Author Topic: 4U Maths Question Thread (Read 665237 times) Tweet Share

RuiAce · « **Reply #705 on:** October 29, 2016, 02:04:24 pm »

Quote from: kiwiberry on October 29, 2016, 01:56:00 pm

how do you do this 😁

if a and b are two non-zero complex numbers, show that if a/b = ik,
(conjugate of a)b + (conjugate of b)a = 0

sorry I'm not sure how to type the conjugate sign hahah

If you're lazy with LaTeX just say conj(z)

Is k a constant? If so, is it real or complex?

kiwiberry · « **Reply #706 on:** October 29, 2016, 02:13:31 pm »

Quote from: RuiAce on October 29, 2016, 02:04:24 pm

If you're lazy with LaTeX just say conj(z)

Is k a constant? If so, is it real or complex?

k is a real constant, oops should've typed the whole question

RuiAce · « **Reply #707 on:** October 29, 2016, 02:19:15 pm »

Yasminpotts1105 · « **Reply #708 on:** November 17, 2016, 06:42:33 pm »

If ODEF is a rhombus, where O, D and F represent complex numbers 0, 4 +7i, and 7 + 4i respectively, find
a) the number represented by the point E
b) the length of the diagonals
The area of the rhombus

RuiAce · « **Reply #709 on:** November 17, 2016, 06:59:34 pm »

Quote from: Yasminpotts1105 on November 17, 2016, 06:42:33 pm

If ODEF is a rhombus, where O, D and F represent complex numbers 0, 4 +7i, and 7 + 4i respectively, find
a) the number represented by the point E
b) the length of the diagonals
The area of the rhombus

$\text{Sketch out a diagram clearly showing the rhombus }ODEF\text{ on the Argand plane}$

$\text{It is clear that the vector addition }\overrightarrow{OD}+\overrightarrow{DE}\text{ results in }\overrightarrow{OE}\\ \text{which represents the complex number we wish to find.}\\ \text{Additionally, }\overrightarrow{DE}=\overrightarrow{OF}\\\text{ because they are opp. sites of a rhombus}$
$\text{Hence }E\text{ is the complex number }(4+7i)+(7+4i)=11+11i$
_____________________________
$\text{Since }OE\textit{ is}\text{ the diagonal}\\ \text{The length }OE\text{ is just the modulus of }11+11i\\ OE=|11+11i|=11\sqrt{2}$
$\text{Then the vector }\overrightarrow{DE}\text{ is given by}\\ \begin{align*}\overrightarrow{DE}&=\overrightarrow{DO}+\overrightarrow{OE}\\ &= \overrightarrow{OE}-\overrightarrow{OD}\end{align*}\\ \text{So the vector }\overrightarrow{DE}\text{ represents }(7+4i)-(4+7i) = 3-3i$
$\text{In a similar way, the length of }DE\text{ is just the modulus of this number}\\ DE=|3-3i|=3\sqrt{2}$
_____________________________

This is just the formula area = xy/2 where x and y are the lengths of the diagonal.

kiwiberry · « **Reply #710 on:** November 21, 2016, 12:01:43 pm »

how would you describe the locus of arg(z-i) - arg(1-iz) = pi/4 geometrically?

Ali_Abbas · « **Reply #711 on:** November 21, 2016, 12:45:34 pm »

Quote from: kiwiberry on November 21, 2016, 12:01:43 pm

how would you describe the locus of arg(z-i) - arg(1-iz) = pi/4 geometrically?

