Question 1074213
<pre>
{{{(a+bi)^2=i }}}

{{{(a+bi)(a+bi)=i }}}

{{{a^2+2abi+b^2i^2=i}}}

Since i<sup>2</sup> = -1

{{{a^2+2abi+b^2(-1)=i}}}

{{{a^2+2abi-b^2=i}}}

Set the real part of the left side equal to the real
part of the right side:

and

Set the imaginary part of the left side equal to the 
imaginary part of the right side:
 
{{{a^2-b^2=0}}}  and  {{{2abi=i}}}
                      {{{b=i/(2ai)}}}
                      {{{b=1/(2a)}}}

Substitute for b in

{{{a^2-b^2=0}}}
{{{a^2-(1/(2a))^2=0}}}
{{{a^2-1/(4a^2)=0}}}
{{{4a^4-1=0}}}
{{{(2a^2-1)(2a^2+1)=0}}}

{{{2a^2-1=0}}}; {{{2a^2+1=01}}}
{{{2a^2=1}}};   {{{2a^2=-1}}}
{{{a^2=1/2}}};  {{{a^2=-1/2}}}
{{{a= "" +- sqrt(1/2);   {{{a^2="" +- sqrt(-1/2)}}}
{{{a= "" +- 1/sqrt(2)}}}; {{{a="" +-i*sqrt(1/2)}}}
{{{a="" +- sqrt(2)/2}}}; ignore the imaginary value for a
Substitute in:
{{{b=1/(2a)}}}
{{{b=1/(2("" +- sqrt(2)/2))}}}
{{{b=1/(cross(2)("" +- sqrt(2)/cross(2)))}}} 
{{{b=1/("" +- sqrt(2))}}} 
{{{b="" +- sqrt(2)/2}}}

So there are two solutions:

{{{a="" +- sqrt(2)/2}}}; {{{b="" +- sqrt(2)/2}}}

One solution is where a and b are both positive,
and the other solution is when they are both
negative.

Edwin</pre>