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I need help with this practiceThe subject is complex numbers and vectors*It asks to fill in the four boxes, *enter the roots in order of increasing angle measure

I need help with this practiceThe subject is complex numbers and vectors*It asks to-example-1
User Gwyn
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1 Answer

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17 votes

ANSWER:


\begin{gathered} \:\sqrt[4]{4}cis\left(-(\pi )/(24)\right)\: \\ \\ \:\sqrt[4]{4}cis\left((11\pi)/(24)\right)\: \\ \\ \:\sqrt[4]{4}cis\left((23\pi)/(24)\right)\: \\ \\ \:\sqrt[4]{4}cis\left((35\pi)/(24)\right)\: \end{gathered}

Explanation:

We have the following expression:


2√(3)-2i

To calculate the 4 roots we must match the equation with x raised to 4, just like this:


x^4=2√(3)-2i

For this case the roots are given as follows:


\begin{gathered} \:z_k=\sqrt[n]\left\left(\cos\left((\arctan\left(\alpha\right)+2k\pi)/(n)\right)+i\sin\left((\arctan\left(\alpha\right)+2k\pi)/(n)\right)\right) \\ \\ \text{ In this case:} \\ \\ n=4 \\ \\ |a|=\sqrt{\left(2√(3)\right)^2+\left(-2\right)^2}=√(12+4)=√(16) \\ \\ a=4 \\ \\ \alpha=\left((-2)/(2√(3))\right) \\ \\ \text{ Therefore:} \\ \\ \arctan\left((-2)/(2√(3))\right)=-(\pi)/(6) \end{gathered}

Taking into account the above, we calculate for each x,

when k = 0,1, 2 3, just like this:


\begin{gathered} x_1=\sqrt[4]{4}\left(\cos\left((-(\pi)/(6)+2\cdot\:0\pi)/(4)\right)+i\sin\left((-(\pi)/(6)+2\cdot\:0\pi)/(4)\right)\right)=\sqrt[4]{4}\left(\cos\left(-(\pi)/(24)\frac{}{}\right)+i\sin\left(-(\pi)/(24)\right)\right)=\sqrt[4]{4}cis\left(-(\pi)/(24)\right) \\ \\ x_2=\sqrt[4]{4}\left(\cos\left((-(\pi)/(6)+2\cdot\:1\pi)/(4)\right)+i\sin\left((-(\pi)/(6)+2\cdot\:1\pi)/(4)\right)\right)=\sqrt[4]{4}\left(\cos\left((11\pi\:)/(24)\right)+i\sin\left((11\pi\:)/(24)\right)\right)=\:\sqrt[4]{4}cis\left((11\pi\:)/(24)\right)\: \\ \\ x_3=\sqrt[4]{4}\left(\cos\left((-(\pi)/(6)+2\cdot\:2\pi)/(4)\right)+i\sin\left((-(\pi)/(6)+2\cdot\:2\pi)/(4)\right)\right)=\sqrt[4]{4}\left(\cos\left((23\pi\:)/(24)\right)+i\sin\left((23\pi\:)/(24)\right)\right)=\sqrt[4]{4}cis\left((23\pi\:)/(24)\right)\: \\ \\ x_4=\sqrt[4]{4}\left(\cos\left((-(\pi)/(6)+2\cdot\:3\pi)/(4)\right)+i\sin\left((-(\pi)/(6)+2\cdot\:3\pi)/(4)\right)\right)=\sqrt[4]{4}\left(\cos\left((35\pi\:)/(24)\right)+i\sin\left((35\pi\:)/(24)\right)\right)=\:\sqrt[4]{4}cis\left((35\pi\:)/(24)\right)\: \end{gathered}

User Andy Noelker
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