Question 394201
<pre><font face = "batangche" color = "indigo" size = 4><b>
6-11i

Rule:

A + Bi is represented by the line segment that goes from the origin to
the point (A,B).  It has length r.

So,
6 - 11i is represented by the line segment that goes from the origin to
the point (6,-11)

So we draw that line segment and label it r in length:

{{{drawing(400,400,-12,12,-15,9, locate(6,-11,"(6,-11)"),
locate(2.5,-6,r), 
graph(400,400,-12,12,-15,9), line(0,0,6,-11) )}}}

Next we'll indicate the angle <font face = "symbol">q</font> starting
at the right hand of the x-axis going around counter-clockwise
to the line we just drew.  I'll indicate <font face = "symbol">q</font> with
a red arc: 

{{{drawing(400,400,-12,12,-15,9, locate(6,-11,"(6,-11)"),
locate(2.5,-6,r),  red(arc(0,0,10,-10,0,298.6104597), locate(-5,4,theta)),
graph(400,400,-12,12,-15,9), line(0,0,6,-11) )}}}

Next we draw a line from that point perpendicular to the x-axis.
I'll draw it in green:

{{{drawing(400,400,-12,12,-15,9, locate(6,-11,"(6,-11)"),
locate(2.5,-6,r),   green(line(6,0,6,-11)), 
red(arc(0,0,10,-10,0,298.6104597), locate(-5,4,theta)),

graph(400,400,-12,12,-15,9), line(0,0,6,-11) )}}}

That makes a right triangle, so we label the horizontal leg
as the x value of the point. That is, x = 6.  We label the
vertical leg the y -value of the point. That is y = -11 

{{{drawing(400,400,-12,12,-15,9, locate(6,-11,"(6,-11)"),
locate(2.5,-6,r),   green(line(6,0,6,-11)), locate(3,1.5,x=6), locate(6.5,-5,y=-11), red(arc(0,0,10,-10,0,298.6104597), locate(-5,4,theta)),

graph(400,400,-12,12,-15,9), line(0,0,6,-11) )}}}

Next we calculate the value of r using the Pythagorean theorem:

{{{r^2=x^2+y^2}}}
{{{r^2=(6)^2+(-11)^2}}}
{{{r^2=36+121}}}
{{{r^2=157}}}
{{{r=sqrt(157)}}}

So we label the r as r = {{{sqrt(157)}}}

{{{drawing(400,400,-12,12,-15,9, locate(6,-11,"(6,-11)"),
green(line(6,0,6,-11)), locate(3,1.5,x=6), locate(6.5,-5,y=-11), locate(.2,-6,r=sqrt(157)), red(arc(0,0,10,-10,0,298.6104597), locate(-5,4,theta)),

graph(400,400,-12,12,-15,9), line(0,0,6,-11) )}}}

Next we calculate <font face = "symbol">q</font>

by first calculating the tangent of the reference angle,

and then placing it in the 4th quadrant.

{{{tan(theta)=y/x}}}
{{{tan(theta)=-11/6}}}
Use the inverse tangent function on the calculator to find
the inverse tangent of {{{+11/6}}} to get the reference angle:

reference angle = 61.38954033°

To get <font face = "symbol">q</font> in the 4th quadrant subtract
the reference angle from 360° and get 

<font face = "symbol">q</font> = 298.6° rounded to the nearest
tenth of a degree.

So we label <font face = "symbol">q</font> = 298.6°

{{{drawing(400,400,-12,12,-15,9, locate(6,-11,"(6,-11)"),
green(line(6,0,6,-11)), locate(3,1.5,x=6), locate(6.5,-5,y=-11), locate(.2,-6,r=sqrt(157)), red(arc(0,0,10,-10,0,298.6104597), 
locate(-9,4,theta="298.6°")),

graph(400,400,-12,12,-15,9), line(0,0,6,-11) )}}}

Now the trigonometric form is

r(cos<font face = "symbol">q</font> + i·sin<font face = "symbol">q</font>)

and upon substituting, the final answer is

{{{sqrt(157)}}}(cos298.6° + i·sin298.6°)

Edwin</pre>