Going through some exercises in complex functions and I get $\frac{1}{2}(ee^{-i\sqrt{3}} + e^{-1}e^{i\sqrt{3}})$ but not sure if I can simplify this further?
Thanks
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Yes, you can, you have to use that $e^{ix}=cos(x)+isin(x)$ and you get to the final result.
S -
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Is it always better to express these functions in terms of sin and cos then? Thanks for the reply – Ashley Dec 07 '15 at 19:38
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@Ashley I wouldn't say better, but for me it's always more natural to imagine the behaviour of the function if I express it in terms of sines and cosines. You're welcome. – S - Dec 07 '15 at 19:45
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Notice, $$\cosh(1-i\sqrt 3)=\frac{e^{1-i\sqrt 3}+e^{-1+i\sqrt 3}}{2}$$ $$=\frac{ee^{-i\sqrt 3}+e^{-1}e^{i\sqrt 3}}{2}$$ $$=\frac{e(\cos(\sqrt 3))-i\sin(\sqrt 3))+e^{-1}(\cos(\sqrt 3)+i\sin(\sqrt 3))}{2}$$ $$=\frac{(e+e^{-1})\cos(\sqrt 3)-i(e-e^{-1})\sin(\sqrt 3)}{2}$$ $$=\left(\frac{e+e^{-1}}{2}\right)\cos(\sqrt 3)-i\left(\frac{e-e^{-1}}{2}\right)\sin(\sqrt 3)$$ $$=\cosh (1)\cos(\sqrt 3)-i\sinh (1)\sin(\sqrt 3)$$
Harish Chandra Rajpoot
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it is $$\cosh \left( 1 \right) \cos \left( \sqrt {3} \right) -i\sinh \left( 1 \right) \sin \left( \sqrt {3} \right) $$
Dr. Sonnhard Graubner
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