Tag: finding nth roots of a complex number

Questions Related to finding nth roots of a complex number

If $z _{1}$ is a root of the equation $a^{n} _{0}z^{n}+a _{1}z^{n-1}+....+a _{n-1^{z}}+a _{n}=3$, where $|a _{i}|<2$ for $i=0,1,....,n.$ Then,

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $|z _{1}|>\dfrac {1}{3}$

  2. $|z _{1}|<\dfrac {1}{4}$

  3. $|z _{1}|>\dfrac {1}{4}$

  4. $|z|<\dfrac {1}{3}$

Show answer
Correct Option: A
Explanation:

According to question,

${l} { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } }=3 \ \Rightarrow \left| { { a _{ 0 } }{ z^{ n } }+{ a _{ 1 } }{ z^{ n-1 } }+.............+{ a _{ n-1 } }z+{ a _{ n } } } \right| =\left| 3 \right|  \ \Rightarrow \left| { { a _{ 0 } } } \right| \, { \left| z \right| ^{ n } }+\left| { { a _{ 1 } } } \right| \, { \left| z \right| ^{ n-1 } }\, +..........+\left| { { a _{ n-1 } } } \right| \, \left| z \right| \, +\left| { { a _{ n } } } \right| \ge 3 \ \Rightarrow 2\, ({ \left| z \right| ^{ n } }+{ \left| z \right| ^{ n-1 } }+...........\left| z \right| +1)\, \, >\, 3 \ \Rightarrow (1+\left| z \right| +{ \left| z \right| ^{ 2 } }+...........+{ \left| z \right| ^{ n } })\, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow \frac { { \, \, \, \, 1-{ { \left| z \right|  }^{ n+1 } } } }{ { 1-\left| z \right|  } } \, \, >\, \frac { 3 }{ 2 }  \ \Rightarrow 2-2{ \left| z \right| ^{ n+1 } }\, >\, 3-3\left| z \right|  \ \Rightarrow 2{ \left| z \right| ^{ n+1 } }<3\left| z \right| \, -1 \ \Rightarrow 3\, \left| z \right| -1\, >0 \ \, \, \, \, \, \therefore \, \, \, \left| z \right| \, >\, \frac { 1 }{ 3 }  \ so\, the\, correct\, option\, is\, \, A$

The number of common roots of the 15th and  of 25th roots of unity are

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. 1

  2. 5

  3. 6

  4. 10

Show answer
Correct Option: B
Explanation:

$15th$ roots of unity are $e^{\dfrac{{i}2{k}\pi}{15}}$ where $k=1,2,\cdots,15$

$25th$ roots of unity are $e^{\dfrac{{i}2{m}\pi}{25}}$ where $m=1,2,3,\cdots,25$
The common roots are $e^{i\dfrac{2{n}}{5}}$ where $n=1,2,3,4,5$
Number of common roots will be $5$

If $\alpha $ is a non- real fifth root of unity, then the value of ${3^{\left[ {1 + a + {a^2} - {a^{ - 1}}} \right]}}$,is

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $9$

  2. $1$

  3. $11/3$

  4. none of these

Show answer
Correct Option: A

Let principle argument of complex number be re-defined between $(\pi,3\pi)$, then sum of principle arguments of roots of equation $z^{n}+z^{2}+1=0$ is

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $0$

  2. $3\pi$

  3. $6\pi$

  4. $12\pi$

Show answer
Correct Option: A

The value of the expression 

$1 \cdot (2 - \omega) (2 - \omega^2) + 2\cdot (3 - \omega) (3 - \omega^2) +$ _____$+ (n - 1)(n - \omega)(n - \omega^2)$

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\dfrac{1}{4}n(n - 1)(n^2 + 3n + 4)$

  2. $n(n - 1)(n^2 + 3n + 4)$

  3. $\dfrac{1}{4}n(n - 1)(n^2 + 3n - 4)$

  4. None of these

Show answer
Correct Option: A
Explanation:
Here, $a _n= (n-1)(n-\omega )(n-\omega ^2)$

$\sum a _n= \sum (n-1)(n-\omega )(n-\omega ^2)$

$ = \sum (n^2-\omega n-n+\omega )(n-\omega ^2)$

$=\sum (n^3-\omega n^2 n^2+\omega ^3 n +\omega ^2n-\omega ^3)$

$= \sum (n^3-(\omega +\omega ^3)n^2-n^2(\omega +\omega ^2)n+n-1) \,\,\,\,\, [\because \omega ^3 =1]$

