6.(Primitive root of unity) is a primitive nth root of unity if it is an nth root of unity and 1;;:::; n 1 are all distinct. 7.(Primitive root of unity v2) = e2ˇik=nis a primitive nth root of unity i gcd(k;n) = 1. 8.(Cyclotomic polynomial) The nth cyclotomic polynomial, n(x), is the polynomial whose roots are the nth primitive roots of unity. 6.(Primitive root of unity) is a primitive nth root of unity if it is an nth root of unity and 1;;:::; n 1 are all distinct. 7.(Primitive root of unity v2) = e2ˇik=nis a primitive nth root of unity i gcd(k;n) = 1. 8.(Cyclotomic polynomial) The nth cyclotomic polynomial, n(x), is the polynomial whose roots are the nth primitive roots of unity.

• 6 Roots Of Unity
• 6 Roots Of Unity Meaning
• 6 Roots Of Unity Lutheran Church

Roots of Complex Numbers

Key Questions

• To evaluate the #nth# root of a complex number I would first convert it into trigonometric form:
  #z=r[cos(theta)+isin(theta)]#
  and then use the fact that:
  #z^n=r^n[cos(n*theta)+isin(n*theta)]#
  and:
  #nsqrt(z)=z^(1/n)=r^(1/n)*[cos((theta+2kpi)/n)+isin((theta+2kpi)/n)]#
  Where #k=0..n-1#

  For example: consider #z=2+3.46i# and let us try #sqrt(z)#;
  #z# can be written as:
  #z=4[cos(pi/3)+isin(pi/3)]#
  So:
  #k=0#
  #sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+0)/2)+isin((pi/3+0)/2)]=#
  #=2[cos(pi/6)+isin(pi/6))]#
  And:
  #k=n-1=2-1=1#
  #sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+2pi)/2)+isin((pi/3+2pi)/2)]=#
  #=2[cos(7pi/6)+isin(7pi/6))]#
  Which gives, in total, two solutions.

• Answer:

  If you express your complex number in polar form as #r(cos theta + i sin theta)#, then it has fourth roots:

  #alpha = root(4)(r)(cos (theta/4) + i sin (theta/4))#, #i alpha#, #-alpha# and #- i alpha#

  Explanation:

  Given #a+ib#, let #r = sqrt(a^2+b^2)#, #theta = 'atan2'(b, a)#

  Then #a + ib = r (cos theta + i sin theta)#

  This has one #4th# root #alpha = root(4)(r)(cos (theta/4) + i sin (theta/4))#

  There are three other #4th# roots: #i alpha#, #-alpha# and #-i alpha#

• A root of unity is a complex number that when raised to some positive integer will return 1.

  It is any complex number #z# which satisfies the following equation:

  #z^n = 1#

  where #n in NN#, which is to say that n is a natural number. A natural number is any positive integer: (n = 1, 2, 3, ...). This is sometimes referred to as a counting number and the notation for it is #NN#.

  For any #n#, there may be multiple #z# values that satisfy that equation, and those values comprise the roots of unity for that n.

  When #n = 1#
  Roots of unity: #1#

  When #n = 2#
  Roots of unity: #-1, 1#

  When #n = 3#
  Roots of unity = #1, (1 + sqrt(3)i)/2, (1 - sqrt(3)i)/2#

  When #n = 4#
  Roots of unity = #-1, i, 1, -i#

• To evaluate the square root (and in general any root) of a complex number I would first convert it into trigonometric form:
  #z=r[cos(theta)+isin(theta)]#
  and then use the fact that:
  #z^n=r^n[cos(n*theta)+isin(n*theta)]#

  Where, in our case, #n=1/2# (remembering that #sqrt(x)=x^(1/2)#).
  To evaluate the #nth# root of a complex number I would write:

  #nsqrt(z)=z^(1/n)=r^(1/n)*[cos((theta+2kpi)/n)+isin((theta+2kpi)/n)]#
  Where #k=0..n-1#

  For example: consider #z=2+3.46i# and let us try #sqrt(z)#;
  #z# can be written as:
  #z=4[cos(pi/3)+isin(pi/3)]#
  So:
  #k=0#
  #sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+0)/2)+isin((pi/3+0)/2)]=#
  #=2[cos(pi/6)+isin(pi/6))]#
  And:
  #k=n-1=2-1=1#
  #sqrt(z)=z^(1/2)=4^(1/2)[cos((pi/3+2pi)/2)+isin((pi/3+2pi)/2)]=#
  #=2[cos(7pi/6)+isin(7pi/6))]#
  Which gives, in total, two solutions.

Questions

• 2Primitive root in modular arithmetic

Primitive root of unity

6 Roots Of Unity

A primitive root of unity of order $m$ in a field $K$ is an element $zeta$ of $K$ such that $zeta^m = 1$ and $zeta^r neq 1$ for any positive integer $r < m$. The element $zeta$ generates the cyclic group $mu_m$ of roots of unity of order $m$.

If in $K$ there exists a primitive root of unity of order $m$, then $m$ is relatively prime to the characteristic of $K$. An algebraically closed field contains a primitive root of any order that is relatively prime with its characteristic. If $zeta$ is a primitive root of order $m$, then for any $k$ that is relatively prime to $m$, the element $zeta^k$ is also a primitive root. The number of all primitive roots of order $m$ is equal to the value of the Euler function $phi(m)$ if $mathrm{hcf}(m,mathrm{char}(K)) = 1$.

In the field of complex numbers, there are primitive roots of unity of every order: those of order $m$ take the form$$cos frac{2pi k}{m} + i sin frac{2pi k}{m}$$where $0 < k < m$ and $k$ is relatively prime to $m$.

Primitive root in modular arithmetic

A primitive root modulo $m$ is an integer $g$ such that$$g^{phi(m)} equiv 1 pmod m text{and} g^gamma notequiv 1 pmod m$$for $1 le gamma < phi(m )$, where $phi(m)$ is the Euler function. For a primitive root $g$, its powers $g^0=1,ldots,g^{phi(m)-1}$ are incongruent modulo $m$ and form a reduced system of residues modulo $m$. Therefore, for each number $a$ that is relatively prime to $m$ one can find an exponent $gamma$, $0 le gamma < phi(m)$ for which $g^gamma equiv a pmod m$: the index of $a$ with respect to $g$.

Primitive roots do not exist for all moduli, but only for moduli $m$ of the form $2,4, p^a, 2p^a$, where $p>2$ is a prime number. In these cases, the multiplicative groups of reduced residue classes modulo $m$ have the simplest possible structure: they are cyclic groups of order $phi(m)$. The concept of a primitive root modulo $m$ is closely related to the concept of the index of a number modulo $m$.

Primitive roots modulo a prime number were introduced by L. Euler, but the existence of primitive roots modulo an arbitrary prime number was demonstrated by C.F. Gauss (1801).

References

6 Roots Of Unity Meaning

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References

6 Roots Of Unity Lutheran Church