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Can anyone give me some clues on how to simplify this expression while eliminating all imaginary terms? he general solution is given as

$$x=A_1e^{(-3+i)t}+A_2e^{(-3-i)t}$$

where

$$A_1=\frac{1-3i}{2} \quad A_2=\frac{1+3i}{2}$$

The question then asks given $e^{it}=\cos{t}+i\sin{t} $ and $ e^{-it}=\cos{t}-i\sin{t}$, rewrite the general solution eliminating all imaginary terms.

So

$$x=\dfrac{1-3i}{2}e^{(-3+i)t}+\frac{1+3i}{2}e^{(-3-i)t}$$

then

$$x=\dfrac{e^{-3}}{2}((1-3i)(\cos{t}+i\sin{t} )+(1+3i)(\cos{t}-i\sin{t} )$$

to be simplified without any leftovers.

Narasimham
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Buddy
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  • I am starting to think I have made an error somewhere up the chain. – Buddy Oct 01 '16 at 03:19
  • Are you sure that it is $(1-3i)$ and $(-3-i)$? check the two terms again. – msm Oct 01 '16 at 03:51
  • The general solution given was – Buddy Oct 01 '16 at 04:25
  • Thanks msm you are right. It should look like this The general solution is given as $x=A_1e^{(-3+i)t}+A_2e^{(-3-i)t}$ where $A_1=\frac{1-3i}{2}$ and $A_2=\frac{1+3i}{2}$

    The question then asks Given $e^{it}=\cos{t}+i\sin{t} $ and $ e^{-it}=\cos{t}-i\sin{t}$, rewrite the general solution eliminating all imaginary terms.

    So

    $x=\frac{1-3i}{2}e^{(-3+i)t}+\frac{1+3i}{2}e^{(-3-i)t}$

    then

    $x=\frac{e^{-3}}{2}((1-3i)(\cos{t}+i\sin{t} )+(1+3i)(\cos{t}-i\sin{t} )$

     – Buddy Oct 01 '16 at 04:46
  • Now when I combine like terms it all cancels out. Thanks for your help – Buddy Oct 01 '16 at 04:48
  • You need to update the question as well. I have given the answer. Ask if you need more details. – msm Oct 01 '16 at 04:50

3 Answers3

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consider $$z=(1-3i)e^{it}$$ then $$z^*=(1+3i)e^{-it}$$

Knowing that $z+z^*=2Re(z)$, your expression is $$\frac{e^{-3}}{2} (z+z^*)=e^{-3}Re(z)=e^{-3}\left(\cos(t)+3\sin(t)\right)$$

msm
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You need to notice $i^2=-1$ $$x=\frac{e^{-3}}2((-2-4i)cost+(2+4i)sint)$$ $$={e^{-3}}((sint-cost)+2(sint+cost)i)$$

Takahiro Waki
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  • Thanks for this. I found my error (see above). I see what you have done although i remains. Sorry to confuse – Buddy Oct 01 '16 at 04:51
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First note that , $i^{2}=-1$

$\frac{e^{-3}}{2}\left ( \left ( 1-3i \right )\left ( \cos t+i\sin t \right )+\left ( -3-i \right ) \left ( \cos t-i\sin t \right )\right )$
$=\frac{e^{-3}}{2}\left (-2\cos t+2\sin t-4i\cos t+4i\sin t \right )$
=$\frac{e^{-3}}{2}\left ( -4\cos t+4i\sin t+4\sin t-4i\cos t+\left ( 2\cos t-2\sin t \right ) \right )$
=$\frac{e^{-3}}{2}\left ( -4\left ( \cos t-i\sin t \right )-4i\left ( \cos t+i\sin t \right ) +2\left ( \cos t-\sin t \right )\right )$
=$\frac{e^{-3}}{2}\left ( -4e^{-it} -4ie^{it}+2\left ( \cos t-\sin t \right ) \right )$
=$\frac{e^{-3}}{2}\left ( -4\left ( e^{-it}+ie^{it} \right )+2\left ( \cos t-\sin t \right ) \right )$

theBookworm
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