Need help with question:

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)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:

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

Code:Item 1:a)Sketch the circle C defined by the Cartesian co-ordinates:(x, y): x2 + (y – 1)2 = 1[IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image001.png[/IMG] 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 = 1Equation 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 = 1Wherehandkare thexandycoordinates of the centre of the circle andris the radius. We can then use the radius, x co-ordinate and y co-ordinate to sketch up the circle ‘C’: [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image003.jpg[/IMG]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]zP [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image006.jpg[/IMG] 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 itsxandycomponents: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] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image018.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image020.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image022.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image024.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image026.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image028.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: [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image042.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image044.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image046.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image048.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image050.png[/IMG] 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 ofCos(-Ѳ) = -Cos(Ѳ)andSin(-Ѳ) = -Sin(Ѳ)[IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image050.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image054.png[/IMG] [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image056.png[/IMG] 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 ofz’isand the argument ofz’is [IMG]file:///C:/Users/Jacksane/AppData/Local/Temp/msohtmlclip1/01/clip_image064.png[/IMG]. The diagram above just explains the inverse (or reciprocal) ofz(which isz’) 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.

Questions:

Thanks!(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.