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Given $A$ and $B$ are complex numbers.

I want to request anyone who might know any formulas for expanding this following expression. $$ |A-B|^{2n}$$ where $n$ is an integer.

The one that I commonly used for an order of $2$, i.e. $n=1$ is $$ |A-B|^{2}= (A-B)(A-B)^*$$ where the $[.]^*$ means the Hermitian adjoint.

So I want to see if I can increase the order from $n=1$ to $n=\{2,3,4,5...\}$ Is there such rule? Maybe like a Pascal triangle theorem and/or Binomial property?

Thanks in advance!

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Specifically for $2n$, you can use what you already stated above: $|A - B|^{2n} = (|A - B|^2)^n = ((A - B)(A - B)^*)^n = (A - B)^n(A - B)^{* n}$. If you have $2n + 1$, then this obviously becomes $|A - B|(A - B)^n(A - B)^{*n}$. –  William Mar 29 '12 at 9:07

2 Answers 2

up vote 3 down vote accepted

$$ \begin{eqnarray} \left|A-B\right|^{2n} &=& \left|(A-B)^n\right|^2 \\ &=& \left|\sum_{k=0}^n\binom nkA^k(-B)^{n-k}\right|^2 \\ &=& \left(\sum_{k=0}^n\binom nkA^k(-B)^{n-k}\right)\left(\sum_{k=0}^n\binom nkA^{*k}(-B^*)^{n-k}\right)\;. \end{eqnarray} $$

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What about $|A-B|^{2n}=(|A-B|^2)^n=\left((A-B)(A-B)^* \right)^n$?

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