Question 389436
<pre><font face = "batangche" color = "indigo"><b>
{{{i^2 = -1}}} by definition of i

By Euler's equation:

{{{e^(i*theta) = cos(theta) + i*sin(theta)}}}

Substitute <font face = "symbol">p</font>
 for <font face = "symbol">q</font>

{{{e^(i*pi) = cos(pi) + i*sin(pi)}}}

{{{e^(i*pi) =  -1 + i*0}}}

{{{e^(i*pi) = -1}}}

Go back to Euler's equation.

{{{e^(i*theta) = cos(theta) + i*sin(theta)}}}

Substitute {{{pi/2}}}
 for <font face = "symbol">q</font>

{{{e^(i*expr(pi/2)) = cos(pi/2) + i*sin(pi/2)}}}

{{{e^(i*pi/2) =  0 + i*1}}}

{{{e^(i*pi/2) = i}}}

Square both sides:

{{{(e^(i*pi/2))^2 = i^2}}}

{{{(e^(i*pi/cross(2)))^cross(2) = i^2}}}

{{{e^(i*pi) = i^2}}}

-----------------------
 From the first part we have

{{{e^(i*pi) = -1}}}

and from the last part we have

{{{e^(i*pi) = i^2}}}

So {{{i^2=-1}}} because things equal to the 

same thing are equal to each other

Edwin</pre>