color {blue}{Solution:}\ z = \frac{1\ -\ i}{1\ +\ i}\ ×\ \frac{1\ -\ i}{1\ -\ i}\ \hspace{18cm}$
$= \frac{1 – i – i – 1}{(1)^2 + (1)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{1 – i – i – 1}{1\ +\ 1}\ \hspace{15cm}$
$= \frac{-\ 2i}{2}\ \hspace{15cm}$
$= \frac{0}{2}\ +\ i\ \frac{-2}{2}\ \hspace{15cm}$
$z\ =\ 0\ -\ i\ \hspace{15cm}$
$a =\ 0,\ b\ = – 1\ \hspace{15cm}$
$\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}$
$|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}$
$= \sqrt{(\sqrt{0})^2 + (-1)^2}\ \hspace{12cm}$
$= \sqrt{0 + 1}\ =\ \sqrt{1}\ =\ 1\ \hspace{12cm}$
$|z| = 1\ \hspace{15cm}$
$\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}$
$θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1}( \frac{- 1}{0})$
$θ =\ -\ 90^0\ \hspace{15cm}$

$\color {purple} {Example\ 16:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ \frac{5 – i}{2- 3i}\ \hspace{18cm}$
$\color {blue}{Solution:} \hspace{22cm}$
$Let\ z = \frac{5 – i}{2- 3i}\ \hspace{18cm}$
$= \frac{5 – i}{2- 3i}\ ×\ \frac{2 + 3i}{2 + 3i}\ \hspace{15cm}$
$= \frac{10 + 15i -2i – 3i^2}{(3)^2 + (2)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]$
$= \frac{10 + 13i + 3 }{9+ 4}\ \hspace{4cm}\ \because [i^2\ =\ -1]$
$= \frac{13 + 13i}{13}\ \hspace{10cm}$
$Z = \frac{13}{13}\ +\ \frac{13i}{13}\ \hspace{15cm}$
$z = 1 + \ i\ = a\ + ib\ \hspace{14cm}$
$a = 1,\ b\ = \ 1\ \hspace{15cm}$
$\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}$
$|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}$
$= \sqrt{(1)^2 + {1}^2}\ \hspace{12cm}$
$= \sqrt{1 + 1}\ =\ \sqrt{2}\ \hspace{12cm}$
$|z| = \sqrt{2}\ \hspace{15cm}$
$\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}$
$θ = tan^{-1} (\frac{b}{a})\ =\ tan^{1} \frac{1}{1}\ =\ tan^{-1} {1}$
$θ = 45^0\ \hspace{15cm}$
$\color {royalblue} {Argand\ diagram}:\ \hspace{20cm}$
$Every\ complex\ number\ a + ib\ can\ be\ considered\ as\ an\ ordered\ pair\ ( a, b)\ of\ real numbers,$
$we\ can\ represent\ such\ number\ by\ a\ point\ in\ xy-plane\ called\ the\ complex\ plane$
$and\ such\ representation\ is\ also\ known\ as\ argand\ diagram.$
$\color {royalblue} {Distance\ between\ two\ Complex\ numbers}:\ \hspace{20cm}$
$If\ Z_1\ = a + ib,\ Z_2\ = c + id\ be\ any\ two\ complex\ numbers,\ then$
$Z_1 Z_2 =\ \sqrt{ (a- c)^2 + (b- d)^2 )}$
$\color {purple} {Example\ 17\ :}\ \color {red} {Find\ the\ distance\ between\ the\ complex\ numbers}\ 3-2i,\ and\ 1+4i\ \hspace{18cm}$
$\color {blue}{Solution:} \hspace{22cm}$
$Let\ z_1 = 3-2i\ = (3, -2)\ \hspace{15cm}$
$z_2 = 1+4i\ = (1, 4)\ \hspace{15cm}$
$a = 3,\ b= -2,\ c = 1,\ d = 4\ \hspace{15cm}$
$Distance\ between\ z_1\ and\ z_2\ is\ =\ \sqrt{ (a- c)^2 + (b- d)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((3- 1)^2 + (-2- 4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ ((2)^2 + (-6)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 36 )}\ \hspace{8cm}$
$=\ \sqrt{ 40}\ \hspace{8cm}$
$\color {royalblue} {a)\ Condition\ for\ collinear\ points}:\ \hspace{20cm}$
$If\ the\ three\ complex\ numbers\ say,\ A(x_1 +iy_1),\ B(x_2 +iy_2),\ and\ C(x_3 +iy_3),\ are\ collinear$
$if\ area\ of\ ∆ABC = 0$
$\frac{1}{2}[\ x_1(y_2 – y_3)\ +\ x_2(y_3 – y_1)\ +\ x_3(y_1 – y_2)] = 0$
$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix}\ = 0\ \hspace{15cm}$
$\color {purple} {Example\ 18\ :}\ \color {red} {Prove\ that\ the\ points\ in\ the\ argand\ plane\ represented\ by\ the\ complex\ numbers}\ \hspace{10cm}$$1+4i,\ 2+7i,\ and\ 3+10i\ \color {red} {are\ collinear}\ \hspace{5cm}$
$\color {blue}{Solution:} \hspace{22cm}$
$Let\ A = 1+4i\ = (1, 4)\ \hspace{15cm}$
$B = 2+7i\ = (2, 7)\ \hspace{15cm}$
$C = 3+10i\ = (3, 10)\ \hspace{15cm}$
$\color {brown} \ {Condition\ for\ collinear}\ \hspace{15cm}$
$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix}\ = 0\ \hspace{15cm}$
