Question 581638
<pre>
{{{(1-4i)/(2+3i)}}}

The denominator is 2+3i.  Its conjugate is 2-3i 

(change the sign of the term containing i)

Multiply by the unit fraction formed by putting the con=jugate
of the denominator over itself: {{{red((2-3i)/(2-3i))}}}

{{{(1-4i)/(2+3i)}}}·{{{red((2-3i)/(2-3i))}}}

{{{((1-4i)(2-3i))/((2+3i)(2-3i))}}}

Multiply (FOIL) out the top and bottom:

{{{(2-3i-8i+12i^2)/(4-6i+6i-9i^2)}}}

Combine the like terms. (the two middle terms in the bottom cancel)  

{{{(2-11i+12i^2)/(4-9i^2)}}}

Substitute (-1) for i²

{{{(2-11i+12(-1))/(4-9(-1))}}}

Simplify

{{{(2-11i-12)/(4+9)}}}

Combine like terms:

{{{(-10-11i)/13}}}

Divide the -10 and the -11 each by 13

{{{(-10)/13 - expr(11/13)i}}}

{{{-10/13 - expr(11/13)i}}}

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

Also, how would you find the equation of a circle with a diameter AB, given that the coordinates of a and B are (-6,1) and (4,-5). What is the answer in standard form:

We need the the radius r and the center (h,k) and then we can 
substitute thes is the equation for a circle:

(x - h)² + (y - k)² = r²

Let's graph the points and draw the diameter:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(-9,1.5,"A(-6,1)"), locate(4.5,-5,"B(4,-5)"),green(line(-6,1,4,-5)),
circle(-6,1,.1), circle(4,-5,.1)  )}}}

We will use the distance formula to find the diameter, which we will take
on half of for the radius. The distance formula is:

d = {{{sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

d = {{{sqrt(((4)-(-6))^2+((-5)-(1))^2))}}} = {{{sqrt((4+6)^2+(-5-1)^2))}}} = {{{sqrt(10^2+(-6)^2))}}} = {{{sqrt(100+36)}}} = {{{sqrt(136)}}}

d = {{{sqrt(4*34)}}} = {{{sqrt(4)sqrt(34)}}} = {{{2sqrt(34)}}}

So the diameter is {{{2sqrt(34)}}} and the radius is one-half of that 
diameter which is {{{sqrt(34)}}}
 
The midpoint of any diameter is the center.  So we use the midpoint formula:

Midpoint = {{{(matrix(1,3,      (x[1]+x[2])/2,   ",", (y[1]+y[2])/2))}}} = 
{{{(matrix(1,3,      ((-6)+(4))/2,   ",", ((1)+(-5))/2))}}} = {{{(matrix(1,3,      (-6+4)/2,   ",", (1-5)/2))}}} = {{{(matrix(1,3,      (-2)/2,   ",", (-4)/2))}}} = {{{(matrix(1,3,      -1,   ",", -2))}}}

So the center is (h,k) = C(-1,-2) and the radius is r = {{{sqrt(34)}}}.

Substituting in

(x - h)² + (y - k)² = r²


(x - (-1))² + (y - (-2))² = ({{{sqrt(34)}}})²

(x + 1)² + (y + 2)² = 34

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(-9,1.5,"A(-6,1)"), locate(4.5,-5,"B(4,-5)"),green(line(-6,1,4,-5)),
locate(-4.5,-1.7,"C(-1,-2)"),

circle(-6,1,.1), circle(4,-5,.1), circle(-1,-2,sqrt(34)), circle(-1,-2,.1)  )}}}


Edwin</pre>