Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

appleibeats · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Find the values of a for which z - ai is factor of the equation

Is it factor theorem ?? Seems tedious?

InteGrand · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Find the values of a for which z - ai is factor of the equation

Is it factor theorem ?? Seems tedious?

What equation?

appleibeats · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Sorry,

z^4 - 2z^3 + 7z^2 - 4z + 10 = 0

Ambility · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Find the values of a for which z - ai is factor of the equation

Is it factor theorem ?? Seems tedious?

appleibeats said:
Sorry,

z^4 - 2z^3 + 7z^2 - 4z + 10 = 0

This is an imaginary number question, why is it in the year 11 3U thread? Anyway, this is how I'd work it out, and yes it does use the factor theorem.

$(ai)^4-2(ai)^3+7(ai)^2-4(ai)+10=0\\a^4i^4-2a^3i^3+7a^2i^2-4ai+10=0\\a^4+2a^3i-7a^2-4ai+10=0\\$Equating imaginary parts:$\\2a^3-4a=0\\2a(a^2-2)=0\\a=0, \pm\sqrt{2}\\$Equating real parts:$\\a^4-7a^2+10=0\\(a^2-5)(a^2-2)=0\\a=\pm\sqrt{5}, \pm\sqrt{2}\\$ai is only a factor if both the imaginary and real parts are equal to each other.$\\a=\pm\sqrt{2}$

appleibeats · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Let z = cisE and w = cisB, that is |z| = |w| = 1. Evaluate z + w in mod-arg form and hence show that arg(z+w) = 1/2 (arg z + arg w)

The hint given is to use sums to products

Answer is z + w = 2cos((E-B)/2) cis((E+B)/2)

kawaiipotato · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Let z = cisE and w = cisB, that is |z| = |w| = 1. Evaluate z + w in mod-arg form and hence show that arg(z+w) = 1/2 (arg z + arg w)

The hint given is to use sums to products

Answer is z + w = 2cos((E-B)/2) cis((E+B)/2)

use the formula: cos(a) + cos(b) = 2cos((a+b)/2)cos((a-b)/2)
and sin(a) + sin(b) = 2cos((a+b)/2)sin((a-b)/2)
when finding z+w

InteGrand · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Let z = cisE and w = cisB, that is |z| = |w| = 1. Evaluate z + w in mod-arg form and hence show that arg(z+w) = 1/2 (arg z + arg w)

The hint given is to use sums to products

Answer is z + w = 2cos((E-B)/2) cis((E+B)/2)

$Let $z=\cos \alpha + i\sin \alpha$ and $w=\cos \beta + i\sin \beta$. Then $z+w = \left( \cos \alpha + \cos \beta \right)+i \left( \sin \alpha + \sin \beta \right)$.$

$Now, the sum-to-product rules (see link at bottom) state that $\cos \alpha + \cos \beta = 2\cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha-\beta}{2} \right)$ and $\sin \alpha + \sin \beta = 2\sin \left( \frac{\alpha + \beta}{2} \right) \cos \left(\frac{\alpha-\beta}{2} \right)$.$

$Therefore,$

$\begin{align*} z+w &= 2\cos \left( \frac{\alpha + \beta}{2} \right)\cos \left( \frac{\alpha-\beta}{2} \right) + i \cdot 2\sin \left( \frac{\alpha + \beta}{2}\right) \cos \left( \frac{\alpha-\beta}{2} \right) \\ &= 2\cos \left( \frac{\alpha-\beta}{2} \right) \left[\cos \left(\frac{\alpha + \beta}{2} \right)+i \sin \left( \frac{\alpha + \beta}{2}\right)\right] \quad (\text{factorising}).\end{align*}$

$This is the polar (mod-arg) form of $z+w$, since $2\cos \left( \frac{\alpha-\beta}{2} \right) $ is just a real number (equal in magnitude to the modulus of $z+w$), so we can read off that $\arg \left(z+w\right) = \frac{\alpha + \beta}{2}=\frac{1}{2}\left(\arg (z) + \arg (w)\right)$. Q.E.D.$

$Sum-to-product rules (also has a link to product-to-sum rules):$

http://www.mathwords.com/s/sum_to_product_identities.htm

$\textbf{Note.} You are not allowed to just quote the sum-to-product and product-to-sum rules in the HSC, so you must derive them first before using them if you are going to use them (or an earlier part of the question will get you to derive it or give it to you). So you should learn how to derive the sum-to-product and product-to-sum identities, as they can come in handy.$

appleibeats · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Let z = 1 + cosE + isinE

a) Show that |z| = 2cosE/2 and argz = E/2

b) Hence show taht z^-1 = 1/2 - 1/2 itanE/2

kawaiipotato · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Let z = 1 + cosE + isinE

a) Show that |z| = 2cosE/2 and argz = E/2

b) Hence show taht z^-1 = 1/2 - 1/2 itanE/2

$(a) $ Recall that $ 1 + \cos \alpha = 2 \cos ^2 \frac{\alpha}{2} $ and $ \sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$

