Question

How do you find the sixth roots of $64i$? \[\]

Answer 1

Hint: We find the sixth root of $i$ using Demoivre’s theorem for $n=6$ ${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta $ and then assume the root as $z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)$. We design the equation ${{z}^{6}}={{64}^{i}}$ . We take $r=2$ and find a solution for angle $\theta $ . We use sine difference of angle formula $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ and cosine difference of angle $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$\[\]

Complete step by step answer:
We know the complex number $z$ can also be represented in the polar form
\[z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)\]
 We know from Demoivre’s theorem that for real angle $\theta $ and real integral exponent $n$ as
\[{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta \]
Let us find six roots of $i$. We take $n=6$ in the above step and equate to $i$ to have
\[\begin{align}
  & {{\left( \cos \theta +i\sin \theta \right)}^{6}}=\cos 6\theta +i\sin 6\theta =i. \\
 & \Rightarrow \cos 6\theta +i\sin 6\theta =0+i\cdot 1 \\
 & \Rightarrow \cos 6\theta +i\sin 6\theta =0\cdot \cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right)....\left( 1 \right) \\
\end{align}\]
We equate imaginary parts from both sides to have;
\[\begin{align}
  & \sin \left( 6\theta \right)=\sin \left( \dfrac{\pi }{2} \right) \\
 & \Rightarrow 6\theta =\dfrac{\pi }{2} \\
 & \Rightarrow \theta =\dfrac{\pi }{12} \\
\end{align}\]
So our principal angle is $\theta =\dfrac{\pi }{12}$. We can find the other angle by using periodicity of sine so that we can add multiples of $\dfrac{2\pi }{6}=\dfrac{\pi }{3}$ to satisfy the equation (1) as
\[\begin{align}
  & \cos \left( \dfrac{\pi }{12}+\dfrac{k\pi }{3} \right)+i\sin \left( \dfrac{\pi }{12}+\dfrac{k\pi }{3} \right)=\cos \left( \dfrac{\pi }{2}+2k\pi \right)+i\sin \left( \dfrac{\pi }{2}+2k\pi \right) \\
 & \Rightarrow \cos \left( \dfrac{\pi }{12}+\dfrac{k\pi }{3} \right)+i\sin \left( \dfrac{\pi }{12}+\dfrac{k\pi }{3} \right)=i,\left( k=1,2,3,4,5,6 \right) \\
\end{align}\]
Let us assume the sixth roots of $64i$ is $z={{e}^{i\theta }}$ which means the solutions of the equation
\[{{z}^{6}}=64i\]
We put $z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)$ in the above step to have
\[\begin{align}
  & {{z}^{6}}=64i \\
 & \Rightarrow {{\left( r\left( \cos \theta +i\sin \theta \right) \right)}^{6}}=64i \\
 & \Rightarrow {{r}^{6}}{{\left( \cos \theta +i\sin \theta \right)}^{6}}=64i \\
\end{align}\]
We see the above equation is valid only for $r=2$ since $r$ is always a positive and real number. We have already obtained values of $\theta $from sixth roots of $i$.So the sixth roots of $64i$ is
\[\begin{align}
  & z=2\left( \cos \left( \dfrac{\pi }{12}+\dfrac{k\pi }{3} \right)+i\sin \left( \dfrac{\pi }{12}+\dfrac{k\pi }{3} \right) \right) \\
 & \Rightarrow z=2\left( \cos \left( \dfrac{1+4k}{12}\pi \right)+i\sin \left( \dfrac{1+4k}{12}\pi \right) \right)\left( k=0,1,2,3,4,5 \right) \\
