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I need to tranform: $$z:=-4e^{{\pi}i/3}$$ to the polar (I know it's almost polar) and cartesian form, i.e. find x and y coordiantes.
$$-4e^{{\pi}i/3}=-4(\cos(\frac\pi3)+\sin(\frac\pi3)i)=4(-1)(\cos(\frac\pi3)+\sin(\frac\pi3)i)=4(-\cos(\frac\pi3)-\sin(\frac\pi3)i)$$ Don't really know how to continue.
So what's the trick behind this complex number?

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    $\begingroup$ Just put $\cos(\pi/3) = 1/2$ and $\sin(\pi/3) = \sqrt{3}/2$. $\endgroup$ – Rodrigo Dias Feb 7 '18 at 15:35
  • $\begingroup$ There's little to continue. $z$ started out in polar form. To finish cartesian, just distribute the $4$ over the parentheses in your last expression to find the real and imaginary parts. You might explicitly evaluate the trig functions in terms of $\sqrt{3}$. $\endgroup$ – Ethan Bolker Feb 7 '18 at 15:36
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    $\begingroup$ it is $$z=-2+2i\sqrt{3}$$ $\endgroup$ – Dr. Sonnhard Graubner Feb 7 '18 at 15:40
  • $\begingroup$ Usually by definision the form $$z:=\rho e^{i\theta}$$ is defined with $ \rho \ge0$ $\endgroup$ – user Feb 7 '18 at 15:41
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    $\begingroup$ @GNUSupporter will remember to do it in the future $\endgroup$ – user3125470 Feb 7 '18 at 15:44
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Note that

  • $\cos(\frac\pi3)=\frac12$
  • $\sin(\frac\pi3)=\frac{\sqrt3}2$

thus

$$4\left(-\cos(\frac\pi3)-\sin(\frac\pi3)i\right)=-2-2\sqrt3\,i$$

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Multiplying a complex number by $-1$ is the same as rotating it (if we think of it as an arrow from the origin) 180 degrees in the complex plane, to make it point in the opposite direction. So you can remove the minus sign and compensate the angle by adding (or subtracting) $\pi$.

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Continue your work by substituting $\sin \frac\pi3 = \frac{\sqrt3}{2}$ and $\cos\frac\pi3 = \frac12$.

$$\begin{aligned} -4e^{{\pi}i/3}&=-4(\cos(\frac\pi3)+\sin(\frac\pi3)i)\\ &=4(-1)(\cos(\frac\pi3)+\sin(\frac\pi3)i)\\ &=4(-\cos(\frac\pi3)-\sin(\frac\pi3)i)\\ &=4\left(-\frac12-\frac{\sqrt3}{2}i\right)\\ &=-2-2\sqrt3 \,i \end{aligned}$$

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