Help with finding the four roots of complex number(Resolved)

Resolved Forum QuestionsCategory: Algebra -IIHelp with finding the four roots of complex number(Resolved)
Imaan asked 1 month ago

z^4=-32/(1+isqr3)

 

To find the 4 roots of a complex number, the first step is to find the  polar  form as

z=r(cos x +i sin x) where “x” is the angle in degrees or radians.

Let us try to represent the -32/(1+isqr3) in the polar form. Let us rationalize the denominator (1+isqr3)

z^4=-32/(1+isqr3)

\dfrac {-32}{\left( 1+i\sqrt {3}\right) }\cdot \dfrac {1-i\sqrt {3}}{1-i\sqrt {3}}

=>\dfrac {-32\left( 1-i\sqrt {3}\right) }{\left( 1+\left( \sqrt {3}\right) ^{2}\right) }=\dfrac {-32\left( 1-i\sqrt {3}\right) }{\left( 1+3\right) }=-8\left( 1-i\sqrt {3}\right)

Multiplying and dividing the side side of “=” by 2 we get

=>z^{4}=-8\cdot \dfrac {2}{2}\left( 1-\sqrt {3}i\right) =-16\left( \dfrac {1}{2}-\dfrac {\sqrt {3}}{2}i\right)

=>z^{4}=+16\left( -\dfrac {1}{2}+\dfrac {\sqrt {3}}{2}i\right) =16\left[ \cos \left( 2n\pi +\dfrac {2\pi }{3}\right) +\sin ( 2n\pi +\dfrac {2\pi }{3}\right]

=>Taking 4th power both sides.

=>\left( z^{4}\right) ^{\dfrac {1}{4}}=[ 16\left[ \cos \left( 2n\pi +\dfrac {2\pi }{3}\right) +i\sin \left( 2n\pi +\dfrac {2\pi }{3}\right) \right] ^{\dfrac {1}{4}}

=>z=\left( 16\right) ^{\dfrac {1}{4}}\left( \cos \left( \dfrac {2n\pi }{4}+\dfrac {\pi }{6}\right) +i\sin \left( \dfrac {2n\pi }{4}+\dfrac {\pi }{6}\right) \right)

Now we plug n=0,1,2,3 to get 4  roots.

When n=0

\begin{aligned}z=2\left( \cos \left( 0+\dfrac {\pi }{6}\right) +i\sin \left( 0+\dfrac {\pi }{0}\right) \right) \\ =2\left( \dfrac {\sqrt {3}}{2}+i\dfrac {1}{2}\right) =\sqrt {3}+i\end{aligned}

when n=1

\begin{aligned}z=2\left( \cos \left( \dfrac {2\left( 1\right) \pi }{4}+\dfrac {\pi }{0}\right) +i\sin \left( \dfrac {2\left( 1\right) \pi }{4}+\dfrac {\pi }{6}\right) \right) \\ =2\left[ \cos \left( \dfrac {\pi }{2}+\dfrac {\pi }{6}\right) +i\sin \left( \dfrac {\pi }{2}+\dfrac {\pi }{6}\right) \right] \\ =2\left( -\dfrac {\sqrt {3}}{2}+i\dfrac {1}{2}\right) =-\sqrt {3}+i\end{aligned}

When n=2

\begin{aligned}\begin{aligned}z=2\left( \cos \left( \dfrac {2\left( 2\right) \pi }{4}+\dfrac {\pi }{6}\right) +i\sin \left( \dfrac {2\left( 2\right) \pi }{4}+\dfrac {\pi }{6}\right) \right) \\ =2( \cos \left( \pi +\dfrac {\pi }{6}\right) +i\sin \left( \pi +\dfrac {\pi }{6}\right) \right] \end{aligned}\\ =2\left( \dfrac {\sqrt {3}}{2}-i\dfrac {1}{2}\right) \\ =\sqrt {3}-i\end{aligned}

when n=3

\begin{aligned}z=2\left[ \cos \left( \dfrac {2\left( 3\right) \pi }{4}+\dfrac {\pi }{6}\right) +i\sin \left( \dfrac {2\left( 3\right) ^{\pi }}{4}+\dfrac {\pi }{6}\right) \right] \\ =2\left( \cos \left( 3\dfrac {\pi }{2}+\dfrac {\pi }{6}\right) +i\sin \left( \dfrac {3\pi }{2}+\dfrac {\pi }{6}\right) \right) \\ =2\left( \dfrac {1}{2}-i\dfrac {\sqrt {3}}{2}\right) =1-i\sqrt {3}\end{aligned}}

Therefore the four  roots of

z^{4}=\dfrac {-32}{1+i\sqrt {3}} are

\begin{aligned}\sqrt {3}+i\\ -\sqrt {3}+i\\ \sqrt {3}-i\\ 1-\sqrt {3}i\end{aligned}

admin Staff replied 1 month ago

Hi!
Thanks for posting the question. Please find the step by step solution. Any queries please let us know.