Problem :

The angle made by the complex number $\frac{1}{(\sqrt{3}+i)^{100}}$ with the positive real axis is ( options)

(a) $240^{\circ}$

(b) $140^{\circ}$

(c) $120^{\circ}$

(d) $260^{\circ}$

My approach :

We can write the given form as :

$(\frac{\sqrt{3} -i}{(2)})^{100}$

Its real part is $\frac{\sqrt{3}}{2}$ and with positive real axis the angle made by cos$\theta = 30^{\circ}$ but this is wrong ..please suggest..


Let $\displaystyle z= \sqrt{3}+i = 2 e^{\dfrac{i\pi}6}$.

Then $\displaystyle {1 \over z^{100}} = {1 \over 2^{100}} e^{\dfrac{-i 100\pi}6}$.

Since your question is using degrees, we get $-100 \cdot 30 ^ \circ = -3000^\circ$, which is equivalent to $240^\circ$ (since $-3000+9 \cdot 360 = 240$).


$\frac{1}{(\sqrt{3}+i)^{100}}=\frac{1}{2^{100} (\frac{\sqrt{3}}{2}+i\frac12)^{100}}=(2 (\cos \frac{\pi}{6} +i\sin \frac{\pi}{6}))^{-100}$

Use de_Moivre's formula http://pl.wikipedia.org/wiki/Wz%C3%B3r_de_Moivre%E2%80%99a

  • $\begingroup$ @copper.hat Thank you. I fixed it $\endgroup$ – nadia-liza Jan 21 '14 at 6:34

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