Question 1075784
<pre>
{{{e^(a + bi) = e^a(cos(b) + i*sin(b))}}}

Just substitute {{{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}}}

Edwin</pre>