# ROOTS OF COMPLEX NUMBERS (Excercise Problems)

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {1.}\ \color {red} {If \ ω\ is\ the\ cube\ roots\ of unity,\ what\ is\ the\ value\ of}\ ω( ω\ +\ 1)\ \hspace{15cm}$
$\color {blue}{Solution:}\ Given\ ω\ is\ the\ cube\ roots\ of unity\ \hspace{20cm}$
$i.\ e\ 1\ +\ ω\ +\ ω ^2\ =\ 0\ ——\ (1)\ \hspace{10cm}$
$ω( ω\ +\ 1)\ =\ ω ^2\ +\ ω\ =\ -\ 1\ (using\ (1))\ \hspace{10cm}$
$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {2\ .}\ \color {red} {Solve}\ x^4 -\ 1\ =\ 0\ \hspace{18cm}$
$x^4\ =\ 1\ \hspace{10cm}$
$\ x\ =\ (1)^\frac{1}{4}\ \hspace{10cm}$
$=\ (cos\ 0\ +\ i\ sin\ 0)^\frac{1}{4}\ \hspace{10cm}$
$=\ (cos\ 2kπ\ +\ i\ sin\ 2kπ)^\frac{1}{4}\ \hspace{10cm}$
$=\ cos\ (\frac{2kπ}{4})\ +\ i\ sin\ (\frac{2kπ}{4})\ \hspace{5cm}$
$=\ cos\ (\frac{kπ}{2})\ +\ i\ sin\ (\frac{kπ}{2})\ where\ k\ =\ 0,\ 1,\ 2,\ 3\ \hspace{5cm}$
$The\ roots\ are$
$When\ k = 0,\ \hspace{2cm}\ x\ =\ cos\ 0\ +\ i\ sin\ 0\ =\ 1$
$When\ k = 1,\ \hspace{2cm}\ x\ =\ cos\ \frac{π}{2}\ +\ i\ sin\ \frac{π}{2}$
$When\ k = 2,\ \hspace{2cm}\ x\ =\ cos\ \frac{2π}{2}\ +\ i\ sin\ \frac{2π}{2}\ =\ cos\ π\ +\ i\ sin\ π$
$When\ k = 3,\ \hspace{2cm}\ x\ =\ cos\ \frac{3π}{2}\ +\ i\ sin\ \frac{3π}{2}$

$\color {purple} {3\ .}\ \color {red} {Solve}\ x^5 +\ x^3\ +\ x^2\ +\ 1\ =\ 0\ \hspace{18cm}$
$\color {blue}{Solution:}\ Given\ x^5 +\ x^3\ +\ x^2\ +\ 1\ =\ 0\ \hspace{15cm}$
$x^3(x^2 +\ 1)\ +\ 1( x^2\ +\ 1)\ =\ 0\ \hspace{10cm}$
$(x^3\ +\ 1)\ ( x^2\ +\ 1)\ =\ 0\ \hspace{10cm}$
$\color {green} {Case\ ( i ) :}\ \hspace{16cm}$
$x^3\ +1 =\ 0\ \hspace{10cm}$
$x^3\ =\ -\ 1\ \hspace{10cm}$
$\ x\ =\ (-\ 1)^\frac{1}{3}\ \hspace{10cm}$
$=\ (cos\ π\ +\ i\ sin\ π)^\frac{1}{3}\ \hspace{10cm}$
$=\ (cos\ (π\ + 2kπ) +\ i\ sin\ (π\ + 2kπ))^\frac{1}{3}\ \hspace{10cm}$
$=\ cos\ (\frac{π\ + 2kπ}{3})\ +\ i\ sin\ (\frac{π\ + 2kπ}{3})\ where\ k\ =\ 0,\ 1,\ 2\ \hspace{5cm}$
$The\ roots\ are$
$When\ k = 0,\ \hspace{2cm}\ x\ =\ cos\ \frac{π}{3}\ +\ i\ sin\ \frac{π}{3}$
$When\ k = 1,\ \hspace{2cm}\ x\ =\ cos\ \frac{3π}{3}\ +\ i\ sin\ \frac{3π}{3}\ =\ -1$
$When\ k = 2,\ \hspace{2cm}\ x\ =\ cos\ \frac{5π}{3}\ +\ i\ sin\ \frac{5π}{3}$
$\color {green} {Case\ ( ii ) :}\ \hspace{16cm}$
$x^2\ +1 =\ 0\ \hspace{10cm}$
$x^2\ =\ -\ 1\ \hspace{10cm}$
$\ x\ =\ (-\ 1)^\frac{1}{2}\ \hspace{10cm}$
$=\ (cos\ π\ +\ i\ sin\ π)^\frac{1}{2}\ \hspace{10cm}$
$=\ (cos\ (π\ + 2kπ) +\ i\ sin\ (π\ + 2kπ))^\frac{1}{2}\ \hspace{10cm}$
$=\ cos\ (\frac{π\ + 2kπ}{2})\ +\ i\ sin\ (\frac{π\ + 2kπ}{2})\ where\ k\ =\ 0,\ 1\ \hspace{5cm}$
$The\ roots\ are$
$When\ k = 0,\ \hspace{2cm}\ x\ =\ cos\ \frac{π}{2}\ +\ i\ sin\ \frac{π}{2}$
$When\ k = 1,\ \hspace{2cm}\ x\ =\ cos\ \frac{3π}{2}\ +\ i\ sin\ \frac{3π}{2}$