# Using polar form find product

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.ams.org/ams/doc/amsmath/short-math-guide.pdf –  Todd Wilcox Dec 20 '12 at 22:36
If $z_1=\rho_1e^{i\theta_1}$ and $z_2=\rho_2e^{i\theta_2}$ then $z_1z_2=\rho_1\rho_2e^{\theta_1+\theta_2}$. –  Sigur Dec 20 '12 at 22:38
@Fernando: In addition to Todd's link, you can find some good starting points on how to format mathematics on the site here and here. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Dec 20 '12 at 22:38
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 $\theta =\tan^{-1}(b/a)$. –  Todd Wilcox Dec 20 '12 at 22:40

## 1 Answer

\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. –  Fernando Martinez Dec 20 '12 at 23:20