$= \sum (n^3-(-1)n^2-n^2+(-1)n+n-1) \,\,\,\,\, [\because 1+\omega +\omega ^2=0]$

$= \sum (n^3+n^2-n^2-n+n-1)$

$=\sum (n^3-1)$

$=\sum n^3-\sum 1$

$= \left [ \dfrac {n(n+1)}{2} \right ]^2-n$

$=\dfrac {n^2(n+1)^2}{4}-n$

$=\dfrac {n^4+n^2+2n^3-4n}{4}$

$=\dfrac {n}{4}(n^3+2n^2+n-4)$

$=\dfrac {n(n-1)(n^2+3n+4)}{4}$

$\therefore 1 \cdot (2 - \omega) (2 - \omega^2) + 2\cdot (3 - \omega) (3 - \omega^2) +$ _____$+ (n - 1)(n - \omega)(n - \omega^2) = \dfrac {1}{4}n (n-1)(n^2+3n+4)$

If $p, q, r, s, t$ are the roots of the equation $x^5-1 = 0$, then $p^{ 10 }+q^{ 10 }+{ r }^{ 10 }+{ s }^{ 10 }+t^{ 10 }=$

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $0$

  2. $1$

  3. $3$

  4. $5$

Show answer
Correct Option: D
Explanation:

$p, q, r, s, t$ are all fifth root of unity
$\Rightarrow p^5=q^5= r^5= s^5= t^5=1$   
$ \Rightarrow p^{10}=q^{10}= r^{10}= s^{10}= t^{10}=1$
Hence, the required sum is $5$

If $1, a _1, a _2, ..a _{n-1}$ are $n^{th}$ roots of unity then $\dfrac{1}{1-a _1}+\dfrac{1}{1-a _2}+....+\dfrac{1}{1-a _{n-1}}$ equals?

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\dfrac{2^n-1}{n}$

  2. $\dfrac{n-1}{2}$

  3. $\dfrac{n}{n-1}$

  4. $\dfrac{n}{n+1}$

Show answer
Correct Option: A

If $1,{z} _{1},{z} _{2},{z} _{n-1}$ are the ${n}^{th}$ roots of unity then the value of $\dfrac{1}{3-z _{1}}+\dfrac{1}{3-z _{2}}+.......+\dfrac{1}{3-z _{n-1}}$ is equal to 

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } +\frac { 1 }{ 2 }$

  2. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } -1$

  3. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } +1$

  4. $\displaystyle\frac { { n.3 }^{ n-1 } }{ { 3 }^{ n }-1 } -\frac { 1 }{ 2 }$

Show answer
Correct Option: D
Explanation:
$z _{1}, z _{2} , z _{3}........ z _{n-1}$ are normal $n^{th}$
$x^{n}-1= (x-1) (x-z _{1}) (x-z _{2}) (x-z _{3})....... (x-z _{n-1})$
ln $(x^{n}-1)= ln (x-1) (x-z _{2}) (x-z _{3})......... (x-z _{n-1})$
$\dfrac{nx^{n-1}}{x^{n}-1}= \dfrac{1}{x-1} + \dfrac{1}{x- z _{1}} + \dfrac{1}{x- _{2}}...... \dfrac{1}{x- z _{n-1}}$
Putting $x=3$
$\dfrac{n3^{n-1}}{3^{n-1}}= \dfrac{1}{2} + \dfrac{1}{3- z _{1}}+ \dfrac{1}{3-z _{2}}........ \dfrac{1}{3-z _{n-1}}$
$\dfrac{1}{3-z _{1}}+ \dfrac{1}{3- z _{2}}.............. \dfrac{1}{3-z _{n-1}}= \dfrac{n(3)^{n-1}}{3^{n}-1}\dfrac{1}{2}$

If $1,{a _1},{a _2},....{a _{n - 1}}$ are ${n^{th}}$ roots of unity then $\frac{1}{{1 - {a _1}}} + \frac{1}{{1 - {a _2}}} + .... + \frac{1}{{1 - {a _{n - 1}}}}$ equals                                                            

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\frac{{{2^n} - 1}}{n}$

  2. $\frac{{n - 1}}{2}$

  3. $\frac{n}{{n - 1}}$

  4. $\frac{n}{{n + 1}}$

Show answer
Correct Option: B

If $1, \alpha _1, \alpha _2, \alpha _3, \alpha _4, \alpha _5, \alpha _6$, are seven, $7^{th}$ root of unity them $|(3-\alpha _1)(3-\alpha _3)(3-\alpha _5)|$ is?

finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

  1. $\sqrt{2186}$

  2. $\sqrt{1093}$

  3. $\sqrt{1023}$

  4. $\sqrt{511}$

Show answer
Correct Option: B
Explanation:
$x^7 -1 = 0$
$(x-1)(x-\alpha _1).....(x-\alpha _6) = x^7-1$
Putting $x=3$, we get
$2.(x-\alpha _1).(x-\alpha _2)...(x-\alpha _6) = 2186$
$|(x-\alpha _1)|^2 |(x- \alpha _3)|^2 |(x-\alpha)|^2 = 1093$
$|(x-\alpha _1)(x- \alpha _3)(x-\alpha _5)| = \sqrt{1093}$