# Ugh, blanking out --- complex numbers, exponents...

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• July 7th 2010, 10:47 PM
jayshizwiz
Ugh, blanking out --- complex numbers, exponents...
solve $(1-i)^{2009}$

Attempt:

r = $\sqrt{2}$

tan $\Theta$ = -1 $\rightarrow$ tan $\Theta$ = $\frac{7\pi}{4}$

so...

$(1-i)^{2009}$ = $\sqrt{2}^{2009}(cis \frac{2009 \cdot 7\pi}{4})$

$\sqrt{2}^{2009} = \sqrt{2} \cdot 2^{1004}$

so... $(1-i)^{2009}$ = $\sqrt{2} \cdot 2^{1004}(cis \frac{2009 \cdot 7\pi}{4})$

can someone please tell me what I do with $(cis \frac{2009 \cdot 7\pi}{4})$

Thanks. I'm blanking out...
• July 7th 2010, 11:01 PM
Also sprach Zarathustra
Try to compute first:

(1-i)^2
• July 7th 2010, 11:18 PM
jayshizwiz
Quote:

Try to compute first:

(1-i)^2
I know the question is much easier and quicker that way, but I want to try and solve the question with De Mouvre's (or wtvr his name is) equation.
• July 7th 2010, 11:26 PM
mr fantastic
Quote:

Originally Posted by jayshizwiz
solve $(1-i)^{2009}$

Attempt:

r = $\sqrt{2}$

tan $\Theta$ = -1 $\rightarrow$ tan $\Theta$ = $\frac{7\pi}{4}$

so...

$(1-i)^{2009}$ = $\sqrt{2}^{2009}(cis \frac{2009 \cdot 7\pi}{4})$

$\sqrt{2}^{2009} = \sqrt{2} \cdot 2^{1004}$

so... $(1-i)^{2009}$ = $\sqrt{2} \cdot 2^{1004}(cis \frac{2009 \cdot 7\pi}{4})$

can someone please tell me what I do with $(cis \frac{2009 \cdot 7\pi}{4})$

Thanks. I'm blanking out...

Note that $\frac{2009 \cdot 7\pi}{4} = 3514 \pi + \frac{7 \pi}{4}$ ....
• July 7th 2010, 11:46 PM
sa-ri-ga-ma
Write the given problem as

$(1-i)^{2009} = (\sqrt{2})^{2009}[cos(-\frac{\pi}{4}) + sin(-\frac{\pi}{4})]^{2009}$

= $(\sqrt{2})^{2009}[cos(-\frac{2009\pi}{4}) + sin(-\frac{2009\pi}{4})]$

= $(\sqrt{2})^{2009}[cos(-\frac{(2008 + 1)\pi}{4}) + sin(-\frac{(2008 + 1)\pi}{4})]$

= $(\sqrt{2})^{2009}[cos(-\frac{\pi}{4}) + sin(-\frac{\pi}{4})]$

Now proceed.