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How would I use polar form to show

$(-1-i\sqrt{3})(-4\sqrt{3}+4i)=8\sqrt{3}+8i$

I tried putting it in polar form. And I got

$2(\cos(225)+i\sin(225))(2\sqrt{7}\cos(150)+i\sin(150))$

But I keep getting an incorrect answer can any kind soul show me how to solve this problem?

  • 2
    I think what Sigur is saying is that polar form for complex numbers is actually $re^{i\theta}$, where $r=\sqrt{a^2+b^2}$ and $\theta =\tan^{-1}(b/a)$.2012-12-20

1 Answers 1

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\begin{gather} -1-i \sqrt{3}=-2\left(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\right)=-2\left(\cos{\dfrac{\pi}{3}}+i\sin{\dfrac{\pi}{3}}\right) =-2e^{\tfrac{\pi i}{3}},\\ -4\sqrt{3}+4i=-8\left(\dfrac{\sqrt{3}}{2}-i\dfrac{1}{2} \right)=-8\left(\cos{\dfrac{11\pi}{6}}+i\sin{\dfrac{11\pi}{6}} \right) =-8e^{\tfrac{11\pi i}{6}}. \end{gather} Therefore, \begin{gather} (-1-i \sqrt{3})(-4\sqrt{3}+4i)=16e^{\tfrac{\pi i}{3}+\tfrac{11\pi i}{6}}=16e^{\tfrac{13\pi i}{6}}=16e^{\tfrac{\pi i}{6}}= \\=16\left(\cos{\dfrac{\pi}{6}} +i\sin{\dfrac{\pi}{6}}\right)=16\left(\dfrac{\sqrt{3}}{2}+i\dfrac{1}{2} \right)= 8(\sqrt{3}+i). \end{gather}

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    thanks your answer has been well constructed and great.2012-12-20