HSC 2015 MX2 Marathon (archive) (4 Viewers)

FrankXie · Dec 29, 2014

Re: HSC 2015 4U Marathon

NEXT QUESTION

$$ If the complex number $ z $ satisfies that $ |z|=|z-4+2i|, $ find the minimum value of $ |z|+|z-2i|.$

porcupinetree · Dec 30, 2014

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$$ If the complex number $ z $ satisfies that $ |z|=|z-4+2i|, $ find the minimum value of $ |z|+|z-2i|.$

Here's my go:

√(x^2 + y^2) = √[(x-4)^2 + (y+2)^2]
Simplification gives:
y = 2x - 5

|z| + |z - 2i| => |2z - 2i| = 2|z - i| (triangle inequality)
Clearly, equality is reached when x = 0
Sub x=0 into y = 2x - 5 = 2*0 - 5 = -5.
|-5| + |-5i+2i| = |-5| + |-3i| = 5 + 3 = 8.

I feel like my third last line ('clearly, equality...') is not very (for want of a better word) good. I tried solving it algebraically but I got nowhere, I just ended up with three square root terms, which, if x = 0 is subbed in, produces a correct result, however it just looked like a big mess. Is there a better way?

dan964 · Dec 30, 2014

Re: HSC 2015 4U Marathon

hmm I am not sure whether that's right consider whether |z| lies on the
x-axis.
You get |z| = 2.5??
|z-2i| approx. 3 point something
which is less than 8 when added together.
|z-2i| is the distance from (0,2) to line not (0,-2)

so you actually consider the distance between
the line Y=2X+5 and (0,0)
and (0,2) added together

braintic · Dec 30, 2014

Re: HSC 2015 4U Marathon

porcupinetree said:
Here's my go:

√(x^2 + y^2) = √[(x-4)^2 + (y+2)^2]
Simplification gives:
y = 2x - 5

|z| + |z - 2i| => |2z - 2i| = 2|z - i| (triangle inequality)
Clearly, equality is reached when x = 0
Sub x=0 into y = 2x - 5 = 2*0 - 5 = -5.
|-5| + |-5i+2i| = |-5| + |-3i| = 5 + 3 = 8.

I feel like my third last line ('clearly, equality...') is not very (for want of a better word) good. I tried solving it algebraically but I got nowhere, I just ended up with three square root terms, which, if x = 0 is subbed in, produces a correct result, however it just looked like a big mess. Is there a better way?

Geometry beats algebra

porcupinetree · Dec 30, 2014

Re: HSC 2015 4U Marathon

dan964 said:
hmm I am not sure whether that's right consider whether |z| lies on the
x-axis.
You get |z| = 2.5??
|z-2i| approx. 3 point something
which is less than 8 when added together.
|z-2i| is the distance from (0,2) to line not (0,-2)

so you actually consider the distance between
the line Y=2X+5 and (0,0)
and (0,2) added together

Oh my goodness, I'm stupid. Yes, I made the mistake that it was from (0,-2) not (0,2)

Kaido · Dec 30, 2014

Re: HSC 2015 4U Marathon

Find the values that the real numbers a and b can take for z=1 to be a root of the complex equation: iz^2+(ia-1)z+(i-b)=0
(starting easy as i'm not sure if everyone has begun the polynomials topic)

Note: Doesn't require '4U polynomial' knowledge

PhysicsMaths · Dec 30, 2014

Re: HSC 2015 4U Marathon

Kaido said:
Find the values that the real numbers a and b can take for z=1 to be a root of the complex equation: iz^2+(ia-1)z+(i-b)=0
(starting easy as i'm not sure if everyone has begun the polynomials topic)

Note: Doesn't require '4U polynomial' knowledge

Let P(z) = iz^2 + (ia-1)z + (i-b)
For z = 1 to be a root, P(1) = 0
∴ P(1) = i + ia-1 + i-b = 0
i(2+a) - (1+b) = 0
∴ a = -2, b = -1

PhysicsMaths · Dec 30, 2014

Re: HSC 2015 4U Marathon

Write in simplest form:
(sinθ - icosθ)^-8

Kaido · Dec 30, 2014

Re: HSC 2015 4U Marathon

(sinθ - icosθ)^-8 = 1/(i(cosθ+isinθ))^8
(then by DeMoivre's theorem) = 1/(cos8θ+isin8θ)

enigma_1 · Dec 30, 2014

Re: HSC 2015 4U Marathon

Kaido said:
(sinθ - icosθ)^-8 = 1/(i(cosθ+isinθ))^8
(then by DeMoivre's theorem) = 1/(cos8θ+isin8θ)

OMG THETA caps

Kaido · Dec 30, 2014

Re: HSC 2015 4U Marathon

Let P(x) be a polynomial with degree of 1996. If P(n) = 1/n for n = 1,2,3...1997, Find the value of P(1998)

Edit: Not sure if it is technically under the 4U syllabus, so just ignore it

(By all means, try it if you wish)

InteGrand · Dec 30, 2014

Re: HSC 2015 4U Marathon

Kaido said:
Let P(x) be a polynomial with degree of 1996. If P(n) = 1/n for n = 1,2,3...1997, Find the value of P(1998)

Edit: Not sure if it is technically under the 4U syllabus, so just ignore it (By all means, try it if you wish)

Let $P(x)=a_0+\sum_{i=1}^{1996}a_i x^i$ .

We are given that $P(x)=\frac{1}{x}\Leftrightarrow xP(x)-1=0$ for $x=1,2,3,\cdots,1997$ .

$xP(x)-1$ is a 1997th degree polynomial, and has its 1997 roots given to us, so can be factorised as $xP(x)-1=A(x-1)(x-2)(x-3)\cdots(x-1997)$ using the given roots, where $A=a_{1996}$ .

