I see the following identity in my book however they don't prove it so I am wondering how to prove that $e^{i \theta_1+i\theta_2}=e^{i\theta_1}e^{i\theta_2}$?
The definition is:
$$e^{i\theta} = \cos(\theta)+i\sin(\theta)$$
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1If you are able to use $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ then double angle formulae will do the trick. – fGDu94 Jan 27 '20 at 03:12
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I guess that this depends on what your definition is of $e^z$ for a complex number $z$. – Lubin Jan 27 '20 at 03:15
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@fGDu94 yes that is the correct defintion. – Robben Jan 27 '20 at 03:25
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@Lubin I have included the definition. – Robben Jan 27 '20 at 03:26
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More precisely, @fGDu94 , it’s the addition formula, not the double-angle formula. – Lubin Jan 27 '20 at 04:05
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double angle formulae is my catch all term for it, but yes, addition formulae is a good name for it – fGDu94 Jan 27 '20 at 04:06
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Hint $$e^{i\theta_1}e^{i\theta_2}=\bigl(\cos(\theta_1)+i\sin(\theta_1) \bigr) \bigl(\cos(\theta_2)+i\sin(\theta_2)\bigr)$$
Just open the brackets.
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@Robben You can also define the exponential function by its power series and then prove exp(z)exp(z')=exp(z+z') for any complex numbers z and z'. Let me know if you would like the details. – P. Lawrence Jan 27 '20 at 03:46