Question 489811
<pre>
It is perfectly OK to use {{{i/i}}}.  You will get the same answer
if you do it correctly.  You just made a mistake.

The reason they tell you to use {{{(-i)/(-i)}}} is because they
want ONE rule to work for all problems.

When you have 

{{{(A+Bi)/(C+Di)}}}

They always tell you to multiply by the conjugate of the denominator
over itself:

{{{(C-Di)/(C-Di)}}}

You problem {{{(2+i)/i}}} is really the same as {{{(2+i)/(0+i)}}},
so to follow that rule you would multiply by {{{(0-i)/(0-i)}}},
which is the same as {{{(-i)/(-i)}}}.

It would also work to use {{{(-C+Di)/(-C+Di)}}} but they don't
want to tell you to multiply by the negative of the conjugate of
the denominator over itself.

Using {{{i/i}}}

{{{expr((2+i)/i)*expr(i/i)}}}

{{{((2+i)i)/(i*i)}}}

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

Since {{{i^2=-1}}},

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

Separate into two fractions: 

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

{{{-2i + 1}}}

{{{1 - 2i}}}

----------------------------------

Or you can use {{{(-i)/(-i)}}}

{{{expr((2+i)/i)*expr((-i)/(-i))}}}

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

{{{(-2i-i^2)/(-i^2)}}}

Since {{{i^2=-1}}},

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

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

{{{-2i+1}}}

{{{1-2i}}}


------------------------

Edwin</pre>