Question 852727
<pre>
{{{x+iy}}}{{{""=""}}}{{{4i / (-1+ sqrt(3)i)}}}

First get it in the form x+iy by multiplying numerator
and denominator by the conjugate of the denominator: 

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

{{{(-4i-4sqrt(3)i^2)/(1-3i^2)}}}

Change both iČ's to (-1):

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

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

Factor 4 out of the numerator:

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

{{{(cross(4)(-i+sqrt(3)))/cross(4)}}}

{{{-i+sqrt(3)}}}

{{{sqrt(3)-i}}}

So the rectangular form is: 

{{{x+iy}}}{{{""=""}}}{{{sqrt(3)-1}}}

That is represented by the vector (line) connecting the origin to the point 
(x,y)= ({{{sqrt(3)}}},-1), We draw a perpendicular to the x-axis, and indicate
the angle {{{theta}}} by a red arc:

{{{drawing(300,300,-3,3,-3,3,

graph(300,300,-3,3,-3,3),red(arc(0,0,1,-1,0,330),locate(-.5,.7,theta)),
locate(sqrt(3)+.1,-.4,y=-1),locate(.7,.5,x=sqrt(3)), locate(.4,-.6
,"r=?"),line(0,0,sqrt(3),-1), line(sqrt(3),0,sqrt(3),-1) )}}}

We find the value of r (the hypotenuse) by the Pythagorean theorem:

{{{r^2}}}{{{""=""}}}{{{x^2+y^2}}}

{{{r^2}}}{{{""=""}}}{{{(sqrt(3))^2+(-1)^2}}}

{{{r^2}}}{{{""=""}}}{{{3+1}}}

{{{r^2}}}{{{""=""}}}{{{4}}}

r is always positive so we take the positive square root:

{{{r}}}{{{""=""}}}{{{2}}}
{{{drawing(300,300,-3,3,-3,3,
graph(300,300,-3,3,-3,3),red(arc(0,0,1,-1,0,330),locate(-.5,.7,theta)),
locate(sqrt(3)+.1,-.4,y=-1),locate(.7,.5,x=sqrt(3)), locate(.4,-.6
,r=2),line(0,0,sqrt(3),-1), line(sqrt(3),0,sqrt(3),-1) )}}}

We find {{{theta}}} by using any trig function, say the cosine;
cos(theta)=adjacent/hypotenuse=x/r=sqrt(3)/2}}}

This tells us that the angle has a refence angle of 30°, but since
it is in quadrant IV, we subtract from 360° and get {{{theta="330°"}}}

{{{r}}}{{{""=""}}}{{{2}}}
{{{drawing(300,300,-3,3,-3,3,
graph(300,300,-3,3,-3,3),red(arc(0,0,1,-1,0,330),locate(-1.5,.7,theta="330°")),
locate(sqrt(3)+.1,-.4,y=-1),locate(.7,.5,x=sqrt(3)), locate(.4,-.6
,r=2),line(0,0,sqrt(3),-1), line(sqrt(3),0,sqrt(3),-1) )}}}

Next we know that

{{{adjacent/hypotenuse=x/r=cos(theta)}}}, so {{{x=r*cos(theta)=2*cos("330°")}}}

Also, we know that

{{{opposite/hypotenuse=y/r=sin(theta)}}}, so {{{y=r*sin(theta)= 2*sin("330°")}}}

So 

x + iy = 2·cos(330°)+i·2·sin(330°) = 2[cos(330°) + i·sin(330°)]

Edwin</pre>