Question 395841
Hi, I have a big test on complex numbers tomorrow and I am having a very hard time with this particular question: 
Directions: On a complex plane, a point z has been graphed. 
If cosθ=(-8)/(22) and (π)/(2)≤θ≤π: 
<pre><font face = "batangche" color = "indigo" size = 4><b>

The other tutor messed up, although he's right about the 2nd quadrant
between {{{pi/2}}} and {{{pi}}}.



Here is the point:

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2)

 )}}}
    
We draw a line from the origin to the point.  The length of that line is
indicated by the letter "r"

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2), green(line(0,0,-8,2sqrt(105)) ),
locate(-3.5,11,r)
 )}}}

Now we indicate &#952; with a red arc that starts on the right
side of the x-axis and swings counter-clockwise to the green
line:

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2), green(line(0,0,-8,2sqrt(105)) ),
red(arc(0,0,8,-8,0,111.3236863)), locate(1.7,4.5,theta),
locate(-3.5,11,r)


 )}}}

Next we draw a perpendicular from the point to the x-axis. The blue
line below, which we label as y.

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2), green(line(0,0,-8,2sqrt(105)) ),
red(arc(0,0,8,-8,0,111.3236863)), locate(1.7,4.5,theta),
blue(line(-8,2sqrt(106),-8,0)), locate(-3.5,11,r),
locate(-9,11,y),locate(-4,1.5,x)


 )}}}

Now we have a right triangle, with the horizontal leg labeled as x,


Since we are given = cos(&#952;) = {{{(-8)/22}}} and we know that
cos(&#952;) = {{{x/r}}}, we will take x to be -8 and r to be 22,
So we label those:

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2), green(line(0,0,-8,2sqrt(105)) ),
red(arc(0,0,8,-8,0,111.3236863)), locate(1.7,4.5,theta),
blue(line(-8,2sqrt(106),-8,0)), locate(-4,12,r=22),
locate(-9,11,y),locate(-6,1.5,x=-8) 


 )}}}

Now we calculate y by the Pythagorean theorem:

     r² = x² + y²

  (22)² = (-8)² + y²

    484 = 64 + y²

    420 = y²
    ___
   &#9143;420 = y
  _____
 &#9143;4*105 = y
    ___
  2&#9143;105 = y

So we label y:

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2), green(line(0,0,-8,2sqrt(105)) ),
red(arc(0,0,8,-8,0,111.3236863)), locate(1.7,4.5,theta),
blue(line(-8,2sqrt(106),-8,0)), locate(-4,12,r=22),
locate(-12,11,y=2sqrt(105)),locate(-6,1.5,x=-8) 


 )}}}
                                                                 ___
Therefore the given point, call it P has the coordinates P(-8, 2&#9143;105).

{{{drawing(429,429, -22,5,-5,22, graph(429,429,-22,5,-5,22),
circle(-8,2sqrt(105),.2), green(line(0,0,-8,2sqrt(105)) ),
red(arc(0,0,8,-8,0,111.3236863)), locate(1.7,4.5,theta),
blue(line(-8,2sqrt(106),-8,0)), locate(-4,12,r=22),
locate(-12,11,y=2sqrt(105)),locate(-6,1.5,x=-8),
locate(-12,22,P(-8,2sqrt(105))) 


 )}}}
                                                ___
Therefore z in standard form is x + yi = -8 + 2&#9143;105*i 

z in trigonometric form is r(cos&#952; + isin&#952;)

But we must find &#952; by using the given cos&#952; = {{{(-8)/22}}}

We use the inverse cosine of the POSITIVE {{{8/22}}} to find the
reference angle of &#952; to be 68.7° or as a whole number of degrees,
69°.  But to get that in the second quadrant we subtract from 180°
and get &#952; = 111°

Now the trig form

r(cos&#952; + isin&#952;)

becomes 
  ___
2&#9143;105(cos111° + i*sin111°)

Edwin</pre>