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

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.

$$1+i\sqrt3=2e^{i\pi/3}$$ 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 $$8$$.

• I am stuck at how you found that there are $4$ arguements, Wont there be $n$ arguments? – Math_Freak Nov 1 '20 at 14:15
• @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. – Parcly Taxel Nov 1 '20 at 16:19