ALGEBRA OF COMPLEX NUMBERS (Excersice problems with solutions)

\[\LARGE{\color {purple} {PART- A}}\]
\[\color {purple}{1.}\ \color {red}{Find\ the\ value\ of}\ i^3\ +\ i^5\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ i^3\ +\ i^5\ \hspace{19cm}\]
\[ =\ i^2\ i\ +\ (i^2)^2\ i\ \hspace{15cm}\]
\[ =\ (-1)\ i\ +\ (-1)^2\ i\ \hspace{15cm}\]
\[ =\ -\ i\ +\ 1\ i\ \hspace{15cm}\]
\[ =\ 0\ \hspace{15cm}\]
\[\color {purple} {2.}\ \color {red} {Find\ the\ value\ of}\ (2i\ +\ i)(i\ +\ 3i)\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ (2i\ +\ i)(i\ +\ 3i)\ \hspace{18cm}\]
\[ =\ (3\ i)\ (4\ i)\ \hspace{15cm}\]
\[ =\ 12\ i^2\ \hspace{15cm}\]
\[ =\ 12\ (-\ 1)\ \hspace{15cm}\]
\[ =\ -12\ \hspace{15cm}\]
\[\color {purple} {3.}\ \color {red} {If}\ Z_1 = 1 + i,\ Z_2 = 3 + 2i,\ \color {red} {find\ 3Z_1 + 4Z_2}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Z_1 = 1 +\ i,\ Z_2 =\ 3 +\ 2i\ \hspace{18cm}\]
\[3Z_1 + 4Z_2 = 3(1 + i) +\ 4( 3 +\ 2i)\ \hspace{10cm}\]
\[ = 3 + i +\ 12 + 8i\ \hspace{10cm}\]
\[= 3 +\ 12 +\ i (1 + 12)\ \hspace{10cm}\]
\[=\ 15\ +\ 13i\ \hspace{10cm}\]
\[\color {purple} {4.} \color {red} {If}\ Z_1 = 2 + 3i,\ Z_2 = 4 – 5i,\ \color {red} {find}\ Z_1 – Z_2\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Z_1 = 2 + 3i,\ Z_2 = 4 – 5i,\ \hspace{18cm}\]
\[Z_1 – Z_2 = 2 +\ 3i -\ (4 -\ 5i)\ \hspace{10cm}\]
\[Z_1 – Z_2 =\ 2 +\ 3i -\ 4 +\ 5i\ \hspace{10cm}\]
\[Z_1 – Z_2 = 2 – 4 +\ 3i +\ 5i\ \hspace{10cm}\]
\[Z_1 – Z_2 = – 2 +\ 8i\ \hspace{10cm}\]
\[\color {purple} {5.} \ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1}{2\ +\ 3i} \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{1}{2\ +\ 3i}\ ×\ \frac{2\ -\ 3i}{2\ -\ 3i}\ \hspace{18cm}\]
\[ = \frac{2\ -\ 3i}{(2)^2 + (3)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{2\ -\ 3i}{4\ +\ 9}\ \hspace{15cm}\]
\[ = \frac{2\ -\ 3i}{13}\ \hspace{15cm}\]
\[ = \frac{2}{13}\ +\ i\ \frac{-3}{13}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{2}{13};\ Im(Z)\ =\ \frac{-3}{13}\ \hspace{15cm}\]
\[\color {purple} {6.}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ 1 + \ i\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = 1 + \ i\ = a\ + ib\ \hspace{15cm}\]
\[a = 1,\ b\ =\ 1\ \hspace{15cm}\]
\[\color {brown} {To\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(1)^2 + (1)^2}\ \hspace{12cm}\]
\[ = \sqrt{1 + 1}\ =\ 2\ \hspace{12cm}\]
\[|z| = 2\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} \frac{1}{1}\ =\ tan^{-1} \ (1)\]
\[θ = 45^0\ \hspace{15cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[\color {purple} {7.}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{1\ -\ i}{1\ +\ i}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{1\ -\ i}{1\ +\ i}\ ×\ \frac{1\ -\ i}{1\ -\ i}\ \hspace{18cm}\]
\[ = \frac{1 – i – i – 1}{(1)^2 + (1)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{1 – i – i – 1}{1\ +\ 1}\ \hspace{15cm}\]
\[ = \frac{-\ 2i}{2}\ \hspace{15cm}\]
\[ = \frac{0}{2}\ +\ i\ \frac{-2}{2}\ \hspace{15cm}\]
\[Re(Z)\ =\ 0\ Im(Z)\ =\ -1\ \hspace{15cm}\]
\[\color {purple} {8.} \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{2\ +\ 5i}{2\ +\ 3i}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{2+ 5i}{2+ 3i}\ ×\ \frac{2\ – 3i}{2\ – 3i}\ \hspace{18cm}\]
\[ = \frac{4 – 6i + 10i + 15}{(2)^2 + (3)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{8 – 10i + 12i + 15}{4\ +\ 9}\ \hspace{15cm}\]
\[ = \frac{19\ +\ 4i}{13}\ \hspace{15cm}\]
\[ = \frac{19}{13}\ +\ i\ \frac{4}{13}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{19}{13};\ Im(Z)\ =\ \frac{4}{13}\ \hspace{15cm}\]
\[\color {purple} {9.}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{2\ +\ 5i}{4\ -\ 3i}\ \hspace{18cm}\]
\[\color {blue}{Solution:}\ z = \frac{2+ 5i}{4 – 3i}\ ×\ \frac{4\ + 3i}{4\ + 3i}\ \hspace{18cm}\]
\[ = \frac{8 + 6i + 20i – 15\ i^2}{(4)^2 + (3)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{8 + 6i + 20i + 15}{16\ +\ 9}\ \hspace{15cm}\]
