how would you use the complex exponential to evaluate:
Thank you.
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1Note that $e^{i\theta} = \cos\theta + i\sin\theta$ – Prahlad Vaidyanathan May 31 '14 at 06:02
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2Hint: $\sin(t) = \frac{e^{it}-e^{-it}}{2i}$. – Christopher K May 31 '14 at 06:03
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Is there a question here? – copper.hat May 31 '14 at 06:17
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Hint
The $124^{th}$ derivative is effectively $$\left(-1+i \sqrt{3}\right)^{124} e^{\left(-1+i \sqrt{3}\right) t}$$ But $$\left(-1+i \sqrt{3}\right)=2\Big(\cos \frac{2\pi}{3}+i~\sin \frac{2\pi}{3}\Big)$$ So, using Moivre theorem $$\left(-1+i \sqrt{3}\right)^{124}=2^{124}\Big(\cos \frac{248\pi}{3}+i~\sin \frac{248\pi}{3}\Big)=2^{123}\left(-1+i \sqrt{3}\right)$$ Multiply this result by $e^{\left(-1+i \sqrt{3}\right) t}$ and expand again. Finally take the imaginary part of the result.
I am sure that you can take from here.
Claude Leibovici
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Observe that $e^{-t} \sin(\sqrt{3}t) = \text{Im}e^{(-1+i\sqrt{3})t}$, where Im is for the imaginary part. Now compute the derivative of this exponential function and then go for the imaginary part.
Anurag A
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I've gotten up to Im [(-1+√3i)^124 x e^(-i+√3i)t], but how do I solve for ((-1+√3i)^124) ? – user151408 May 31 '14 at 06:51