Question

How do I find the fifth root of a complex number?

Answer 1

Hint: Assume a complex number in its general form given as, \[z=a+ib\], where ‘a’ is the real part and ‘b’ is the imaginary part. Now, write the complex number in polar form as: - \[{{z}^{5}}=r{{e}^{i\theta }}\], where \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}=\left| z \right|\] and \[\theta \] = argument (z) given as: - \[Arg\left( z \right)={{\tan }^{-1}}\left( \dfrac{b}{a} \right)\]. Take the fifth root of the expression. On both the sides and use the relation: - \[{{e}^{in\theta }}=\left( \cos n\theta +i\sin n\theta \right)\] to convert in trigonometric form. Finally, write the complex number in the form, \[z={{r}^{\dfrac{1}{5}}}\left( \cos \left( \dfrac{\theta +2k\pi }{5} \right)+i\sin \left( \dfrac{\theta +2k\pi }{5} \right) \right)\] and substitute k = 0, 1, 2, 3 and 4 to get the fifth roots.

Complete step by step answer:
Here, we have been asked to find the fifth roots of a complex number. Since no particular complex number is given so let us consider the general form of the complex number.
Now, we know that the general form of a complex number is given as, \[z=a+ib\], where ‘a’ is the real part denoted as (Re(z)) and ‘b’ is the imaginary part denoted as (Im(z)). So, we have,
\[\Rightarrow z=a+ib\]
Now, in polar form the complex number is given as \[r{{e}^{i\theta }}\], where \[r=\sqrt{{{a}^{2}}+{{b}^{2}}}=\left| z \right|\] and \[\theta \] = argument (z) = \[Arg\left( z \right)={{\tan }^{-1}}\left( \dfrac{b}{a} \right)\]. Since, we have to find the fifth root so we will substitute \[r{{e}^{i\theta }}\] with \[{{z}^{5}}\]. So, we have,
\[\Rightarrow {{z}^{5}}=r{{e}^{i\theta }}\]
Taking fifth root both the sides, we get,
\[\begin{align}
  & \Rightarrow z={{\left( r{{e}^{i\theta }} \right)}^{\dfrac{1}{5}}} \\
 & \Rightarrow z={{r}^{\dfrac{1}{5}}}.{{e}^{\dfrac{i\theta }{5}}} \\
\end{align}\]
Using the formula: - \[{{e}^{in\theta }}=\cos n\theta +i\sin n\theta \], we get,
\[\Rightarrow z={{r}^{\dfrac{1}{5}}}.\left( \cos \dfrac{\theta }{5}+i\sin \dfrac{\theta }{5} \right)\]
Since, the value of cosine and sine functions repeat itself after an interval of \[2\pi \], so we can have the fifth roots as: -
\[\Rightarrow z={{r}^{\dfrac{1}{5}}}.\left( \cos \left( \dfrac{\theta +2k\pi }{5} \right)+i\sin \left( \dfrac{\theta +2k\pi }{5} \right) \right)\], where k = 0, 1, 2, 3, 4.
\[\Rightarrow z={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{10}}}.\left( \cos \left( \dfrac{\theta +2k\pi }{5} \right)+i\sin \left( \dfrac{\theta +2k\pi }{5} \right) \right)\], where k = 0, 1, 2, 3, 4.
Substituting the values of k one – by – one, we get,
\[\begin{align}
  & \Rightarrow {{z}_{1}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{10}}}.\left( \cos \left( \dfrac{\theta }{5} \right)+i\sin \left( \dfrac{\theta }{5} \right) \right) \\
 & \Rightarrow {{z}_{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{10}}}.\left( \cos \left( \dfrac{\theta +2\pi }{5} \right)+i\sin \left( \dfrac{\theta +2\pi }{5} \right) \right) \\
 & \Rightarrow {{z}_{3}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{10}}}.\left( \cos \left( \dfrac{\theta +4\pi }{5} \right)+i\sin \left( \dfrac{\theta +4\pi }{5} \right) \right) \\
 & \Rightarrow {{z}_{4}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{10}}}.\left( \cos \left( \dfrac{\theta +6\pi }{5} \right)+i\sin \left( \dfrac{\theta +6\pi }{5} \right) \right) \\
 & \Rightarrow {{z}_{5}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\dfrac{1}{10}}}.\left( \cos \left( \dfrac{\theta +8\pi }{5} \right)+i\sin \left( \dfrac{\theta +8\pi }{5} \right) \right) \\
\end{align}\]
Here, \[\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)\].

Note:
 One may note that \[\theta \] lies in the interval \[\left[ 0,2\pi \right]\] and this is the reason that the value of ‘k’ cannot be taken as 5 or more. If we would have been asked to find \[{{n}^{th}}\] roots then we would have taken the values of k = 0, 1, 2, ……., n – 1. Remember that except 0 all other complex numbers have ‘n’ roots. You must know the general forms of a complex number in algebraic form, poplar form, trigonometric form, to solve the question. Take the values of ‘k’ carefully and according to the information provided in the question.