Question 167192
(√2 + √-2)^2
______________
(2)^2 
the line represents division
<pre><font size = 4 color = "indigo"><b>
That's:

{{{   (sqrt(2)+sqrt(-2))^2/2^2 }}}

I suspect that you didn't copy something right.

Why? Because since you are in advanced enough math
to be studying about imaginary numbers, I doubt that
a book or worksheet would make you do an unnecessary 
elementary operation like squaring 2, that is, 
multiplying 2 by itself and getting 4. That's 3rd 
grade math.  The authors wouldn't be putting that 
elementary an operation into a problem. They'd just 
have given you this:

{{{   (sqrt(2)+sqrt(-2))^2/4 }}}

The only purpose the author could have had to put {{{2^2}}}
instead of {{{4}}} was to test you to see if you knew the 
3rd grade math question "What is {{{2^2}}}".  But if you 
didn't know 3rd grade math then you couldn't possibly be 
where you are in math, all the way to studying imaginary numbers.
So no teacher would need to test you on that.

Anyway, if this is the problem:

 {{{   (sqrt(2)+sqrt(-2))^2/2^2 }}}

Then we replace the {{{2^2}}} with {{{4}}}.

 {{{   (sqrt(2)+sqrt(-2))^2/4 }}} 

Now instead of dividing by {{{4}}}, let's change
that to multiplying by {{{1/4}}} which is the same
thing:

{{{1/4}}}{{{((sqrt(2)+sqrt(-2))^2) }}}

Now to the side:
Write the term {{{sqrt(-2)}}} as {{{sqrt(-1*2)}}}.
Then write that as {{{sqrt(-1)*sqrt(2)}}} and
finally write that as {{{i*sqrt(2)}}}.

So we replace {{{sqrt(-2)}}} by {{{i*sqrt(2)}}}
in:

 {{{1/4}}}{{{   ((sqrt(2)+sqrt(-2))^2)}}}

and get:

{{{1/4}}}{{{((sqrt(2)+i*sqrt(2))^2) }}}

Now replace {{{(sqrt(2)+i*sqrt(2))^2}}} by {{{(sqrt(2)+i*sqrt(2))(sqrt(2)+i*sqrt(2))}}}

{{{1/4}}}{{{((sqrt(2)+i*sqrt(2))(sqrt(2)+i*sqrt(2))) }}}

Now out of each of those parentheses we can factor out {{{sqrt(2)}}}

{{{1/4}}}{{{sqrt(2)(1+i)sqrt(2)(1+i) }}}

Repositioning the factors:

{{{1/4}}}{{{sqrt(2)sqrt(2)(1+i)(1+i)}}}

Replace {{{sqrt(2)sqrt(2)}}} by {{{2}}}

{{{1/4}}}{{{2(1+i)(1+i)}}}

Now we use FOIL on {{{(1+i)(1+i)}}}

{{{1/4}}}{{{2(1+i+i+i^2)}}}

Combining the two middle terms:

{{{1/4}}}{{{2(1+2i+i^2)}}}

Replacing {{{i^2}}} by {{{-1}}}

{{{1/4}}}{{{2(1+2i-1)}}}

The {{{1}}} and {{{-1}}} cancel:

{{{1/4}}}{{{2(2i)}}}

or

{{{1/4}}}{{{(4i)}}}

or 

{{{1/cross(4)}}}{{{(cross(4)i)}}}

or just

{{{i}}}

Edwin</pre>