$\text{As far as I'm concerned, attempting this with no algebraic guidance} \\ \text{i.e. purely geometrically is a bit of a stretch. What we should be doing} \\ \text{is deriving the Cartesian equation of the locus first, and that will reveal} \\ \text{the geometric representation by glance. We do this as follows:} \\ \text{Arg}(z - i) - \text{Arg}(1-iz) = \text{Arg}(z-i) + \text{Arg}(\overline{1-iz}) \\ (\text{the Argument of a complex number is the negation of its conjugate}) \\ \text{Arg}(z-i) - \text{Arg}(1-iz) = \text{Arg}[(z-i)(1+i\overline{z})] \\ \text{(the sum of two Arguments is the Argument of their product)} \\ \text{Let } z = x + iy. \text{ The RHS becomes:} \\ \text{Arg}(z+\overline{z} + i|z|^{2} - i) = \text{Arg}(2x + i(x^{2}+y^{2}-1) \\ = \tan^{-1}(\frac{x^{2}+y^{2}-1}{2x}) \\ \text{Since this equals to } \frac{\pi}{4}, \text{ we must be a little bit insightful} \\ \text{and observe that if we define a complex number } w \text{ by:} \\ w = 2x + i(x^{2}+y^{2}-1, \\ \text{then for its Argument to equal } \frac{\pi}{4}, \text{ it must lie in the} \\ \text{first quadrant. This means both its real and imaginary components} \\ \text{must be positive. So this gives the restriction on our locus:} \\ x^{2} + y^{2} > 1, x > 0 \\ \text{Continuing on with our derivation:} \\ \frac{x^{2}+y^{2}-1}{2x} = 1 \\ \text{(taking tan of both sides)} \\ x^{2} + y^{2} - 1 = 2x \\ (x-1)^{2} + y^{2} = 2 \\ \text{Thus, the locus of } z \text{ is the major arc of the circle} \\ \text{centred at (1,0) of radius } \sqrt{2} \text{ occurring for} \\ \text{ positive } x \text{ values.}$

RuiAce · « **Reply #712 on:** November 21, 2016, 12:57:04 pm »

$\text{This is how I would've done it.}\\ \text{First, tearing apart the }\arg(1-iz)\text{ gives}\\ \begin{align*}\arg(1-iz)&=\arg(-i(i+z))\\ &= \arg(-i)+\arg(i+z)\\ &= \arg(z+i)-\frac{\pi}{2}\end{align*}$
$\text{So our equation of the locus rearranges to become}\\ \arg (z-i) - \arg (z+i) = -\frac{\pi}{4}$
$\text{Which then becomes}\\ \arg(z+i)-\arg(z-i)=\frac{\pi}{4}$
I might just drop this question here for now. I'm probably gonna write a little guide on the arg(z-z₁) = arg(z-z₂) = theta locus. I hate it because it's too much effort to explain.

Yasminpotts1105 · « **Reply #713 on:** November 21, 2016, 08:37:12 pm »

Prove |z1z2| = |z1||z2|

Find |3-2i|

RuiAce · « **Reply #714 on:** November 21, 2016, 08:47:33 pm »

Quote from: Yasminpotts1105 on November 21, 2016, 08:37:12 pm

Prove |z1z2| = |z1||z2|

Find |3-2i|

$\text{To prove from first principles }|z_1z_2|=|z_1||z_2|\text{ let:}\\ \begin{align*}z_1&=r(\cos \alpha + i \sin \alpha)\\ z_2 &= w(\cos \beta + i \sin \beta)\end{align*}\\ \text{The reason for doing so is that now we have}\\ |z_1|=r\text{ and }|z_2|=w$
Note: Where applicable, you are permitted to use 'cis' notation, but I will never use it.
$\text{Now, evaluating }z_1z_2\text{, from our compound angle identities:} \\ \begin{align*}z_1z_2&=\left[r(\cos \alpha + i \sin \alpha)\right]\left[w(\cos \beta + i \sin \beta)\right]\\ &=rw(\cos\alpha\cos\beta+i\sin\alpha\cos\beta+i\cos\alpha\sin\beta+i^2\sin\alpha\sin\beta)\\ &= rw[(\cos\alpha\cos\beta-\sin\alpha\sin\beta)+i(\sin\alpha\cos\beta+\cos\alpha\sin\beta)]\\ &= rw(\cos(\alpha+\beta)+i\sin(\alpha+\beta))\end{align*}$
$\text{So because the complex number }z_1z_2\text{ is in mod-arg form}\\ \text{We see that }|z_1z_2|=rw\\ \text{Hence, using the earlier result}\\ |z_1z_2|=|z_1||z_2|$
That second question is just finding the modulus of a given complex number. Is there anything problematic about using a formula?
(Answer is √13)

(Note: You do NOT need to use |z1z2|=|z1||z2| here. Just chuck straight into the formula.)

hinakamishiro · « **Reply #715 on:** November 22, 2016, 04:35:39 pm »

Hey guys could someone please help me with question 2? Thank you so much!

