Question

How to find the cube roots of $\sqrt{2}+i\sqrt{2}$? Exhibit terms as vertices of a certain regular polygon and identify the principal root.

Answer 1

In this question, we are given complex number and we will first suppose it as z = a+ib then we will convert it into polar form using the transformation $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}},\cos \theta =\dfrac{a}{\left| z \right|}\text{ and }\sin \theta =\dfrac{b}{\left| z \right|}$ to convert z into $\left| z \right|$. After that we will take the cube root of z. We will use De-Moivre's theorem and find the three roots of the given complex number. These three in the polar form will be the vertices of the regular polygon and then we will find the principal root.

Complete step-by-step answer:
Here we are given the complex numbers as $\sqrt{2}+i\sqrt{2}$. Let us suppose this complex number as z. We know that, a complex number of the form a+ib can be converted into polar form $z=\left| z \right|\left( \cos \theta +\sin \theta \right)$ using the transformation $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}},\cos \theta =\dfrac{a}{\left| z \right|}\text{ and }\sin \theta =\dfrac{b}{\left| z \right|}$. So here we have $\sqrt{2}+i\sqrt{2}$.
Comparing it with a+ib we get $a=\sqrt{2}\text{ and }b=\sqrt{2}$.
Let us convert $z=\sqrt{2}+i\sqrt{2}$ into the polar coordinates using above transformation we have $\left| z \right|=\sqrt{{{\left( \sqrt{2} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}$.
We know that ${{\left( \sqrt{2} \right)}^{2}}=2$ so we get $\left| z \right|=\sqrt{2+2}=\sqrt{4}=2$ so we get $\left| z \right|=2$.
$\cos \theta =\dfrac{a}{\left| z \right|}\Rightarrow \cos \theta =\dfrac{\sqrt{2}}{2}$.
2 can be written as $\sqrt{2}\times \sqrt{2}$ so we get $\cos \theta =\dfrac{1}{\sqrt{2}}$.
Similarly $\sin \theta =\dfrac{b}{\left| z \right|}=\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}$.
Hence $\sin \theta =\dfrac{1}{\sqrt{2}}$.
Using $z=\left| z \right|\left( \cos \theta +\sin \theta \right)$ we get $z=2\left( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right)$.
Now we know that $\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ so we can write the value of z as $z=2\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)$.
Now let us find the general value of z in polar form. We know that, period of both $\cos \theta \text{ and }\sin \theta$ is $2\pi$ so let us add $2n\pi$ in the angle of $\cos \theta \text{ and }\sin \theta$ to get general value of z where $n\in z$ (Set of integers) we get $z=2\left( \cos \left( \dfrac{\pi }{4}+2n\pi \right)+i\sin \left( \dfrac{\pi }{4}+2n\pi \right) \right)$ where $n\in z$.
Now let us take cube root we get,
${{z}^{\dfrac{1}{3}}}={{2}^{\dfrac{1}{3}}}{{\left( \cos \left( \dfrac{\pi }{4}+2n\pi \right)+i\sin \left( \dfrac{\pi }{4}+2n\pi \right) \right)}^{\dfrac{1}{3}}}$.
From De-moivers theorem we know that, ${{\left( \cos \theta +\sin \theta \right)}^{n}}=\left( \cos n\theta +\sin n\theta \right)$ so we get,

& {{z}^{\dfrac{1}{3}}}={{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{1}{3}\left( \dfrac{\pi }{4}+2n\pi \right) \right)+i\sin \left( \dfrac{1}{3}\left( \dfrac{\pi }{4}+2n\pi \right) \right) \right) \\\ & \Rightarrow {{z}^{\dfrac{1}{3}}}={{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{\pi }{12}+\dfrac{2}{3}n\pi \right)+i\sin \left( \dfrac{\pi }{12}+2n\pi \right) \right) \\\ \end{aligned}$$. For finding the three roots we know that the value of n will be less than 3 i.e. n = 0, 1, 2. So the three cube roots of z will be given by (i) putting n = 0 we get, $${{z}_{1}}={{2}^{\dfrac{1}{3}}}\left( \cos \dfrac{\pi }{12}+i\sin \dfrac{\pi }{12} \right)$$. (ii) putting n = 1, we get $${{z}_{2}}={{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{\pi }{12}+\dfrac{4}{3}n\pi \right)+i\sin \left( \dfrac{\pi }{12}+\dfrac{4}{3}n\pi \right) \right)$$. Solving the angles we get $${{z}_{2}}={{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{3}{4}\pi \right)+i\sin \left( \dfrac{3}{4}\pi \right) \right)$$. (iii) putting n = 2 we get $${{z}_{3}}={{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{\pi }{4}+2\pi \right)+i\sin \left( \dfrac{\pi }{4}+2\pi \right) \right)$$. Solving the angles we get $${{z}_{3}}={{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{17}{4}\pi \right)+i\sin \left( \dfrac{17}{4}\pi \right) \right)$$. Hence the required cube root of $\sqrt{2}+i\sqrt{2}$ are, $${{2}^{\dfrac{1}{3}}}\left( \cos \dfrac{\pi }{12}+i\sin \dfrac{\pi }{12} \right)$$, $${{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{3}{4}\pi \right)+i\sin \left( \dfrac{3}{4}\pi \right) \right)$$, $${{2}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{17}{4}\pi \right)+i\sin \left( \dfrac{17}{4}\pi \right) \right)$$. In general ${{n}^{th}}$ roots of a complex number are the roots of a regular n-gon. Here we have cube roots of complex numbers so these represent the vertices of 3-gon which is a triangle. The principal root is the root in which period is added i.e. the root with n = 0 gives us the principal root. So $${{2}^{\dfrac{1}{3}}}\left( \cos \dfrac{\pi }{12}+i\sin \dfrac{\pi }{12} \right)$$ is the principal root. **Note:** Students should note that, while finding ${{n}^{th}}$ roots we have to take value of n as less than n only i.e. 0, 1, 2, . . . . . . . n-1. Take care while changing into polar form. As $\left| z \right|$ is the magnitude so while taking square root we need not consider the negative sign.