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Mock AIME 1 2007-2008 Problems/Problem 5 - Revision history
2024-03-28T17:43:54Z
Revision history for this page on the wiki
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Azjps: correction
2008-04-03T01:13:12Z
<p>correction</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:13, 3 April 2008</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>\begin{align*}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>\begin{align*}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>&= 2^{17/2}\left<del class="diffchange diffchange-inline">(</del>\text{cis}\<del class="diffchange diffchange-inline">,</del>\frac{\pi}{4} - \text{cis}\<del class="diffchange diffchange-inline">,</del>\frac{<del class="diffchange diffchange-inline">3</del>\pi}{4}\right) \\</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>&= 2^{17/2}\left<ins class="diffchange diffchange-inline">[</ins>\text{cis}\<ins class="diffchange diffchange-inline">left(</ins>\frac{\pi}{4}<ins class="diffchange diffchange-inline">\right) </ins>- \text{cis}\<ins class="diffchange diffchange-inline">left(-</ins>\frac{\pi}{4}\right)<ins class="diffchange diffchange-inline">\right] </ins>\\</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>&= 2^{17/2}\left(<del class="diffchange diffchange-inline">2</del>\<del class="diffchange diffchange-inline">cos </del>\frac{\pi}{4}\right) \\</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>&= 2^{17/2}\left(<ins class="diffchange diffchange-inline">2i</ins>\<ins class="diffchange diffchange-inline">sin </ins>\frac{\pi}{4}\right) \\</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>&= 2^{17/2} \cdot 2 \cdot 2^{-1/2} = 2^<del class="diffchange diffchange-inline">9 </del>= \boxed{512}</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>&= 2^{17/2} \cdot 2 \cdot 2^{-1/2}<ins class="diffchange diffchange-inline">i </ins>= 2^<ins class="diffchange diffchange-inline">9i </ins>= \boxed{512}<ins class="diffchange diffchange-inline">\,i</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\end{align*}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\end{align*}</cmath></div></td></tr>
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Azjps
https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2007-2008_Problems/Problem_5&diff=24473&oldid=prev
Azjps: solution, hope i didn't mess up signs
2008-04-02T21:53:49Z
<p>solution, hope i didn't mess up signs</p>
<p><b>New page</b></p><div>== Problem 5 ==<br />
Let <math>S = (1+i)^{17} - (1-i)^{17}</math>, where <math>i=\sqrt{-1}</math>. Find <math>|S|</math>.<br />
<br />
== Solution == <br />
Rewriting the [[complex number]]s in polar notation form, <math>1+i = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}</math> and <math>1-i = \sqrt{2}\,\text{cis}\,-\frac{\pi}{4}</math>, where <math>\text{cis}\,\theta = \cos \theta + i\sin \theta</math>. By [[De Moivre's Theorem]], <br />
<cmath>\begin{align*}<br />
\left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\<br />
&= 2^{17/2}\left(\text{cis}\,\frac{\pi}{4} - \text{cis}\,\frac{3\pi}{4}\right) \\<br />
&= 2^{17/2}\left(2\cos \frac{\pi}{4}\right) \\<br />
&= 2^{17/2} \cdot 2 \cdot 2^{-1/2} = 2^9 = \boxed{512}<br />
\end{align*}</cmath><br />
<br />
== See also ==<br />
{{Mock AIME box|year=2007-2008|n=1|num-b=4|num-a=6|source=196980}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>
Azjps