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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?

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    May I suggest looking through this for future questions: ftp://ftp.ams.org/ams/doc/amsmath/short-math-guide.pdf2012-12-20
  • 3
    If $z_1=\rho_1e^{i\theta_1}$ and $z_2=\rho_2e^{i\theta_2}$ then $z_1z_2=\rho_1\rho_2e^{\theta_1+\theta_2}$.2012-12-20
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    @Fernando: In addition to Todd's link, you can find some good starting points on how to format mathematics on the site [here](http://meta.math.stackexchange.com/a/464/264) and [here](http://meta.stackexchange.com/a/70559/161783). If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own [TeX.SE](http://tex.stackexchange.com/) site.2012-12-20
  • 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