\end{align}\]
So the possible value of $\theta $ are
\[\theta =\dfrac{\pi }{12},\dfrac{5\pi }{12},\dfrac{9\pi }{12},\dfrac{13\pi }{12}=\pi +\dfrac{\pi }{12},\dfrac{17\pi }{12}=\pi +\dfrac{5\pi }{12},\dfrac{21\pi }{12}=\pi +\dfrac{9\pi }{12}\]
So by periodicity the solutions are \[\pm 2\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right),\pm 2\left( \cos \left( \dfrac{5\pi }{12} \right)+i\sin \left( \dfrac{5\pi }{12} \right) \right),\pm 2\left( \cos \left( \dfrac{9\pi }{12} \right)+i\sin \left( \dfrac{9\pi }{12} \right) \right)\]
We use the difference of angle formula to find the cosine value of $\dfrac{\pi }{12},\dfrac{5\pi }{12}$. We have
\[\begin{align}
  & \sin \left( \dfrac{\pi }{12} \right)=\sin \left( \dfrac{\pi }{3}-\dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{3}\cos \dfrac{\pi }{4}-\cos \dfrac{\pi }{3}\sin \dfrac{\pi }{4}=\dfrac{\sqrt{3}}{2}\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\dfrac{\sqrt{2}}{2}=\dfrac{1}{4}\left( \sqrt{6}-\sqrt{2} \right) \\
 & \cos \left( \dfrac{\pi }{12} \right)=\cos \left( \dfrac{\pi }{3}-\dfrac{\pi }{4} \right)=\cos \dfrac{\pi }{3}\cos \dfrac{\pi }{4}+\sin \dfrac{\pi }{3}\sin \dfrac{\pi }{4}=\dfrac{1}{2}\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\dfrac{\sqrt{2}}{2}=\dfrac{1}{4}\left( \sqrt{6}+\sqrt{2} \right) \\
\end{align}\]
We use complementary angle relation between sine and cosine to have;
\[\begin{align}
  & \sin \left( \dfrac{5\pi }{12} \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{12} \right)=\cos \dfrac{\pi }{12}=\dfrac{1}{4}\left( \sqrt{6}+2 \right) \\
 & \cos \left( \dfrac{5\pi }{12} \right)=\cos \left( \dfrac{\pi }{2}-\dfrac{\pi }{12} \right)=\sin \dfrac{\pi }{12}=\dfrac{1}{4}\left( \sqrt{6}-2 \right) \\
\end{align}\]
We know from trigonometric table that
\[\begin{align}
  & \sin \left( \dfrac{9\pi }{12} \right)=\sin \left( \dfrac{3\pi }{4} \right)=\dfrac{1}{\sqrt{2}} \\
 & \cos \left( \dfrac{9\pi }{12} \right)=\cos \left( \dfrac{3\pi }{4} \right)=\dfrac{-1}{\sqrt{2}} \\
\end{align}\]
So the sixth roots of $64i$ are
\[\begin{align}
  & \pm 2\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right)=\pm \dfrac{1}{2}\left( \left( \sqrt{6}+\sqrt{2} \right)+i\left( \sqrt{6}-\sqrt{2} \right) \right) \\
 & \pm 2\left( \cos \left( \dfrac{5\pi }{12} \right)+i\sin \left( \dfrac{5\pi }{12} \right) \right)=\pm \dfrac{1}{2}\left( \left( \sqrt{6}-\sqrt{2} \right)+i\left( \sqrt{6}+\sqrt{2} \right) \right) \\
 & \pm 2\left( \cos \left( \dfrac{9\pi }{12} \right)+i\sin \left( \dfrac{9\pi }{12} \right) \right)=\pm \sqrt{2}\left( 1+i \right) \\
\end{align}\]

Note:
We note that if all the roots of equation ${{z}^{n}}=a$ where $a$ is any complex number and $n$is an integer, are complex then they occur in $\dfrac{n}{2}$ pairs of conjugates in the form $\left( a+ib \right),\left( a-ib \right)$. Here we obtained $\dfrac{6}{2}=3$pairs of conjugate roots. We should also note that two complex numbers are equal if their real and imaginary parts are equal.