$z = 1 + \cos E + i \sin E$

$z = 2 \cos ^2 \frac{E}{2} + 2 i \sin \frac{E}{2} \cos \frac{E}{2}$

$= 2 \cos \frac{E}{2} \left[ \cos \frac{E}{2} + i \sin \frac{E}{2}\right]$

$|z| = 2 \cos \frac{E}{2} $ and $ \arg z = \frac{E}{2}$

$(b) z^{-1} = \frac{1}{2}* \frac{\left[ \cos \frac{-E}{2} + i \sin \frac{-E}{2}\right]}{\cos \frac{E}{2}} ( $ de Moivre's Theorem$)$

$$ and using the trigonometric identities: \cos (- \alpha) = \cos (\alpha) $and$ \sin (-\alpha) = - \sin(\alpha)$

$= \frac{1}{2} - \frac{1}{2} i \tan \frac{E}{2}$

leehuan · Nov 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

kawaiipotato said:
Solutions

I was watching you continuously edit the whole time

appleibeats · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Prove that the point z1 ,z2, z3 are collinear if (z3 - z1)/ ( z2 - z1 ) is real

Hence show that the points representing 5 = 8i , 13 + 20i, and 19 + 29i are collinear

InteGrand · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Prove that the point z1 ,z2, z3 are collinear if (z3 - z1)/ ( z2 - z1 ) is real

Hence show that the points representing 5 = 8i , 13 + 20i, and 19 + 29i are collinear

$Let $z_1,z_2,z_3 \in \mathbb{C}$. Let $\frac{z_3 - z_1}{z_2 - z_1}=\lambda$ for some real number $\lambda$ ($\lambda \neq 0$ as $z_3 \neq z_1$, since we have three distinct points). Then we have $z_3 - z_1 = \lambda (z_2 - z_1)$. Hence the vector joining $z_1$ to $z_3$ is a real non-zero scalar multiple of the vector joining $z_1$ to $z_2$, which means these vectors have the same direction (possibly rotated by $180^\circ$, so maybe they do not have the same ``sense'', but this does not affect collinearity), so the points are collinear.$

$Since$

$\begin{align*}\frac{(19+29i) - (5+8i)}{(13 + 20i)-(5+8i)}& = \frac{14 + 21 i}{8 + 12 i} \\ &= \frac{7\left(2+3i\right)}{4\left(2+3i\right)} \\ &= \frac{7}{4}\in \mathbb{R}, \end{align*}$

$the points representing these complex numbers are collinear.$

appleibeats · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

If z1 = 4 - 1 and z2 = 2i , find in each case the two possible values of z3 so that the points representing z1, z2,z3 form an isosceles right angled triangle with right angle at :

z1

appleibeats · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Can't you just multiply the point z1 by i. Why is this wrong???

appleibeats · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Answers are 1 - 5i, and 7 + 3i

InteGrand · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Can't you just multiply the point z1 by i. Why is this wrong???

Note that if we have two vertices of an isosceles right-angled triangle in the plane and we specify which vertex is the right angle, we'll have two possible locations for the third vertex.

Furthermore, multiplying z₁ by i would rotate z₁ by 90º counter-clockwise about the origin, which is not what we want to do.

kawaiipotato · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Why is this wrong???

InteGrand said:
Note that if we have two vertices of an isosceles right-angled triangle in the plane and we specify which vertex is the right angle, we'll have two possible locations for the third vertex.

Furthermore, multiplying z₁ by i would rotate z₁ by 90º counter-clockwise about the origin, which is not what we want to do.

We should be multiplying the side joining z2 and z1 by +/- i instead, to find the side joining z3 and z1.
The side joining z2 and z1 is vector z2 - z1, with the rotated side being z3 - z2
So (z2-z1)i = z3 - z1, expand and rearrange to solve for z3
Similarly, -(z2-z1)i = z3 - z1

kawaiipotato · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Why is this wrong???

InteGrand said:
Note that if we have two vertices of an isosceles right-angled triangle in the plane and we specify which vertex is the right angle, we'll have two possible locations for the third vertex.

Furthermore, multiplying z₁ by i would rotate z₁ by 90º counter-clockwise about the origin, which is not what we want to do.

So we should be multiplying the side joining z2 and z1 by +/- i instead, to find the third vertex.
The side joining z2 and z1 is vector z2 - z1

appleibeats · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Cheers I worked it out

appleibeats · Nov 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

How would you solve:

Given that z1 = 1 + i , z2 = 2 + 6i , z3 = -1 + 7i , find the three possible values of z4 so that the points representing z1, z2, z3 and z4 form a parallelogram.

Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

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