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| rev | Admin | (Created page with "'''Solution: B''' <math>K a_{\overline{12 \mid}}=2 K a_{\overline{4} \mid}</math> so <math>a_{\overline{12 \mid}}=2 a_{\overline{4} \mid}</math> and <math>\frac{1-(1+i)^{-12}}{i}=2 \frac{1-(1+i)^{-4}}{i}</math>. Let <math>x=(1+i)^{-4}</math>. Then <math>1-x^3=2(1-x)</math> so <math>x^3-2 x+1=0</math>. One root is clearly <math>x=1</math> so <math>(x-1)</math> is a factor. Thus <math>(x-1)\left(x^2+x-1\right)=0</math>. By the quadratic formula, <math>x=\frac{-1 \pm \sqr...") | Nov 26'23 at 21:41 | +908 |