Question 205028
Expressed in {{{a + bi}}} form, {{{5/(3 + i)}}} is equivalent to: 

A) 15 - 5i,  B) (3/2) - (1/2i),  C) (5/3) - 5i,  D) (15/8) - (5/8i)
<pre><font size = 4 color = "indigo"><b>
{{{5/((3 + i))}}}

Form the conjugate of the denominator by changing the
sign of the term containing {{{i}}}.  The conjugate of
{{{(3+i)}}} is {{{(3-i)}}} so we multiply by the fraction
{{{((3-i))/((3-i))}}} which just equals {{{1}}} and will not
change the value:

{{{5/((3 + i))}}}×{{{((3-i))/((3 - i))}}}

Indicate the multiplication of the tops and bottoms:

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

Distribute the top, FOIL out the denominator:

{{{(5(3-i))/(9-3i+3i-i^2)}}}

{{{(5(3-i))/(9-cross(3i)+cross(3i)-i^2)}}}

{{{(5(3-i))/(9-i^2)}}}

Now we substitute {{{(-1)}}} for {{{i^2}}}

{{{(5(3-i))/(9-(-1))}}}

{{{(5(3-i))/(9+1)}}}

{{{(5(3-i))/10}}}

{{{(cross(5)(3-i))/cross(10[2])}}}

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

Make two fractions:

{{{3/2-i/2}}}

Write {{{i/2}}} as {{{1/2}}}{{{i}}}

{{{3/2-1/2}}}{{{i}}}

Edwin</pre>