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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 4"
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Notice that <math>(1+i)^{2}=2i</math>. We then have <math>(2i)^{50}=-2^{50}</math>.
 
Notice that <math>(1+i)^{2}=2i</math>. We then have <math>(2i)^{50}=-2^{50}</math>.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 3|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 5|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problems|Back to Exam]]
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 13:36, 24 November 2016

Problem

If $(1 + i)^{100}$ is expanded and written in the form $a + bi$ where $a$ and $b$ are real numbers, then $a =$

$\mathrm{(A) \ } -2^{50} \qquad \mathrm{(B) \ } 20^{50} - \frac{100!}{50!50!} \qquad \mathrm{(C) \ } \frac{100!}{(25!)^2 50!} \qquad \mathrm{(D) \ } 100! \left(-\frac 1{50!50!} + \frac 1{25!75!}\right) \qquad \mathrm{(E) \ } 0$

Solution

Notice that $(1+i)^{2}=2i$. We then have $(2i)^{50}=-2^{50}$.

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