Need help with question:
Show that, whatever the position of P on the circle C, the point P’ representing z’ lies on a certain line, and determine the equation of this line:
I'm fairly clueless, other than the fact that I understand the line is probably linear below the x axis, because it's P'. (1/P)

Any help or ideas to get me moving further? P.s the questions before this one are as follows, with my answers as well:

Item 1: 
  a)                  Sketch the circle C defined by the Cartesian co-ordinates:    
  (x, y):               x2 + (y – 1)2     =    1
  From our knowledge of circles, we can express the co-ordinate equation in terms of the equation of a circle. (Ref:              Equation of a Circle)
  Circle C co-ordinates:                        x2 + (y - 1)2 = 1
  Equation of a Circle:              (x - h)2 + (y - k)2 = r2
  (x – 0)2 + (y – 1)2   =   12
  It can be seen that the following variables can be found simply by recognising their positions in comparison to the original circle equation.
  h = 0    k = 1    and      r = 1
  Where h and k are the x and y coordinates of the centre of the circle and r is the radius. We can then use the radius, x co-ordinate and y co-ordinate to sketch up the circle ‘C’:
  b)   The Point P(x,y) representing the non-zero complex number ‘z = x + iy’, lies on C. Express |z| in terms of Ѳ, the argument of z.
  Sketch the diagram:
                        z   =       x  +  iy
  [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png[/IMG]                             Ѳ
                     [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image005.png[/IMG]                             z
  We need to find the Mod-arg form of |z| which is just the mod, so we’re finding the modulus through argument of z with use of imaginary number vector.
  Since we know:     [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image008.png[/IMG]    can be written as:            [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image010.png[/IMG]
  Where the complex number is expressed in polar (mod-arg) form.
  We can then break the number up into its x and y components:
  x component:    [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image012.png[/IMG]       y component:               [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image014.png[/IMG]
  Now we can then go back and substitute each axis value into the original equation of the circle, this should represent the complex number as a form of the equation:
              Circle equation:    [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image016.png[/IMG]  
  |z| expressed in terms of Ѳ is:       [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image030.png[/IMG]
  Sub       [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image032.png[/IMG]     into       [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image034.png[/IMG] POSS REMOVE!
  c)      Given that z’ =  [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image036.png[/IMG], find the modulus and argument of z’ in terms of Ѳ.
  Since we know:
  [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image038.png[/IMG]          and         [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image040.png[/IMG]
  We can then substitute one into the other:
  Now that we have it in Mod-arg form, we can now express its argument and modulus singularly. But before we do this, it is important to put it into common form: ‘rCis[IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image052.png[/IMG]’ to make it clearer. We can do this with our knowledge of the unit circle, where the value of Cos(-Ѳ) = -Cos(Ѳ) and Sin(-Ѳ) = -Sin(Ѳ)
  We can now see that we have expressed z’ in Mod-arg form and can easily identify both the argument and th

Since we can also state that the modulus of z’ isand the argument of z’ is [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image064.png[/IMG].
  The diagram above just explains the inverse (or reciprocal) of z (which is z’) and where about they are on the plane with the conjugate of z as an extra vector to give extra reference inside the diagram.

IN FACT, that didn't turn out so well, you can ignore it if you wish... or decipher it.

(iv) Sketch the curve C with Cartesian equation x2 + (y − 1)2 = 1. The point P,
representing the nonzero complex number z, lies on C.
(a) Express |z| in terms of θ, the argument of z.
(b) Given that z = 1
z find the modulus and argument of z in terms of θ.
Show that, whatever the position of P on the circle C, the point P representing z
lies on a certain line and determine the equation of this line.

p.s this editing form is really scrambled on my computer and it's hard to make this a neat post sorry in advance.