My Math Forum Express in a+bi form

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 September 7th, 2010, 08:50 PM #1 Senior Member   Joined: Sep 2009 Posts: 251 Thanks: 0 Express in a+bi form Define: $\cos(z) = \frac{e^{iz}+e^{-iz}}{2} \\ e^z=e^x e^{iy}$ Evaluate: $\cos(\frac{\pi}{4}i+c)$ My answer (is there a way to make these symbols larger?): $\cos(\frac{\pi}{4}i+c) = \frac{e^{i(\frac{\pi}{4}i+c)}+e^{-i(\frac{\pi}{4}i+c)}}{2} = \frac{e^{ci-\frac{\pi}{4}}+e^{-ci+\frac{\pi}{4}}}{2} = \\ \hspace{40 mm} \frac{e^{ci}e^{-\frac{\pi}{4}}+e^{-ci}e^{\frac{\pi}{4}}}{2} = \frac{e^{2ci}+e^{\frac{\pi}{4}}}{2e^{\frac{\pi}{4} }e^{ci}} = ...$ Where the ellipses (...) means "What comes next?" or "Is this right so far?" Thanks.
 September 8th, 2010, 03:49 AM #2 Senior Member   Joined: Sep 2009 Posts: 251 Thanks: 0 Re: Express in a+bi form Same exact post, but a little bit more readable. Define: $\cos(z) = \frac12 \left( e^{iz}+e^{-iz} \right) \\ e^z=e^x e^{iy}$ Evaluate: $\cos(\frac{\pi}{4}i+c)$ My answer (is there a way to make these symbols larger?): $\cos(\frac{\pi}{4}i+c) = \frac12 \left( e^{i(\frac{\pi}{4}i+c)}+e^{-i(\frac{\pi}{4}i+c)} \right) = \frac12 \left(e^{ci-\frac{\pi}{4}}+e^{-ci+\frac{\pi}{4}} \right) = \\ \hspace{40 mm} \frac12 \left( e^{ci}e^{-\frac{\pi}{4}}+e^{-ci}e^{\frac{\pi}{4}} \right) = \frac{e^{2ci}+e^{\frac{\pi}{4}}}{2e^{\frac{\pi}{4} }e^{ci}} = ...$ Where the ellipses (...) means "What comes next?" or "Is this right so far?" Thanks.
 September 10th, 2010, 06:49 PM #3 Member   Joined: Aug 2010 Posts: 49 Thanks: 0 Re: Express in a+bi form Hi, I am just thinking about the word "evaluate"... does it mean simplify or something? In which sence? I found the original expression pretty simple... Best regards, Jens
 September 10th, 2010, 07:04 PM #4 Member   Joined: Aug 2010 Posts: 49 Thanks: 0 Re: Express in a+bi form Replying myself... just an idea. Why you don't use the normal decomposition $cos(\alpha+\beta)=cos(\alpha) cos(\beta) - sin(\alpha) sin(\beta)$ I mean, looks not that bad ...and again, me idiot: I should have read the header in this case. Damn... Best regards, Jens
September 11th, 2010, 10:22 AM   #5
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Re: Express in a+bi form

Quote:
 Originally Posted by MSUMathGuy Same exact post, but a little bit more readable. Define: $\cos(z) = \frac12 \left( e^{iz}+e^{-iz} \right) \\ e^z=e^x e^{iy}$ Evaluate: $\cos(\frac{\pi}{4}i+c)$ My answer (is there a way to make these symbols larger?): $\cos(\frac{\pi}{4}i+c) = \frac12 \left( e^{i(\frac{\pi}{4}i+c)}+e^{-i(\frac{\pi}{4}i+c)} \right) = \frac12 \left(e^{ci-\frac{\pi}{4}}+e^{-ci+\frac{\pi}{4}} \right) = \\ \hspace{40 mm} \frac12 \left( e^{ci}e^{-\frac{\pi}{4}}+e^{-ci}e^{\frac{\pi}{4}} \right) = \frac{e^{2ci}+e^{\frac{\pi}{4}}}{2e^{\frac{\pi}{4} }e^{ci}} = ...$ Where the ellipses (...) means "What comes next?" or "Is this right so far?" Thanks.
The goal was to keep the answer as a complex number in the form of a+bi.

The trick was to use Euler's formula, which says:
$e^{i \theta}= \cos(\theta) + i \cdot \sin(\theta)$

The other trick was to work out this case of Euler's formula:
$e^{-i \theta}= \cos(\theta) - i \cdot \sin(\theta)$

like this (if you can read this teeny print):
$e^{-i \theta}= \frac{1}{e^{i \theta}} = \frac{1}{\cos(\theta) + i \cdot \sin(\theta)} = \frac{1}{\cos(\theta) + i \cdot \sin(\theta)} \cdot \frac{\cos(\theta) - i \cdot \sin(\theta)}{\cos(\theta) - i \cdot \sin(\theta)} = \frac{\cos(\theta) - i \cdot \sin(\theta)}{\cos^2(\theta) + \sin^2(\theta)} = \cos(\theta) - i \cdot \sin(\theta)$

or even easier, using trig identites:
$e^{-i \theta}= \cos( -\theta ) + i \cdot \sin( -\theta) = \cos( \theta ) - i \cdot \sin( \theta)$

