1974 AHSME Problems/Problem 17

Problem

If $i^2=-1$ , then $(1+i)^{20}-(1-i)^{20}$ equals

$\mathrm{(A)\ } -1024 \qquad \mathrm{(B) \ }-1024i \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 1024 \qquad \mathrm{(E) \ }1024i$

Notice that $(1+i)^2=2i$ and $(1-i)^2=-2i$ . Therefore,

$(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \boxed{\text{C}}.$

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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