HSC Mathematics Marathon (5 Viewers)

[Drongoski]

If z is a complex number such that z = r(cosθ +isinθ), where r is real, show that arg(z+r) = (1/2)θ .

Geometrically, on an Argand diagram, you have a position vector, from origin of length r and with angle . Drawing a vector from end of this vector z to the right of length r, you now have 2 adjacent sides of a rhombus (sides of length r); z+r is then represented by the diagonal of this rhombus from the origin to the tip of vector z+r. By property of rhombus, this vector bisects the angle .

 

[HyperComplexxx]

Polynomial x^4 + qx^2 +rx + s = 0 has roots α, β, γ, δ. Find the value of the constant

term in the polynomial with roots 1-α^2 , 1-β^2 , 1-γ^2 , 1- δ^2.

 

[jyu]

Polynomial x^4 + qx^2 +rx + s = 0 has roots α, β, γ, δ. Find the value of the constant

term in the polynomial with roots 1-α^2 , 1-β^2 , 1-γ^2 , 1- δ^2.

(1-α^2)(1-β^2)(1-γ^2)(1-δ^2)
=(1-α)(1-β)(1-γ)(1-δ)(1+α)(1+β)(1+γ)(1+δ)
=(1+q+r+s)(1+q-r+s)

 

[HyperComplexxx]

(1-α^2)(1-β^2)(1-γ^2)(1-δ^2)
=(1-α)(1-β)(1-γ)(1-δ)(1+α)(1+β)(1+γ)(1+δ)
=(1+q+r+s)(1+q-r+s)

 

[rawrence]

(1-α^2)(1-β^2)(1-γ^2)(1-δ^2)
=(1-α)(1-β)(1-γ)(1-δ)(1+α)(1+β)(1+γ)(1+δ)
=(1+q+r+s)(1+q-r+s)

I got (q+s+1)²-r² which is equivalent but by first finding the equation which has roots (1-α²)(1-β²)(1-γ²)(1-δ²)

I like your method of finding products of roots but could you explain how you got from
HERE
=(1-α)(1-β)(1-γ)(1-δ)(1+α)(1+β)(1+γ)(1+δ)
TO
=(1+q+r+s)(1+q-r+s)
THERE?

Cheers

 

[UnrealAnchovies]

Expand P(x) = (x-1)(x+2)(x-3)(x+1)

And for the sake of the exercise, please don't just expand.

 

[hscishard]

Expand P(x) = (x-1)(x+2)(x-3)(x+1)

And for the sake of the exercise, please don't just expand.

just use the roots and to find the coefficients

 

[Drongoski]

How about checking out the following.

 

[ohexploitable]

 

[Drongoski]

Looks deceptively easy.

Please look at my post above re n-th roots of 1.

 

[rawrence]

Using t-result substitution and partial fractions I get down to:
2 [int 1/(t²-t+1) dt - int 1/t² +1 dt]
2 [int dt/((t-1/2)² + 3/4)] - 2[int dt/(t²+1)]
2 [2/(√3)*tan-1 (t-1/2)/(√3/2) - tan-1 t] + C

which is both unsimplified and illegible because I can't latex

 

[ohexploitable]

Using t-result substitution and partial fractions I get down to:
2 [int 1/(t²-t+1) dt - int 1/t² +1 dt]
2 [int dt/((t-1/2)² + 3/4)] - 2[int dt/(t²+1)]
2 [2/(√3)*tan-1 (t-1/2)/(√3/2) - tan-1 t] + C

which is both unsimplified and illegible because I can't latex

You're correct.

This should make it easier for anyone who wants the solution:



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[Drongoski]

Answer looks like one I got few days ago.

 

[rawrence]

I typed on word LOL!

my bad should be tan(x/2) not inverse

 

[UnrealAnchovies]

Pretty good question tbh.

 

[Drongoski]

 

[ohexploitable]

 

[Drongoski]

Very good. Maybe should have posted under Ext 1.

 

[hscishard]

Differentiate x+y+z = 0

 

[rawrence]

^ cool story