4 u half yearly exam (1 Viewer)

[Rahul]

got the answer thanx to monster.....

 

[SoFTuaRiaL]

Originally posted by Rahul
got the answer thanx to monster.....

HEY its not monstEr, its monstAR, as in mon-star !!

 

[ND]

Da Monstar, just a suggestion: why don't you post the solution on the board, for anyone else that may be having similar problems?

 

[SoFTuaRiaL]

Originally posted by ND
Da Monstar, just a suggestion: why don't you post the solution on the board, for anyone else that may be having similar problems?

sure .........


z = x+iy
z-i/z+1 = (x+iy-i)/(x+1+iy)
multiply both numerator n denominator by (x+1-iy) which is the denominator's conjugate.
u end up with (x^2 + xyi - xi + x + yi - i - xyi + y^2 + y) in the numerator, the denominator doesnt matter, so dont bother abt it.
now since this is sposed to be purely imaginary, the real part is zero. (x^2 + x + y^2 + y)/denominator = 0. now, den. gets cancelled.
therefore, x(x+1) + y(y+1) = 0
TADAA !!

 

[spice girl]

Also, an alternate solution, for people who know how to graph:

arg(z-z1) - arg(z-z2) = alpha (where z1, z2 are complex numbers and -pi < alpha <= pi)

for those who don't know, it's an arc where the angle subtended to the arc by the interval A(z1)B(z2) is the angle alpha.

Anyway, if z-1/z+i is imaginary then
arg(z-1) - arg(z+i) = +- pi/2

so it's 2 half-circles, thus it's a circle, with diameter XY (X represents 1, Y represents -i)

 

[Rahul]

hey exam on friday...shit thats tmw....can some one help me out wid a few q's:

1. factorize z^3 -1. if z is one of teh 3 cube roots of unity find the two possible values of z^2+z+1.
is the answer cis 2pi/3, cis -2pi/3 ?
the cambridge book doesnt have the answers...so i am not sure

2. use de moivre's theom. to find in mod/arg form the cube roots of -2-2i?
i got sq rt2 cis -pi/4, sq rt2 cis 5pi/12, sq rt2 cis -pi/12 as my solution....not sure abt this question.

3. if |z| = r and arg z = @, show that z/z^2 + r^2 is real and give its value

this is about it....

cheers

 

[Neon-Frog]

I have my 4unit maths exam and 3unit english exam tomorrow

Im having alot of trouble fitting all the study in god damn it.

Oh well my test is on complex, graphing and conics, and im pretty familiar with them. Im hoping my teacher sets the test, should be easier then grabbing random questions out of past tests

 

[ND]

Originally posted by Rahul
hey exam on friday...shit thats tmw....can some one help me out wid a few q's:

1. factorize z^3 -1. if z is one of teh 3 cube roots of unity find the two possible values of z^2+z+1.
is the answer cis 2pi/3, cis -2pi/3 ?
the cambridge book doesnt have the answers...so i am not sure

2. use de moivre's theom. to find in mod/arg form the cube roots of -2-2i?
i got sq rt2 cis -pi/4, sq rt2 cis 5pi/12, sq rt2 cis -pi/12 as my solution....not sure abt this question.

3. if |z| = r and arg z = @, show that z/z^2 + r^2 is real and give its value

this is about it....

cheers

1. z^3 - 1 = (z-1)(z^2+z+1)
find two possible values for z^2+z+1:
when the root is 1, z^2+z+1 = 3
when the root does not = 1, z^2+z+1 = 0 (from above)
so the two possible values are 3 and 0.

2. i got the same answer, except for the 3rd root i got cis(-11pi/12) instead of cis(-pi/12). (this isn't in the syllabus btw.)

3. That doesn't look to real to me.

 

[McLake]

3. z = r@
so z/z^2 + r^2 = r@/(r@)^2 + r^2
= r@ / r^2(2@ + 1)
= @/ r(2@ + 1) which is real ....

 

[spice girl]

Originally posted by Rahul
3. if |z| = r and arg z = @, show that z/z^2 + r^2 is real and give its value

Using r^2 = |z|^2 = z*zbar

z/ (z^2 + r^2) = z / (z^z + z*zbar)

= 1 / (z + zbar)

= 1 / 2Re(z)

 

[ND]

Originally posted by spice girl

Using r^2 = |z|^2 = z*zbar

z/ (z^2 + r^2) = z / (z^z + z*zbar)

= 1 / (z + zbar)

= 1 / 2Re(z)

So it's z/ (z^2 + r^2) not (z/z^2) - r^2. I should have deduced that from that fact that the way i saw it wasn't real.