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Math1901 (1 Viewer)
- Thread starter MyLuv
- Start date
freaking_out
Saddam's new life
a is obviously easy- just ya typical 4u maths question: its a circle with equation (x-1)<sup>2</sup> + y<sup>2</sup> = 1Originally posted by MyLuv
Did u guys do the assignment?What you got for 1b?-Its impossible to find the equation so I did tracing and got a Heart...???
Btw,this s the question:
a/ sketch lz-1l=1
b/ sketch w=z^2
as with b- i got a semi circle with diameter -4 and 4.
Re: Re: Math1901
Its not right coz it didnt cover the case where lzl<2(hence has mostly invalid pts)
If u shaded the thing then u ll have too many extra pts coz for any modulus of z theres only fixed 1 Arg(z).
u mean a circle with x-intercep +/- 4 and center O(0;0)?
Its not right coz it didnt cover the case where lzl<2(hence has mostly invalid pts)
If u shaded the thing then u ll have too many extra pts coz for any modulus of z theres only fixed 1 Arg(z).
freaking_out
Saddam's new life
hmm...yeah, i get what u mean, so i guess, i'll have to start tracing then.
from now on, google will do my uni assignments
http://www.math.ucla.edu/~tao/java/Complex.html
wait for the thigo to load, change it to z^2, draw a circle
http://www.math.ucla.edu/~tao/java/Complex.html
wait for the thigo to load, change it to z^2, draw a circle
i have a process for doing it, but the complications are just too insane to consider.
if you let z = x + yi, square that,
w= (x^2 - y^2) + (2xy)i
then from the cartesian equation of the curve in a, get y in terms of x, so you can get w solely in terms of x.
then you have an Re(z) and an Im(z), ie an X and a Y, and you can eliminate x from that to get a relationship between X and Y. but to actually do that is ridiculously difficult.
ie; suggestions as to a simpler, achievable method would be appreciated
if you let z = x + yi, square that,
w= (x^2 - y^2) + (2xy)i
then from the cartesian equation of the curve in a, get y in terms of x, so you can get w solely in terms of x.
then you have an Re(z) and an Im(z), ie an X and a Y, and you can eliminate x from that to get a relationship between X and Y. but to actually do that is ridiculously difficult.
ie; suggestions as to a simpler, achievable method would be appreciated
freaking_out
Saddam's new life
yeah, i just tried for a whole hour to find the equation using z=rcis(theta) and cartesian way, and its ridicoulously difficult as pointed out above...i mean we don't know abt. the "love heart" function do we??
so i guess i'm just gonna have to trace it.
so i guess i'm just gonna have to trace it.
You only have to sketch the locus, so you can get get the general shape, and then join up dots. Two things make this easier.
1. When you square a complex number, square its modulus and double its argument.
2. The circle |z - 1| = 1 is symmetric across the real axis, and so is w, so you only have to figure out the part when Im(z) => 0, then reflect the answer across the real axis.
Start with the easy points. z = 0 transforms to w = 0, and z = 2 transforms to w = 4. These are the only point on the real axis.
Now, what about the max point, z = 1 + i. Well, |z| = sqrt(2), arg z = pi / 4, so w = [sqrt(2)]<sup>2</sup> * e<sup>i*2*pi/4</sup> = 2i - the intercept on the imaginary axis.
Clearly, for the top half, arg z varies from 0 to pi /2 - that is 0 <= arg z < pi / 2.
It follows that 0 <= arg w < pi
Near z = 2, as you travel along the circle, the argument is slowly increasing, so the argument of w = z<sup>2</sup> is only slowly increasing. Thus, w is nearly vertical near w = 4.
Near z = 0, as you travel along the circle in towards the origin, the argument is changing rapidly, so the arg w is approaching pi rapidly - thus, the curve approaches w = 0 from the negative direction with a horizontal gradient.
Testing values of z slightly greater than z = 1 + i will show that Im(w) can be > 2, and thus there is a stationary point somewhere with Re(w) > 0. There must also be a point with a vertical tangent somewhere with Re(w) < 0.
Thus, the curve is sort of a heart shape on its side, except that the 'bottom' is flat, not V shaped.
1. When you square a complex number, square its modulus and double its argument.
2. The circle |z - 1| = 1 is symmetric across the real axis, and so is w, so you only have to figure out the part when Im(z) => 0, then reflect the answer across the real axis.
Start with the easy points. z = 0 transforms to w = 0, and z = 2 transforms to w = 4. These are the only point on the real axis.
Now, what about the max point, z = 1 + i. Well, |z| = sqrt(2), arg z = pi / 4, so w = [sqrt(2)]<sup>2</sup> * e<sup>i*2*pi/4</sup> = 2i - the intercept on the imaginary axis.
Clearly, for the top half, arg z varies from 0 to pi /2 - that is 0 <= arg z < pi / 2.
It follows that 0 <= arg w < pi
Near z = 2, as you travel along the circle, the argument is slowly increasing, so the argument of w = z<sup>2</sup> is only slowly increasing. Thus, w is nearly vertical near w = 4.
Near z = 0, as you travel along the circle in towards the origin, the argument is changing rapidly, so the arg w is approaching pi rapidly - thus, the curve approaches w = 0 from the negative direction with a horizontal gradient.
Testing values of z slightly greater than z = 1 + i will show that Im(w) can be > 2, and thus there is a stationary point somewhere with Re(w) > 0. There must also be a point with a vertical tangent somewhere with Re(w) < 0.
Thus, the curve is sort of a heart shape on its side, except that the 'bottom' is flat, not V shaped.
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freaking_out
Saddam's new life
thanx there CM...i only hope that these sorta questions, don't come up on the exams.
freaking_out
Saddam's new life
btw, can someone post up their sketch here....i will post mine, as soon as i get down to doing it.
freaking_out
Saddam's new life
well with part a, once u proved the case for real axis, then any "line" is just a rotation of the real axis situation- i.e u just need to multiply by cis@.
as with b- know that 1/z= cis(-@)...u should b able to do it from here...other wise its just too easy.
freaking_out
Saddam's new life
i've drawn a sketch for 1b- can someone tell me if this looks right??
cheers
cheers
freaking_out
Saddam's new life
na, u just need to sketch it showing essential features.