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:
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 = 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’:
[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] z
P
[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 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]
[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 of Cos(-Ѳ) = -Cos(Ѳ) and Sin(-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 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.
Questions:
(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.
Thanks!
p.s this editing form is really scrambled on my computer and it's hard to make this a neat post sorry in advance.