Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 5"

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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\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) \\
 
\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) \\
&= 2^{17/2}\left(\text{cis}\,\frac{\pi}{4} - \text{cis}\,\frac{3\pi}{4}\right) \\
+
&= 2^{17/2}\left[\text{cis}\left(\frac{\pi}{4}\right) - \text{cis}\left(-\frac{\pi}{4}\right)\right] \\
&= 2^{17/2}\left(2\cos \frac{\pi}{4}\right) \\
+
&= 2^{17/2}\left(2i\sin \frac{\pi}{4}\right) \\
&= 2^{17/2} \cdot 2 \cdot 2^{-1/2} = 2^9 = \boxed{512}
+
&= 2^{17/2} \cdot 2 \cdot 2^{-1/2}i = 2^9i = \boxed{512}\,i
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  

Latest revision as of 21:13, 2 April 2008

Problem 5

Let $S = (1+i)^{17} - (1-i)^{17}$, where $i=\sqrt{-1}$. Find $|S|$.

Solution

Rewriting the complex numbers in polar notation form, $1+i = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}$ and $1-i = \sqrt{2}\,\text{cis}\,-\frac{\pi}{4}$, where $\text{cis}\,\theta = \cos \theta + i\sin \theta$. By De Moivre's Theorem, \begin{align*} \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) \\ &= 2^{17/2}\left[\text{cis}\left(\frac{\pi}{4}\right) - \text{cis}\left(-\frac{\pi}{4}\right)\right] \\ &= 2^{17/2}\left(2i\sin \frac{\pi}{4}\right) \\ &= 2^{17/2} \cdot 2 \cdot 2^{-1/2}i = 2^9i = \boxed{512}\,i \end{align*}

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15