Question 1075983
{{{ e^(a+bi) = (e^a)*(e^(bi)) = (e^a)*(cos(b) + i*sin(b)) }}}

Let a=0 so {{{e^a = e^0 = 1}}}
Let {{{b=pi}}} so {{{e^(bi) = e^((pi)i) = cos(pi) + i*sin(pi) = -1 + 0*i = -1 }}}

Put the two together and the  result is the famous Euler's Identity:   
{{{e^((pi)i) + 1 = 0 }}}