1974 AHSME Problems/Problem 17

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Problem

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

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

Solution

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

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

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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