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Algebra Difficulty 4.7 AIME Find the answer

2. Let the complex number z=1+i1−iz=\frac{1+\mathrm{i}}{1-\mathrm{i}} (where i\mathrm{i} is the imaginary unit). Then C20140+C20141z+⋯+C20142014z2014=()\mathrm{C}_{2014}^{0}+\mathrm{C}_{2014}^{1} z+\cdots+\mathrm{C}_{2014}^{2014} z^{2014}=(\quad).

Pick one

Solution

2. B.

From the problem, we have z=iz=\mathrm{i}.
Notice that, (1+i)2=2i(1+\mathrm{i})^{2}=2 \mathrm{i}.
Thus, the required expression is
(1+z)2014=(1+i)2014=(2i)1007=(2i)2×503×2i=−21007i. \begin{array}{l} (1+z)^{2014}=(1+\mathrm{i})^{2014}=(2 \mathrm{i})^{1007} \\ =(2 \mathrm{i})^{2 \times 503} \times 2 \mathrm{i}=-2^{1007} \mathrm{i} . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.