RuiAce · « **Reply #716 on:** November 22, 2016, 04:49:09 pm »

Quote from: hinakamishiro on November 22, 2016, 04:35:39 pm

Hey guys could someone please help me with question 2? Thank you so much!

Pointers on part a), because you can't do the other two parts without it.

z₁ is obviously any complex number, so put it wherever you want it to be put. Just don't put it at the origin or things become way less obvious for the later parts.

What is z₂? The significant part is that z₂ involves i times z₁. What does multiplication by i do? Rotate the vector π/2 (i.e. 90 degrees) counterclockwise

And what does k do? Well that's just a constant, so it stretches (or shrinks) the vector. So it just means the length might be different. The important thing, is the rotation.

Post up working out if you want further advice.

Extra tip on part b) - lol the typo on the question though

Inevitably, you're gonna have to plot z₁+z₂

Extra tip on part c)

I suspect that a rectangle is involved... WHY?

Wales · « **Reply #717 on:** December 03, 2016, 04:03:00 pm »

Bit unsure how to do the question. I've expanded the binomial and now I'm a bit lost.

i) Use the binomial theorem to expand (CosX+iSinX)^3

ii) Use DeMoivre's theorem and your result from part (i) to prove that

cos^3(X)=1/4Cos3X+3/4CosX

Where X is theta ( not sure how to type that

)

Thanks, Wales

RuiAce · « **Reply #718 on:** December 03, 2016, 04:25:32 pm »

Quote from: Wales on December 03, 2016, 04:03:00 pm

Bit unsure how to do the question. I've expanded the binomial and now I'm a bit lost.

i) Use the binomial theorem to expand (CosX+iSinX)^3

ii) Use DeMoivre's theorem and your result from part (i) to prove that

cos^3(X)=1/4Cos3X+3/4CosX

Where X is theta ( not sure how to type that )

Thanks, Wales

$\text{From the binomial theorem}\\ \begin{align*}(\cos x + i\sin x)^3&=\binom30\cos^3x+\binom31\cos^2xi\sin x+\binom32\cos x i^2\sin^2x+\binom33i^3\sin^3x\\ &= (\cos^3 x - 3\cos x \sin^2x)+i(3\cos^2x \sin x-\sin^3 x)\end{align*}$
$\text{From De Moivre's theorem}\\ (\cos x + i \sin x)^3 = \cos 3x + i\sin 3x$
$\text{Since the above expressions strictly equal each other}\\ \text{Upon equating real parts:}\\ \begin{align*}\cos 3x&= \cos^3x - 3\cos x\sin^2 x\\ &= \cos^3x - 3\cos x(1-\cos^2x)\end{align*}\\ \text{which upon rearranging becomes the desired result.}$

Wales · « **Reply #719 on:** December 03, 2016, 11:46:33 pm »

Quote from: RuiAce on December 03, 2016, 04:25:32 pm

$\text{From the binomial theorem}\\ \begin{align*}(\cos x + i\sin x)^3&=\binom30\cos^3x+\binom31\cos^2xi\sin x+\binom32\cos x i^2\sin^2x+\binom33i^3\sin^3x\\ &= (\cos^3 x - 3\cos x \sin^2x)+i(3\cos^2x \sin x-\sin^3 x)\end{align*}$
$\text{From De Moivre's theorem}\\ (\cos x + i \sin x)^3 = \cos 3x + i\sin 3x$
$\text{Since the above expressions strictly equal each other}\\ \text{Upon equating real parts:}\\ \begin{align*}\cos 3x&= \cos^3x - 3\cos x\sin^2 x\\ &= \cos^3x - 3\cos x(1-\cos^2x)\end{align*}\\ \text{which upon rearranging becomes the desired result.}$

Cheers, I didn't realise you had to equate them. Appreciate it

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