# Complex number calculation

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?

• what you have tried ? – user162343 Dec 9 '14 at 19:00
• Have you considered the polar form of the number? – abiessu Dec 9 '14 at 19:03
• Do you really mean $\sqrt{3}i$ instead of $\sqrt{3i}$? – Michael Albanese Dec 9 '14 at 19:04
• @MichaelAlbanese The latter is what is says. Did you mean to swap these terms? – AlexR Dec 9 '14 at 19:04
• @AlexR: Presumably what Michael Albanese means is something along the lines of "Maybe you really mean $\sqrt{3}i$ instead of $\sqrt{3i}$?"? – ruakh Dec 10 '14 at 4:53

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 $10\varphi\!\!\! \mod\! 2\pi$.

hint

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

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.

• I believe $(z^2)^2*z^2$ would be $z^6$. Unless you could clarify what you meant by "collapse". – Teepeemm Dec 9 '14 at 21:36
• Typo: $\left( \left( z^2 \right)^2 \right)^2 \cdot z^2$, yes? – Charles Baker Dec 9 '14 at 23:17
• Yep, sorry typo. By collapse I meant cancel/simplify. – Rick Dec 10 '14 at 12:45

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…)