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Find the product of all values of $(1+i\sqrt 3)^{\frac{3}{4}}$.

My try:

$(1+i\sqrt 3)^{\frac{3}{4}}=\exp (\frac{3}{4}(Log(1+i\sqrt 3)))=\exp(\frac{3}{4}(\log2+i\frac{\pi}{3}+2n\pi i))$.

I am kinda stuck on how to find the product of all values of the above expression. Can someone please help me out.

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$$1+i\sqrt3=2e^{i\pi/3}$$ The fourth roots of this number have common magnitude $2^{1/4}$ and arguments $\pi/12,7\pi/12,-5\pi/12,-11\pi/12$. The $\frac34$-powers thus have common magnitude $2^{3/4}$ and arguments $\pi/4,7\pi/4\equiv-\pi/4,-5\pi/4\equiv3\pi/4,-11\pi/4\equiv-3\pi/4$. The producf of all the four possible values of $(1+i\sqrt3)^{3/4}$ therefore has magnitude equal to the product of the magnitudes of each possible value, which is $8$, and argument equal to the sum of the arguments of each possible value, which is $0$. The final answer is $8$.

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  • $\begingroup$ I am stuck at how you found that there are $4$ arguements, Wont there be $n$ arguments? $\endgroup$ – Math_Freak Nov 1 '20 at 14:15
  • $\begingroup$ @Math_Freak $n=4$ here. A rational $p/q$-th root of a complex number (where the fraction is lowest-terms) has $q$ possible values. $\endgroup$ – Parcly Taxel Nov 1 '20 at 16:19

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