Now, $xP(x)-1=x(a_{1996}x^{1996}+a_{1995}x^{1995}+\cdots+a_0)-1$ , so the leading coefficient is $a_{1996}$ , constant term is –1.

Therefore, using product of roots of $xP(x)-1$ (which is $1\cdot2\cdot3\cdot\cdots\cdot1997=1997!$ ), we have

$1997!=-\frac{-1}{a_{1996}}\Rightarrow a_{1996}=\frac{1}{1997!} = A$ , so

$xP(x)-1=\frac{1}{1997!}(x-1)(x-2)(x-3)\cdots(x-1997)$ .

Substituting x = 1998,

$1998\cdot P(1998)-1=\frac{1}{1997!}(1998-1)(1998-2)(1998-3)\cdots(1998-1997)$

$=\frac{1}{1997!}\cdot1997\cdot1996\cdot1995 \cdot \cdots \cdot 1$

$=\frac{1}{1997!}\cdot1997!$

$=1$ .

Solving for P(1998) yields $P(1998)=\frac{1}{999}$ .

porcupinetree · Dec 31, 2014

Re: HSC 2015 4U Marathon

NEXT QUESTION

The equation x^3 + 3x^2 - 4x + 5 = 0 has roots A, B, Y.

Evaluate A^3 + B^3 + Y^3

Kaido · Dec 31, 2014

Re: HSC 2015 4U Marathon

so since A.B.Y are the roots of x^3 + 3x^2 - 4x + 5 = 0, then: A,B,Y are values of x
-> Rearranging: x^3=-3x^2 +4x -5
so, A^3 + B^3 + Y^3 = -3A^2 +4A -5 - 3B^2 +4B -5 - 3C^2 +4C -5 = -3(A^2+B^2+C^2) +4(A+B+C) -15

->A^2+B^2+C^2 = (A+B+Y)^2 - 2(AB +BC +AC) = 17

therefore A^3 + B^3 + Y^3 = -3(17) +4(-3) -15
=-78 (not good at subtraction)

(Prob mistake somewhere, hella hard to type in this shifty font)

Edit: corrected the mistake, thanks braintic.

braintic · Dec 31, 2014

Re: HSC 2015 4U Marathon

Kaido said:
so since A.B.Y are the roots of x^3 + 3x^2 - 4x + 5 = 0, then: A,B,Y are values of x
-> Rearranging: x^3=-3x^2 +4x -5
so, A^3 + B^3 + Y^3 = -3A^2 +4A -5 - 3B^2 +4B -5 - 3C^2 +4C -5 = -3(A^2+B^2+C^2) +4(A+B+C) -15

->A^2+B^2+C^2 = (A+B+Y)^2 - 2(AB +BC +AC) = 1
-
therefore A^3 + B^3 + Y^3 = -3(1) +4(-3) -15
=-30

(Prob mistake somewhere, hella hard to type in this shifty font)

Yep, slight error.

AB+AC+BC = -4
So A^2+B^2+C^2 = 3^2 -2(-4) = 17

Kaido · Dec 31, 2014

Re: HSC 2015 4U Marathon

braintic said:
Yep, slight error.

AB+AC+BC = -4
So A^2+B^2+C^2 = 3^2 -2(-4) = 17

wallah im stoned on chocolates

dan964 · Dec 31, 2014

Re: HSC 2015 4U Marathon

Next question
(I) Using Demoivre's find expressions. for sin 5x and cos 5x

Kaido · Dec 31, 2014

Re: HSC 2015 4U Marathon

-> (cosθ+isinθ)^5: (c=cosθ, s=sinθ. too lazy bro)
=(by Binomial expansion/Pascals) c^5+5c^4(is)+10c^3(is)^2+10c^2(is)^3+5c(is)^4+(is)^5
(grouping real and im.) = c^5 -10c^3s^2 +5cs^4 + yeah
therefore equating: cos5x=c^5-10c^3(1-c^2)+5c(1-c^2)^2
=c^5-10c^3+10c^5+5c-10c^3+5c^5
=16c^5-20c^3+5c
Same thing for sin5x but group the sin and yeah,

(as beforementioned, am stoned, home alone, ceebs)

dan964 · Jan 1, 2015

Re: HSC 2015 4U Marathon

hence show that cos 9 and sin 9 satisfy the equation
16x^5-20x^3+5x-1/sqrt(2) = 0
angles are in degrees please convert to radians

FrankXie · Jan 1, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
Here's my go:

√(x^2 + y^2) = √[(x-4)^2 + (y+2)^2]
Simplification gives:
y = 2x - 5

|z| + |z - 2i| => |2z - 2i| = 2|z - i| (triangle inequality)
Clearly, equality is reached when x = 0
Sub x=0 into y = 2x - 5 = 2*0 - 5 = -5.
|-5| + |-5i+2i| = |-5| + |-3i| = 5 + 3 = 8.

I feel like my third last line ('clearly, equality...') is not very (for want of a better word) good. I tried solving it algebraically but I got nowhere, I just ended up with three square root terms, which, if x = 0 is subbed in, produces a correct result, however it just looked like a big mess. Is there a better way?

good try but i'm afraid it is invalid.
$|z|+|z-2i|=|z-4+2i|+|z-2i|=|4-2i-z|+|z-2i|\geq|4-2i-z+z-2i|=|4-4i|=4\sqrt2, $ where equality holds true when the vectors $ 4-2i-z $ and $ z-2i $ are in the same direction. By solving simultaneous equations ( to find the point of intersection of two lines), the minimum $ 4\sqrt2 $ occurs when $ z=\frac73-\frac13i.$

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