Question 133075
I need help with a few problems that I'm having difficulty with. 
1) {{{(1+i)/i - 3/(4-i)}}}
For the common denominator, I got {{{i(4-i)}}} = {{{4i-1}}}
<pre><font size = 4 color = "blue"><b>
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That's wrong. {{{i(4-i)}}} = {{{4i-i^2}}} = 
Replace {{{i^2}}} by {{{-1}}}
{{{4i-(-1)}}} = {{{4i+1}}} = {{{1+4i}}}
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But you shouldn't have tried to multiply it out yet anyway.
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</pre></font></b>
Do you multiply this normally or do you need a conjugate? 
<pre><font size = 4 color = "blue"><b>
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Here's the whole thing, starting from scratch:

{{{(1+i)/i - 3/(4-i)}}}

The LCD = i(4-i)

DO NOT MULTIPLY THAT OUT YET!  

The denominator of the first fraction needs the factor {{{(4-i)}}}
in order to become the LCD, so we multiply it by {{{(4-i)/(4-i)}}}
which is just 1, so it won't affect the value to multiply the first
fraction by that:

{{{ ( (1+i)/i )( (4-i)/(4-i) ) - 3/(4-i)  }}}

The denominator of the second fraction needs the factor {{{i}}}
in order to become the LCD, so we multiply it by {{{i/i}}}
which is just 1, so it won't affect the value to multiply the second
fraction by that:

{{{ ( (1+i)/i )( (4-i)/(4-i) ) - (3/(4-i))(i/i)}}}

Indicate the multiplication of the numerators and the denominators:

{{{ ((1+i)(4-i))/(i(4-i)) - (3i)/((4-i)i)}}}

Now use FOIL on the left top:

{{{(1+i)(4-i)}}} = {{{4-i+4i-i^2}}} = {{{4+3i-i^2}}} = 
Now replace {{{i^2}}} by {{{-1}}}
{{{4+3i-(-1)}}} = {{{4+3i+1}}} = {{{5+3i}}} and replace the left top by that:

{{{ (5+3i)/(i(4-i)) - (3i)/((4-i)i)}}}
 
Distribute the left bottom

{{{i(4-i)}}} = {{{4i-i^2}}} = 
Replace {{{i^2}}} by {{{-1}}}
{{{4i-(-1)}}} = {{{4i+1}}} = {{{1+4i}}} and
replace the left bottom by that:

{{{ (5+3i)/(1+4i) - (3i)/((4-i)i)}}}

The right bottom is the same as the left bottom:

{{{(4-i)i}}} = {{{i(4-i)}}} = {{{4i-i^2}}} =
Replace {{{i^2}}} by {{{-1}}}
{{{4i-(-1)}}} = {{{4i+1}}} = {{{1+4i}}} 

So replace the right bottom by that:

{{{ (5+3i)/(1+4i) - (3i)/(1+4i)}}}

Now combine the numerators over the common denominator:

{{{ (5+3i-3i)/(1+4i)}}}

Collect like terms on top:

{{{ 5/(1+4i)}}}

Now we are ready to use the conjugate of the bottom.
We multiply by {{{(1-4i)/(1-4i)}}} which again just
equals to 1 so it will not affect the value:

{{{ (5/(1+4i))((1-4i)/(1-4i))}}}

Indicate the multiplication of the numerators and
denominators:

{{{ (5(1-4i))/((1+4i)(1-4i))}}}

Distribute the top and use FOIL on the bottom:

{{{ (5-20i)/(1-4i+4i-16i^2))}}}

Combine like terms:

{{{ (5-20i)/(1-16i^2))}}}

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

{{{ (5-20i)/(1-16(-1)))}}}

{{{ (5-20i)/(1+16)}}}

Combine like terms:

{{{ (5-20i)/(17)}}}

Make two fractions:

{{{5/17-(20i)/17}}}

and write the answer in form {{{A+Bi}}}

{{{5/17-20/17}}}{{{i}}}

</pre></font></b>

2) {{{(sqrt(-15)) * (sqrt(-15))}}} 
I really do not understand this problem. 
Any help would be greatly appreciated! Thank you!
<pre><font size = 4 color = "blue"><b>
Write {{{-15}}} as {{{(15)(-1)}}}

{{{ sqrt((15)(-1)) * sqrt((15)(-1))}}} 

Take individual square roots of factors under radicals

{{{ sqrt(15)sqrt(-1) * sqrt(15)sqrt(-1)}}}

Replace {{{sqrt(-1)}}} by {{{i}}}

{{{ sqrt(15)*i * sqrt(15)*i}}}

{{{(sqrt(15))^2i^2}}}

Replace {{{(sqrt(15))^2}}} by 15 and {{{i^2}}} by {{{-1}}}

{{{(15)*(-1)}}}

{{{-15}}}

Edwin</pre>