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I'm supposed to show $ z^{10} $, when z = $ \frac{1+ \sqrt{3i} }{1- \sqrt{3i} } $

I can work it out to $ \frac{(1+\sqrt{3}\sqrt{i})^{10}}{(1-\sqrt{3}\sqrt{i})^{10}} $

However this is inconclusive because I need to show $ z^{10} $ in the form x+yi, and I can't figure out the real and imaginary parts from from this answer because of the exponent.

What must I do?

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    $\begingroup$ what you have tried ? $\endgroup$ – user162343 Dec 9 '14 at 19:00
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    $\begingroup$ Have you considered the polar form of the number? $\endgroup$ – abiessu Dec 9 '14 at 19:03
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    $\begingroup$ Do you really mean $\sqrt{3}i$ instead of $\sqrt{3i}$? $\endgroup$ – Michael Albanese Dec 9 '14 at 19:04
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    $\begingroup$ @MichaelAlbanese The latter is what is says. Did you mean to swap these terms? $\endgroup$ – AlexR Dec 9 '14 at 19:04
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    $\begingroup$ @AlexR: Presumably what Michael Albanese means is something along the lines of "Maybe you really mean $\sqrt{3}i$ instead of $\sqrt{3i}$?"? $\endgroup$ – ruakh Dec 10 '14 at 4:53
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Hint
First you should obtain the polar form of $z$, $$z = r e^{i\varphi}$$ For $\varphi\in[0,2\pi)$ and $r>0$ real. Then $$z^{10} = r^{10} e^{i10\varphi}$$ Where you can simplify $10\varphi\!\!\! \mod\! 2\pi$.

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hint

$z^n=(re^{i\theta})^n=r^ne^{in\theta}$

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This is similar to an AMIE problem which was something along the lines:

$$z=\frac{1+\sqrt{3}i}{1-\sqrt{3}i}$$

What is $z^{10}$?

Which was most easily solved as $z^{10}=((z^2)^2)^2*z^2$ As the terms collapsed so you only needed 4 terms for any multiplication. A similar strategy would probably work here.

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  • $\begingroup$ I believe $(z^2)^2*z^2$ would be $z^6$. Unless you could clarify what you meant by "collapse". $\endgroup$ – Teepeemm Dec 9 '14 at 21:36
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    $\begingroup$ Typo: $\left( \left( z^2 \right)^2 \right)^2 \cdot z^2$, yes? $\endgroup$ – Charles Baker Dec 9 '14 at 23:17
  • $\begingroup$ Yep, sorry typo. By collapse I meant cancel/simplify. $\endgroup$ – Rick Dec 10 '14 at 12:45
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It seems to me that @MichaelAlbanese is right. It should be $ \frac{1+\sqrt{3} i}{1-\sqrt{3} i} $. After removing any complex part from denominator we get $$ \frac{1+\sqrt{3} i}{1-\sqrt{3} i} = \frac{1+\sqrt{3} i}{1-\sqrt{3} i} \frac{1+\sqrt{3} i}{1+\sqrt{3} i} = \frac{-2 + 2 \sqrt{3} i}{4} = -\frac{1 - \sqrt{3} i}{2} $$

We can transfer it into polar form, which is $ {e}^{\frac{2}{3} \pi i} $

And now the answer is quite easy to find $$ {{e}^{\frac{2}{3} \pi i}}^{10} = {e}^{10 \frac{2}{3} \pi i} = {e}^{\frac{20}{3} \pi i} = {e}^{\frac{18 + 2}{3} \pi i} = {e}^{{6 \pi i} + {\frac{2}{3} \pi i}} = {e}^{6 \pi i} {e}^{\frac{2}{3} \pi i} = {e}^{\frac{2}{3} \pi i} $$ So the final answer is that $ z^{10} = z = -\frac{1 - \sqrt{3} i}{2}$

BTW, if really $ z = \frac{1+\sqrt{3 i}}{1-\sqrt{3 i}} $ you have big troubles because its polar form is something like that: $$ z = \frac{\sqrt{{\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }^{2}+\frac{3}{2}}\,{e}^{i\,\left( \mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }\right) +\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }\right) +\pi \right) }}{\sqrt{{\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }^{2}+\frac{3}{2}}} $$ and the answer is $$ z^{10} = \frac{{\left( {\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }^{2}+\frac{3}{2}\right) }^{5}\,{e}^{i\,\left( \mathrm{atan}\left( \frac{\mathrm{sin}\left( 10\,\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }\right) \right) }{\mathrm{cos}\left( 10\,\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( \frac{\sqrt{3}}{\sqrt{2}}+1\right) }\right) \right) }\right) -\mathrm{atan}\left( \frac{\mathrm{sin}\left( 10\,\left( -\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }\right) -\pi \right) \right) }{\mathrm{cos}\left( 10\,\left( -\mathrm{atan}\left( \frac{\sqrt{3}}{\sqrt{2}\,\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }\right) -\pi \right) \right) }\right) \right) }}{{\left( {\left( 1-\frac{\sqrt{3}}{\sqrt{2}}\right) }^{2}+\frac{3}{2}\right) }^{5}} $$

(thanks Maxima…)

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