$L.H.S=\begin{vmatrix} 1 & 4 & 1 \\ 2 & 7 & 1 \\ 3 & 10 & 1 \\ \end{vmatrix}\ \hspace{15cm}$
$=1\begin{vmatrix} 7 & 1 \\ 10 & 1 \\ \end{vmatrix}\ -\ 4\begin{vmatrix} 2 & 1 \\ 3 & 1 \\ \end{vmatrix}\ +\ 1\begin{vmatrix} 2 & 7\\ 3 & 10 \\ \end{vmatrix}\ \hspace{10cm}$
$=1(7\ -\ 10)\ – 4 (2\ -\ 3) + 1(20\ -\ 21)\ \hspace{9cm}$
$=1(-3)\ – 4 (-1) + 1(-1)\ \hspace{13cm}$
$= -3\ + 4 – 1\ \hspace{13cm}$
$= -4\ + 4 \ \hspace{13cm}$
$= 0\ \hspace{13cm}$
$∴\ The\ given\ complex\ numbers\ are\ collinear$
$\color {royalblue} {b)\ Condition\ for\ Square}:\ \hspace{20cm}$
$If\ A,\ B,\ C\ and\ D\ are\ any\ four\ complex\ numbers\ representing\ the\ vertices\ of\ a\ square$$then\ the\ required\ condition\ is$
$( i )\ AB = BC = CD = DA$
$i.e\ all\ sides\ are\ equal$
$( ii )\ AC = BD$
$i.e\ the\ diagonals\ are\ equal$
$\color {purple} {Example\ 19\ :}\ \color {red} {Show\ that\ the\ complex\ numbers}\ 2+i,\ 4+3i,\ 2+5i\ and\ 3i\ \color {red}{form\ a\ square}\ \hspace{10cm}$
$\color {blue}{Solution:} \hspace{22cm}$
$Let\ A = 2+i\ = (2, 1)\ \hspace{15cm}$
$B = 4+3i\ = (4, 3)\ \hspace{13cm}$
$C = 2+5i\ = (2, 5)\ \hspace{13cm}$
$D = 3i\ = (0, 3)\ \hspace{13cm}$
$AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (2- 4)^2 + (1- 3)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-2)^2 + (-2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 4 )}\ \hspace{8cm}$
$\boxed{AB = \sqrt {8}}\ \hspace{8cm}$
$BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (4- 2)^2 + (3- 5)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((2)^2 + (-2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 4 )}\ \hspace{8cm}$
$\boxed{BC = \sqrt{8}}\ \hspace{8cm}$
$CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (2- 0)^2 + (5- 3)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((2)^2 + (2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 4 )}\ \hspace{8cm}$
$\boxed{CD = \sqrt{8}}\ \hspace{8cm}$
$DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (0- 2)^2 + (3- 1)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-2)^2 + (2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 4 )}\ \hspace{8cm}$
$\boxed{DA = \sqrt{8}}\ \hspace{8cm}$
$∴\ AB = BC = CD = DA$
$AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (2- 2)^2 + (1- 5)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((0)^2 + (-4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (0 + 16 )}\ \hspace{8cm}$
$\boxed{AC = \sqrt{16}\ =\ 4}\ \hspace{8cm}$
$BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (4- 0)^2 + (3- 3)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((4)^2 + (0)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (0 + 16 )}\ \hspace{8cm}$
$\boxed{BD = \sqrt{16}\ =\ 4}\ \hspace{8cm}$
$∴\ AC = BD$
$∴\ The\ given\ complex\ numbers\ form\ a\ square$
$\color {royalblue} {c)\ Condition\ for\ rhombus}:\ \hspace{20cm}$
$If\ A,\ B,\ C\ and\ D\ are\ any\ four\ complex\ numbers\ representing\ the\ vertices\ of\ a\ rhombus$$then\ the\ required\ condition\ is$
$( i )\ AB = BC = CD = DA$
$i.e\ all\ sides\ are\ equal$
$( ii )\ AC \neq BD$
$i.e\ the\ diagonals\ are\ not\ equal$
$\color {purple} {Example\ 20}\ :\ \color {red} {Show\ that\ the\ complex\ numbers}\ 3-2i,\ 7+6i,\ -1+2i\ and\ -5-6i\ \color {red}{form\ a\ rhombus}\ \hspace{10cm}$
$\color {blue}{Solution:} \hspace{22cm}$
$Let\ A = 3-2i\ = (3, 2)\ \hspace{15cm}$
$B = 7+6i\ = (7, 6)\ \hspace{13cm}$
$C = -1+2i\ = (-1, 2)\ \hspace{13cm}$
$D = -5 – 6i\ = (-5, -6)\ \hspace{13cm}$
$AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (3- 7)^2 + (-2- 6)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-4)^2 + (-8)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (16 + 64 )}\ \hspace{8cm}$
$\boxed{AB = \sqrt{80}}\ \hspace{8cm}$
$BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (7+ 1)^2 + (6- 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((8)^2 + (4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (64 + 16 )}\ \hspace{8cm}$
$\boxed{BC = \sqrt{80}}\ \hspace{8cm}$
$CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-1+ 5)^2 + (2+ 6)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((4)^2 + (8)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (16 + 64 )}\ \hspace{8cm}$
$\boxed{CD = \sqrt{80}}\ \hspace{8cm}$
$DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-5- 3)^2 + (-6 + 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-8)^2 + (-4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (64 + 16 )}\ \hspace{8cm}$
$\boxed{DA = \sqrt{80}}\ \hspace{8cm}$
$∴\ AB = BC = CD = DA$
$AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (3 + 1)^2 + (-2 – 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((4)^2 + (-4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (16 + 16 )}\ \hspace{8cm}$
$\boxed{AC = \sqrt{32}}\ \hspace{8cm}$
$BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (7 + 5)^2 + (6 + 6)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((12)^2 + (12)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (144 + 144 )}\ \hspace{8cm}$
$\boxed{BD = \sqrt{288}}\ \hspace{8cm}$
$∴\ AC \neq BD$
$∴\ The\ given\ complex\ numbers\ form\ a\ rhombus$
$\color {royalblue} {c)\ Condition\ for\ rectangle}:\ \hspace{20cm}$
$If\ A,\ B,\ C\ and\ D\ are\ any\ four\ complex\ numbers\ representing\ the\ vertices\ of\ a\ rectangle$$then\ the\ required\ condition\ is$
$( i )\ AB = CD\ and\ BC= DA$
$i.e\ opposite\ sides\ are\ equal$
$( ii )\ AC = BD$
$i.e\ the\ diagonals\ are\ equal$
$\color {purple} {Example\ 21\ :}\ \color {red}{Show\ that\ the\ complex\ numbers}\ 1+2i,\ -2+5i,\ 7i\ and\ 3+4i\ \color {red} {form\ a\ rectangle}\ \hspace{10cm}$
$\color {blue}{Solution:} \hspace{22cm}$
$Let\ A = 1+2i\ = (1, 2)\ \hspace{15cm}$
$B = -2+5i\ = (-2, 5)\ \hspace{13cm}$
$C = 7i\ = (0, 7)\ \hspace{13cm}$
$D = 3 + 4i\ = (3, 4)\ \hspace{13cm}$
$AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (1 + 2)^2 + (2- 5)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((3)^2 + (-3)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (9 + 9)}\ \hspace{8cm}$
$\boxed{AB = \sqrt{18}}\ \hspace{8cm}$
$BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-2 – 0)^2 + (5- 7)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-2)^2 + (-2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 4)}\ \hspace{8cm}$
$\boxed{BC = \sqrt{8}}\ \hspace{8cm}$
$CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (0 – 3)^2 + (7- 4)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-3)^2 + (3)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (9 + 9)}\ \hspace{8cm}$
$\boxed{CD = \sqrt{18}}\ \hspace{8cm}$
$DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (3 – 1)^2 + (4- 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((2)^2 + (2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 4)}\ \hspace{8cm}$
$\boxed{DA = \sqrt{8}}\ \hspace{8cm}$
$∴\ AB = CD\ and\ BC= DA$
$AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (1 – 0)^2 + (2- 7)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((1)^2 + (-5)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (1 + 25)}\ \hspace{8cm}$
$\boxed{AC = \sqrt{26}}\ \hspace{8cm}$
$BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-2 – 3)^2 + (5- 4)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-5)^2 + (1)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (25 + 1)}\ \hspace{8cm}$
$\boxed{BD = \sqrt{26}}\ \hspace{8cm}$
$∴\ AC = BD$
$∴\ The\ given\ complex\ numbers\ form\ a\ rectangle$
$\color {royalblue} {c)\ Condition\ for\ parallelogram}:\ \hspace{20cm}$
$If\ A,\ B,\ C\ and\ D\ are\ any\ four\ complex\ numbers\ representing\ the\ vertices\ of\ a\ parallelogram$$then\ the\ required\ condition\ is$