Then, you can get the final answer from this point:
$\frac12 \left( e^{ci} \cdot e^{-\frac{\pi}{4}}+e^{-ci} \cdot e^{\frac{\pi}{4}} \right)$

by substituting for all four exponentials, and then simplifying (I'm not sure I did all of this right):
$\frac12 \left( e^{ci} \cdot e^{-\frac{\pi}{4}}+e^{-ci} \cdot e^{\frac{\pi}{4}} \right) =
\frac12 \left[ \left(\cos(c) + i \cdot \sin(c) \right) \cdot
\left(\cos(\frac{\pi}{4}) - i \cdot \sin(\frac{\pi}{4}) \right) +
\left(\cos(c) - i \cdot \sin(c) \right) \cdot
\left(\cos(\frac{\pi}{4}) + i \cdot \sin(\frac{\pi}{4}) \right) \right] = \\
\hspace{40 mm} \frac12 \left[ \left(\cos(c) + i \cdot \sin(c) \right) \cdot (-i) +
\left(\cos(c) - i \cdot \sin(c) \right) \cdot (i) \right] =
\frac12 \left[ \left( -i \cdot \cos(c) + \sin(c) \right) +
\left(i \cdot \cos(c) - \sin(c) \right) \right] = 0$

 September 12th, 2010, 06:01 PM #6 Senior Member   Joined: Sep 2009 Posts: 251 Thanks: 0 Re: Express in a+bi form I posted too soon. There is an error in that last post. You can't expand $e^{\frac{\pi}{4}}$ using Euler's formula. This is the whole problem with, I think, no mistakes: Define: $\cos(z) = \frac12 \left( e^{iz}+e^{-iz} \right) \\ e^z=e^x e^{iy}$ Evaluate: $\cos(\frac{\pi}{4}i+c)$ Use Euler's formula: $e^{i \theta}= \cos(\theta) + i \cdot \sin(\theta)$ $e^{-i \theta}= \cos(\theta) - i \cdot \sin(\theta)$ Solution (I think): $\cos(\frac{\pi}{4}i+c) = \frac12 \left( e^{i(\frac{\pi}{4}i+c)}+e^{-i(\frac{\pi}{4}i+c)} \right) = \frac12 \left(e^{ci-\frac{\pi}{4}}+e^{-ci+\frac{\pi}{4}} \right) = \frac12 \left( e^{ci}e^{-\frac{\pi}{4}}+e^{-ci}e^{\frac{\pi}{4}} \right) = \\ \hspace{40 mm} \frac12 \left[ e^{-\frac{\pi}{4}} \cdot \left(\cos(c) + i \cdot \sin(c) \right) + e^{\frac{\pi}{4}} \cdot \left(\cos(c) - i \cdot \sin(c) \right) \right] = \frac12 \left[ (e^{\frac{\pi}{4}} + e^{\frac{-\pi}{4}}) \cdot \cos(c) \right]$ That's as simplified as I can get it. There is no imaginary component.
 October 6th, 2010, 07:18 PM #7 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Re: Express in a+bi form let c = a + i*b where a and b are real. then $cos(\frac{\pi}{4}i + c) \,= \, cos(a + i*(\frac{\pi}{4} + 1)) \, = \, \frac{e^{i(a + i*(\frac{\pi}{4} + 1))} + e^{-i(a + i*(\frac{\pi}{4} + 1))}}{2} \, = \, \frac{1}{2}(cos(a) + isin(a))e^{-(\frac{\pi}{4} + 1)} + \frac{1}{2}(cos(a) - isin(a))e^{(\frac{\pi}{4} + 1)}$ $\,= \, cos(a)$$\frac{e^{-(\frac{\pi}{4} + 1)} + e^{(\frac{\pi}{4} + 1)}}{2}$$ + i*(-sin(a)) $$\frac{e^{(\frac{\pi}{4} + 1)} - e^{-(\frac{\pi}{4} + 1)}}{2}$$$ Giving us the more aesthetically pleasing answer (in z = Re(z) + i*Im(z) form): $\,= \, cos(a)\cosh(\frac{\pi}{4} + 1 ) \, + \, i \, (\, -sin(a) \sinh{(\frac{\pi}{4} + 1)} \, )$ The cosh and sinh are hyperbolic functions, which are defined by the exponential substitutions I utilized to get to the last step. http://en.wikipedia.org/wiki/Hyperbolic_function
 October 7th, 2010, 07:06 AM #8 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Re: Express in a+bi form I just noticed I somehow forgot the b in the expansion $c \,= \, a + i*b$ (where a and b are real valued) , so wherever you see $(\frac{\pi}{4} + 1)$ in the above it should be $(\frac{\pi}{4} + 1 + b)$. The final answer is thus supposed to be $cos(\frac{\pi}{4}i + c) \,= \, cos(a + i\, (\frac{\pi}{4} + 1 + b)) \, = \, cos(a)\cosh(\frac{\pi}{4} + 1 + b) + i\, ( \, -sin(a)\sinh(\frac{\pi}{4} + 1 + b) \, )$

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