\[ = \frac{23\ +\ 26i}{25}\ \hspace{15cm}\]
\[ = \frac{23}{25}\ +\ i\ \frac{26}{25}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{23}{25};\ Im(Z)\ =\ \frac{26}{25}\ \hspace{15cm}\]
\[\color {purple} {10:}\ \color {red} {Find\ the\ modulus\ and\ amplitude\ of}\ \sqrt{3} – \ i\ \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = \sqrt{3} – \ i\ = a\ + ib\ \hspace{15cm}\]
\[a = \sqrt{3},\ b\ = – 1\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(\sqrt{3})^2 + (-1)^2}\ \hspace{12cm}\]
\[ = \sqrt{3 + 1}\ =\ \sqrt{4}\ =\ 2\ \hspace{12cm}\]
\[|z| = 2\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1}( \frac{- 1}{\sqrt{3}})\]
\[θ =\ -\ 30^0\ \hspace{15cm}\]
\[\LARGE{\color {purple} {PART- C}}\]
\[\color {purple} {11:}\ \color {red} {Find\ the\ Real\ and\ Imaginary\ parts\ of}\ \frac{(1+ i)(2 – i)}{1+ 3i} \hspace{18cm}\]
\[\color {blue}{Solution:} \hspace{20cm}\]
\[Let\ z = \frac{(1+ i)(2 – i)}{1+ 3i}\ \hspace{18cm}\]
\[ = \frac{2 – i + 2i – i^2}{1+ 3i}\ \hspace{15cm}\]
\[ = \frac{2+ i – 1}{1+ 3i}\ \hspace{15cm}\]
\[ = \frac{1+ i}{1+ 3i}\ \hspace{15cm}\]
\[= \frac{1+i}{1+ 3i}\ ×\ \frac{1 – 3i}{1 – 3i}\ \hspace{15cm}\]
\[ = \frac{1 – 3i + i – 3i^2}{(1)^2 + (3)^2}\ \hspace{3cm}\ \because [(a +ib)(a – ib)\ =\ (a)^2 + (b)^2]\]
\[ = \frac{1 – 2i + 3 }{1+ 9}\ \hspace{4cm}\ \because [i^2\ =\ -1]\]
\[ = \frac{4 – 2i}{10}\ \hspace{10cm}\]
\[ = \frac{2 – i}{5}\ \hspace{10cm}\]
\[ Z = \frac{2}{5}\ -\ i\ \frac{1}{5}\ \hspace{15cm}\]
\[Re(Z)\ =\ \frac{2}{5};\ Im(Z)\ =\ \frac{-1}{5}\ \hspace{15cm}\]
\[\color {purple} {12\ :}\ \color {red} {Show\ that\ the\ points}\ 3 + 2i,\ 5 + 4i,\ 3 + 6i\ and\ 1 + 4i\ \color {red}{in\ an\ Arand\ diagram\ form\ a\ square}\ \hspace{10cm}\]
\[\color {blue}{Solution:} \hspace{22cm}\]
\[Let\ A = 3 + 2i\ = (3, 2)\ \hspace{15cm}\]
\[ B = 5 + 4i\ = (5, 4)\ \hspace{13cm}\]
\[ C = 3 + 6i\ = (3, 6)\ \hspace{13cm}\]
\[ D = 1 + 4i\ = (1, 4)\ \hspace{13cm}\]
\[AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (3 – 5)^2 + (2 – 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((-2)^2 + (-2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[\boxed{AB = \sqrt {8}}\ \hspace{8cm}\]
\[BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (5 – 3)^2 + (4 – 6)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((2)^2 + (-2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[\boxed{BC = \sqrt{8}}\ \hspace{8cm}\]
\[CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (3 – 1)^2 + (6 – 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((2)^2 + (2)^2 )}\ \hspace{8cm}\]
\[\boxed{CD = \sqrt{8}}\ \hspace{8cm}\]
\[DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (1 – 3)^2 + (4 – 2)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((-2)^2 + (2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 4 )}\ \hspace{8cm}\]
\[\boxed{DA = \sqrt{8}}\ \hspace{8cm}\]
\[∴\ AB = BC = CD = DA\]
\[AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (3 – 3)^2 + (2 – 6)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((0)^2 + (-4)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (0 + 16 )}\ \hspace{8cm}\]
\[\boxed{AC = \sqrt{16}\ =\ 4}\ \hspace{8cm}\]
\[BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (5 – 1)^2 + (4 – 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((4)^2 + (0)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (16 + 0)}\ \hspace{8cm}\]
\[\boxed{BD = \sqrt{16}\ =\ 4}\ \hspace{8cm}\]
\[∴\ AC = BD\]
\[∴\ The\ given\ complex\ numbers\ form\ a\ square\]
\[\color {purple} {13.}\ \color {red} {Prove\ that\ the\ complex\ numbers}\ 3\ +\ 4\ i,\ 9\ +\ 8\ i,\ 5\ +\ 2\ i\ and\ -\ 1\ -\ 2\ i\ \color {red} {form\ a\ rhombus}\ \hspace{10cm}\]\[\color {red} {in\ the\ argand\ diagram}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ Part\ -\ C\ of Question\ 2\ \hspace{20cm}\]

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