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Given a complex number $z_j$ such that $$z_j\in\{a_1+b_1 i,\ a_2+b_2i, \ ...\ ,a_n+b_ni\}$$ is there formula for calculating $$z_1 \cdot z_2 \cdot \dots \cdot z_n =\prod_j z_j?$$ For two complex numbers you use the distributive law, $$(a_1+b_1i)\cdot(a_2+b_2i)=a_1a_2+a_1b_2i+a_2b_1i-b_1b_2$$ yet it seems to become infinitely more complicated as the number of complex numbers increases. Any help would be appreciated.

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  • $\begingroup$ You can rewrite the complex numbers in the form $re^{i\theta}$. Multiplying them should be easier now. $\endgroup$ – Pranav Arora Jun 30 '14 at 5:27
  • $\begingroup$ I am aware, I was just wondering if there is a formula for when the complex numbers are in Cartesian or rectangular form. Pointless question but I am curious. $\endgroup$ – user124862 Jun 30 '14 at 5:29
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Let $A=(a_1,a_2,\dots,a_n)$, $B=(b_1,b_2,\dots,b_n)$, and $[n]=\{1,2,\dots,n\}$. It appears that $$ \text{Re}\prod_{j} z_j = \sum_{S \subseteq [n], n-|S|\text{ even}} (A_S^*)(B_{[n]\setminus S}^*) (-1)^{\frac{n-|S|}{2}} $$ and $$ \text{Im}\prod_{j} z_j = \sum_{S \subseteq [n], n-|S|\text{ odd}} (A_S^*) (B_{[n]\setminus S}^*) (-1)^{\frac{n-|S|-1}{2}}, $$ where $A_S^*$ denotes the product of the elements in $\{a_i \in A: i \in S\}$. I verified the equations for $n\le 4$, so they can probably be easily proved with induction.

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  • $\begingroup$ I had a similar approach. Thank you for your efforts. $\endgroup$ – user124862 Jun 30 '14 at 7:00
  • $\begingroup$ You're welcome. It was a nice exercise. $\endgroup$ – fahrbach Jun 30 '14 at 7:01
  • $\begingroup$ A lecturer told me it could be solved using a combination of the multinomial and binomial expansions. This is going to be a fun night. $\endgroup$ – user124862 Jun 30 '14 at 7:40
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in the $(a_1+b_1i)\cdot(a_2+b_2i)\cdot ...\cdot(a_2+b_2i)$ expansion we will have $2^n$ terms like $a_1b_2b_3...a_n$ and $a_1a_2a_3...b_n$ and others. terms that are in this representation are from two category. the first category($c_1$) is the category of terms that number of 'b's in it, is even and the second category($c_2$) is the category of terms that number of 'b's in it, is ood. now, it is simple to show that

$$(a_1+b_1i)\cdot(a_2+b_2i)\cdot ...\cdot(a_2+b_2i)=(\Sigma_{s\in c_1} (-1)^{\frac{n(s)}{2}}s)+i\cdot(\Sigma_{s\in c_2} (-1)^{\frac{n(s)-1}{2}}s)$$

that $n(s)$ is a function that counts the number of 'b's in every term like $a_1b_2b_3...a_n$.

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