MATH1901 Assignment (1 Viewer)

[antarctic]

Hi everyone:

I've got a question about the Math1901 assignment. Are we allowed to assume the "Fundamental theorem of algebra", i.e. that if a polynomial of degree of n has only got real coefficients, then it must have n roots (either real or complex)?
Also, for question 2, should we prove that if a polynomial of degree n has only real coefficients, then complex roots of the polynomial (if any exist) exist in conjugate pairs? (if we're gonna use that theorem, that is)

Thanks in advance

Antarctic

 

[zeropoint]

Hi everyone:

I've got a question about the Math1901 assignment. Are we allowed to assume the "Fundamental theorem of algebra", i.e. that if a polynomial of degree of n has only got real coefficients, then it must have n roots (either real or complex)?

I did. Note also that the fundamental theorem of algebra holds for polynomials with real or complex coefficients.

Also, for question 2, should we prove that if a polynomial of degree n has only real coefficients, then complex roots of the polynomial (if any exist) exist in conjugate pairs? (if we're gonna use that theorem, that is)

I sort of proved it implicity when I showed that 1 - i is a solution. I think this is satisfactory.

 

[acmilan]

If you want to be safe you can always use the proof give in the lecture notes to prove the complex conjugate pairs thing. But i dont think youd need to prove it.

 

[antarctic]

Okay, cool, thanks everyone.

Antarctic

 

[Templar]

I doubt the fundamental theorem of algebra could be proved so easily, although I do know someone who proved it.

 

[xiao1985]

I doubt the fundamental theorem of algebra could be proved so easily, although I do know someone who proved it.

lolz arnold gives a simplified version of thsi proof by induction...

 

[gordo]

errr,
wat assignment?

 

[nit]

The fundamental theorem of algebra states that every polynomial of degree at least one has at least one zero over C. Arnold assumes this result. Using this, he goes on to prove that every polynomial of degree n has exactly n zeroes over C.

 

[gordo]

did we find out about this in thursday's lecture?
cause that's the only one i barred and i know nothing about an assignment

 

[acmilan]

did we find out about this in thursday's lecture?
cause that's the only one i barred and i know nothing about an assignment

Go to the MATH 1901 site, its in the resource section and is due next week. Its not too hard, just on complex numbers

 

[gordo]

got it,
doesn't seem too difficult

 

[gordo]

how do u do 2b
i thought u just multiplied the conjugates and divided them into the polynomial but u get a remainder of 10
i presume the other 2 roots are a conjugate pair since their are no real roots, but how do u find them?

 

[Xayma]

Try your long division again. It should divide nicely.

 

[gordo]

yes yes it does
haha in multiplying the two i mistook a z for a 2, with devastating consequences

stuck on 2c now

really should ahve listened in the lecturs

any hints?

 

[gordo]

can u take the modulus of both sides, cause wats | 2e^z - 4 | ?

|e^2z| = |2e^z - 4|
e^2x = 2e^x - 4
e^2x - 2e^x = -4
e^x(e^x - 2) = -4
e^x - 2 = -4e^-x

logs????

helppppppppppppppppppp

 

[gordo]

i don't have a clue how to do anything from 2c onwards

should i be dropping this subject

 

[Xayma]

2c.

You have done similar things in the HSC. e^2x=(e^x)²

3. Is just complex images again.

Find the equation of the relationship in terms of x and y's. Then look at the lines of the triangle to map them.

 

[gordo]

ok i think i got 2c out
it was a pretty cimple quadratic with changin e^z to cis' at the end

part 3 is still screwed though
i basically did it like this and got some wier shapes
took the complex co ordinate of each point (q,p,r) in z plane, then pluggesd that in x+iy format into the w eqn e^iz for eg and then plotted the new points on the w planes and connect p to q then q tor and r back to p

i got 2 triangles for i and a circle with a vertical line through its diameter for ii

i need help its due tommorow
just help me, u'll cane me in the exam anyway
let me get my 5% here

 

[Xayma]

Try plotting the mid points of the lines and the verticies. It will give you a rough shape. If necessary get the equations of the line and see other points.

Why do you have three coordinates?

The complex plane is just that a plane. You have a real axis and an imaginary one.

 

[Templar]

For the second image you can actually derive the formula of the new curve in terms of x and y quite easily if you let z=x+iy.

Otherwise, plot more points like Xayma said.