[Font]

how do you do these questions in the heinnmann book?
ex 2c Q10 (the locus of z+1/z for which the imaginary part is 0)

ex 2c Q12

no idea how to draw arg((z-2)/(z-2i)) = 3pi/4. and that it represents an arc of a circle and that |z-4|/|z-1| is constant on this circle.

also, Find the values of z coerresponding to the points in which this circle is cut by the curve given by: |z-1|+|z-4| =5and ques 14 (a)

Also having probs on ex 2D ques 14,15,16 and 18

sorry for all the questions! - just spent hours revising complex numbers, and have no idea how to do these sort of problems

Phil

[dvs]

2C
10. Just set Im(z + 1/z) = 0.

12. http://www.thestudentroom.co.uk/show....php?p=2407614

14a. Read RichE's post in the thread above.

2D
14. Set Re(w)=0, then see what that might inspire.

15. http://www.thestudentroom.co.uk/t87130.html

16 & 18. What parts are you having trouble with here?

[Gaz031]

Originally Posted by Phil23

how do you do these questions in the heinnmann book?
ex 2c Q10 (the locus of z+1/z for which the imaginary part is 0)

Let z=x+iy.
z+(1/z)=(x+iy)+[1/(x+iy)
=(x+iy)+[(x-iy)/(x+iy)(x-iy)]
=(x+iy)+[(x-iy)/(x^2+y^2)]
=x+iy+x/(x^2+y^2)-iy/(x^2+y^2)
=[x+x/(x^2+y^2)]+i[y-y/(x^2+y^2)]
Imaginary part is zero then:
y=y/(x^2+y^2)
x^2+y^2=1
That is, we have a circle centre the origin and radius 1.

[Soton Hopeful]

Help im a bit stuck with a matricies question.

Could anyone help me with how to prove that 9 is an eigenvalue of

1 0 4
0 5 4
4 4 3

Its question 5b) of the june 02 paper. iv read the answer scheme but still cant make head or tail of it.

ta

[dvs]

Call the matrix A, then k is an eigenvalue if det(A - kI) = 0. So plug in 9 in the LHS and show that it equals zero.

[Font]

Originally Posted by Gaz031

Let z=x+iy.
z+(1/z)=(x+iy)+[1/(x+iy)
=(x+iy)+[(x-iy)/(x+iy)(x-iy)]
=(x+iy)+[(x-iy)/(x^2+y^2)]
=x+iy+x/(x^2+y^2)-iy/(x^2+y^2)
=[x+x/(x^2+y^2)]+i[y-y/(x^2+y^2)]
Imaginary part is zero then:
y=y/(x^2+y^2)
x^2+y^2=1
That is, we have a circle centre the origin and radius 1.

i did that then i was trying to sub back into the Re eqtn - dont i do that and why? I havte complex numbers soooo much

[Font]

14a on 2C and,16 and 18 from exercise 2D - anyone know how to do these?

[dvs]

I think 14a is now taken care of. So again:

Originally Posted by dvs

16 & 18. What parts are you having trouble with here?

[lesser weevil]

the whole COURSE!

[Font]

Originally Posted by dvs

I think 14a is now taken care of. So again:

i did something random and got |z-1.5i|<=|z-2.5i| but not sure whether this is right - got no confidence you see - i got the shaded bit being the part under the line of y=2 (is this correct)

for 18 - i dont get the 2nd-4th bits - i.e. from show that if the point representing z lies..bla bla, to teh indicating which region the positive real axis of the z plane is mapped. Also, how do you guys know what method to use - i just have no idea where to start on some of these questions - on the other question you posted somewhere you took u and v to be the subject, but i made x and y to be my subject and plugged in Im(w) and Re(w) into that - i'm so lost here - can you use both methods - and the three methods they use in the book - i've only come across where they use the making the u and v in terms of x and y - never seen anything else

also, anyone know how to do question 14 on 2d too? - the showing it lies between -i and i think?

pk

[Gaz031]

anyone know how to do question 14 on 2d too? - the showing it lies between -i and i think?

I'll assume you know how to do the first part, so:
P represents w=u+iv.
Q represents z=x+iy.
[Mimetex cannot convert this formula]
P describes the portion of the imaginary axis between -i and i.
Re(w)=u=0 so x=0.
Hence [Mimetex cannot convert this formula]
We have -1<Im(w)<1 so -1<[Mimetex cannot convert this formula]<1.
If we were to break this into two inequalities we need:
(1) -y^2-2y-1<y^2-1 so 0<2y^2+2y so 0<2y(y+2) so either y>0 or y<-2.
(2) y^2-1<y^2+2y+1 so 0<2y+2 so 0<2(y+1) so y>-1.
Hence, putting the inequalities together we have y>0.
But x=0 and z=x+iy is represented by Q, so Q describes the whole of the positive imaginary axis.

Sorry about mixing tex and non tex. Fractions written with tex are so much better but it seems to write -1<Im(w)<1 as [Mimetex cannot convert this formula] for some very strange reason