Question 1075983
{{{e^(a+bi)=e^a*e^(bi)}}} 
OK so when {{{a=0}}} and {{{b=pi}}},
{{{e^(0+pi*i)=e^0*(cos(pi)+i*sin(pi))}}} 
{{{e^(pi*i)=1(-1+i*0)}}}
{{{e^(pi*i)=-1}}}
{{{e^(pi*i)+1=0}}}