$( i )\ AB = CD\ and\ BC= DA$
$i.e\ opposite\ sides\ are\ equal$
$( ii )\ AC \neq BD$
$i.e\ the\ diagonals\ are\ not\ equal$
$\color {purple} {Example\ 22\ :}\ \color {red}{Show\ that\ the\ complex\ numbers}\ 1-2i,\ -1+4i,\ 5 +8i\ and\ 7+2i\ \color {red} {form\ a\ parallelogram}\ \hspace{10cm}$
$\color {blue}{Solution:} \hspace{20cm}$
$Let\ A = 1-2i\ = (1, -2)\ \hspace{15cm}$
$B = -1+4i\ = (-1, 4)\ \hspace{13cm}$
$C = 5 + 8i\ = (5, 8)\ \hspace{13cm}$
$D = 7 + 2i\ = (7, 2)\ \hspace{13cm}$
$AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (1 + 1)^2 + (-2- 4)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((2)^2 + (-6)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 36)}\ \hspace{8cm}$
$AB = \sqrt{40}\ \hspace{8cm}$
$BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-1 – 5)^2 + (4- 8)^2 }\ \hspace{8cm}$
$=\ \sqrt{ ((-6)^2 + (-4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (36 + 16)}\ \hspace{8cm}$
$BC = \sqrt{52}\ \hspace{8cm}$
$CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (5 – 7)^2 + (8- 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-2)^2 + (6)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (4 + 36)}\ \hspace{8cm}$
$CD = \sqrt{40}\ \hspace{8cm}$
$DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (7 – 1)^2 + (2+ 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (6)^2 + (4)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (36 + 16)}\ \hspace{8cm}$
$DA = \sqrt{52}\ \hspace{8cm}$
$∴\ AB = CD\ and\ BC= DA$
$AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (1 – 5)^2 + (-2 – 8)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-4)^2 + (-10)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (16 + 100)}\ \hspace{8cm}$
$AC = \sqrt{116}\ \hspace{8cm}$
$BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-1 – 7)^2 + (4 – 2)^2 }\ \hspace{8cm}$
$=\ \sqrt{ (-8)^2 + (2)^2 )}\ \hspace{8cm}$
$=\ \sqrt{ (64 + 4)}\ \hspace{8cm}$
$DA = \sqrt{68}\ \hspace{8cm}$
$\ AC \neq BD$
$∴\ The\ given\ complex\ numbers\ form\ a\ parallelogram$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$\color {purple}{1.}\ \color {red}{Find\ the\ value\ of}\ i^3\ +\ i^5\ \hspace{18cm}$
$\color {purple} {2.}\ \color {red} {Find\ the\ value\ of}\ (2i\ +\ i)(i\ +\ 3i)\ \hspace{15cm}$
$\color {purple} {3.}\ \color {red} {If}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i,\ \color {red} {find\ 3Z_1 + 4Z_2}\ \hspace{15cm}$
$\color {purple} {4.} \color {red} {If}\ Z_1 = 2 + 3i,\ Z_2 = 4 – 5i,\ \color {red} {find}\ Z_1 – Z_2\ \hspace{15cm}$
$\color {purple} {5.} \ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1}{2\ +\ 3i} \hspace{18cm}$
$\color {purple} {6.}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ 1 + \ i\ \hspace{18cm}$
$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {7.}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1\ -\ i}{1\ +\ i}\ \hspace{18cm}$
$\color {purple} {8.} \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{2\ +\ 5i}{2\ +\ 3i}\ \hspace{18cm}$
$\color {purple} {9.}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{2\ +\ 5i}{4\ -\ 3i}\ \hspace{18cm}$
$\color {purple} {10:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ \sqrt{3} – \ i\ \hspace{18cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {11:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{(1+ i)(2 – i)}{1+ 3i} \hspace{18cm}$
$\color {purple} {12\ :}\ \color {red} {Show\ that\ the\ points}\ 3 + 2i,\ 5 + 4i,\ 3 + 6i\ and\ 1 + 4i\ \color {red}{in\ an\ Arand\ diagram\ form\ a\ square}\ \hspace{10cm}$
$\color {purple} {13.}\ \color {red} {Prove\ that\ the\ complex\ numbers}\ 3\ +\ 4\ i,\ 9\ +\ 8\ i,\ 5\ +\ 2\ i\ and\ -\ 1\ -\ 2\ i\ \color {red} {form\ a\ rhombus}\ \hspace{10cm}$$\color {red} {in\ the\ argand\ diagram}\ \hspace{10cm}$